Mathematical Principles
- Identify different types of fractions and apply mathematical principles to solve fractions and percentages
- Apply algebra to solve simple equations and transpose formulae
- Identify and apply laws of indices and powers of ten
- Apply Pythagoras' theorem and trigonometry to solve right-angled triangles
- Use statistical methods to collect, sort, analyse and present numerical data
This chapter covers the essential mathematical foundations that underpin electrotechnical practice. The topics are treated as a refresher — building on prior knowledge rather than introducing concepts from scratch. A solid grasp of these mathematical tools is essential for an electrician: you will use fractions when calculating cable fill percentages, apply algebra when rearranging Ohm's Law, and use trigonometry when finding cable lengths across obstructions. Every topic here has a direct practical counterpart in your day-to-day installation and maintenance work.
Fractions
A fraction represents a part of a whole. The numerator (top number) tells you how many equal parts you have; the denominator (bottom number) tells you how many equal parts make up the whole. For example, \(\frac{3}{4}\) means the whole has been divided into 4 equal parts and you are referring to 3 of them.
Understanding fractions is fundamental in electrical work. Conductor cross-sectional areas are often expressed as fractions of a standard size, voltage divider ratios use fractional relationships, and transformer turns ratios are effectively fractions that set the output voltage. For instance, a 230 V secondary winding on a 400 V transformer uses \(\frac{230}{400} = \frac{23}{40}\) of the full winding turns.
- Proper fraction – numerator is less than denominator, so its value is less than 1, e.g. \(\frac{1}{2},\; \frac{5}{8},\; \frac{3}{4}\). Electrical example: a voltage divider that drops \(\frac{3}{4}\) of the supply gives \(\frac{3}{4} \times 230 = 172.5\text{ V}\).
- Improper fraction – numerator is greater than denominator, so its value is greater than 1, e.g. \(\frac{9}{4},\; \frac{16}{5},\; \frac{22}{7}\). These arise in step-up transformer calculations where output voltage exceeds input.
- Mixed fraction – a whole number plus a proper fraction, e.g. \(8\frac{1}{2}\). You might express a conduit run as \(3\frac{3}{4}\) metres, or a load current as \(8\frac{1}{2}\text{ A}\) before converting to a decimal for calculator use.
- Decimal fractions – fractions whose denominator is a power of ten (10, 100, 1 000 etc.), written using a decimal point. Cable resistance values from BS 7671 are given as decimal fractions of an ohm per metre — for example, 1.5 mm² copper conductor has a resistance of approximately \(0.012\text{ Ω/m}\).
Decimal fractions table — each row divides by a further factor of ten:
| Fraction | Decimal | Electrical context |
|---|---|---|
| \(\frac{1}{10}\) | 0.1 | 0.1 Ω — resistance of a short run of 2.5 mm² copper at room temperature |
| \(\frac{1}{100}\) | 0.01 | 0.01 A = 10 mA — threshold for ventricular fibrillation; justifies 30 mA RCD protection |
| \(\frac{1}{1\,000}\) | 0.001 | 1 mA — the milliampere; minimum operating current of many RCDs |
| \(\frac{1}{10\,000}\) | 0.0001 | 100 µF = 0.0001 F — a large smoothing electrolytic capacitor in a power supply |
| \(\frac{1}{100\,000}\) | 0.00001 | 10 µF — typical motor run capacitor for a single-phase induction motor |
| \(\frac{1}{1\,000\,000}\) | 0.000001 | 1 µF (microfarad) — standard unit for capacitors in lighting control gear |
Percentages
A percentage is a fraction with a denominator of 100. Writing something as a percentage makes it easy to compare proportions regardless of the original quantity. The symbol % means "out of 100", so \(58\% = \frac{58}{100} = 0.58\). To convert any fraction to a percentage, multiply it by 100.
Percentages appear throughout electrical installation work, particularly in efficiency calculations, voltage drop limits, and cable fill rules. BS 7671 and IET Guidance Note 1 specify that voltage drop in most circuits must not exceed 3% for lighting and 5% for other uses of the supply voltage. An immediate, practical grasp of percentage calculation helps you design compliant installations without reaching for a calculator every time.
Maximum permitted volt drop = 3% of 230 V:
\[\text{Volt drop} = 230 \times \frac{3}{100} = 6.9\text{ V}\]So the voltage at the end of the circuit must be at least \(230 - 6.9 = 223.1\text{ V}\).
Maximum permitted volt drop = 5% of 400 V:
\[\text{Volt drop} = 400 \times \frac{5}{100} = 20\text{ V}\]A motor consumes 5 kW from the supply and delivers 4.25 kW of mechanical output. Find its efficiency:
\[\text{Efficiency} = \frac{\text{output}}{\text{input}} \times 100 = \frac{4.25}{5} \times 100 = 85\%\]A 25 mm conduit has a cross-sectional area of 491 mm². Three 4 mm² cables (each csa = 4 mm²) occupy: \(3 \times 4 = 12\text{ mm}^2\). The fill percentage is:
\[\text{Fill}\% = \frac{12}{491} \times 100 = 2.44\%\]This is well within BS 7671's space factor guidelines, confirming the conduit is adequately sized.
Lowest Common Multiple (LCM) and Lowest Common Denominator (LCD)
The Lowest Common Multiple (LCM) of two or more numbers is the smallest number that is exactly divisible by each of them. To find it, list the multiples of each number and identify the first one they share.
- Multiples of 3: 3, 6, 9, 12, 15, 18 …
- Multiples of 4: 4, 8, 12, 16 …
- LCM of 3 and 4 = 12
The Lowest Common Denominator (LCD) is simply the LCM applied to the denominators of the fractions you wish to combine. This concept is relevant in electrical work when combining fractions representing parallel resistances or when adding currents expressed as fractional parts of a total.
Adding fractions — step-by-step method
- Identify the LCD of all the denominators.
- Convert each fraction: divide the LCD by the original denominator, then multiply both numerator and denominator by that result.
- Add the numerators (the denominators stay the same).
- Simplify the result by cancelling common factors.
\(\frac{1}{3} + \frac{1}{5}\) — LCD = 15:
\[\frac{1\times5}{3\times5} + \frac{1\times3}{5\times3} = \frac{5}{15} + \frac{3}{15} = \frac{8}{15}\]\(\frac{1}{2} + \frac{1}{3} + \frac{1}{6}\) — LCD = 6:
\[\frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1\]Two resistors in parallel have conductances of \(\frac{1}{4}\text{ S}\) and \(\frac{1}{6}\text{ S}\). Total conductance (LCD = 12):
\[\frac{1}{4}\text{ S} + \frac{1}{6}\text{ S} = \frac{3}{12}\text{ S} + \frac{2}{12}\text{ S} = \frac{5}{12}\text{ S}\]Total resistance \(= \frac{12}{5} = 2.4\text{ Ω}\) — consistent with the parallel resistance formula.
Subtracting fractions
The method is identical to addition, but you subtract the numerators after finding the LCD.
A cable circuit distributes voltage: \(\frac{3}{5}\) of supply goes to one load, \(\frac{1}{4}\) to another. The fraction remaining for the third load (LCD = 20):
\[1 - \frac{3}{5} - \frac{1}{4} = \frac{20}{20} - \frac{12}{20} - \frac{5}{20} = \frac{3}{20}\]Multiplying fractions
To multiply fractions, multiply the numerators together and the denominators together, then simplify. There is no need to find a common denominator. You can simplify before multiplying by cancelling any numerator with any denominator — this keeps the numbers small and reduces arithmetic errors.
- Cancel any common factors diagonally (numerator of one fraction with denominator of the other).
- Multiply the remaining numerators together.
- Multiply the remaining denominators together.
- Simplify further if possible.
Or cancel first: the 9 and 3 share a factor of 3, giving \(\frac{1}{1} \times \frac{3}{16} = \frac{3}{16}\) directly.
A transformer has a turns ratio of \(\frac{2}{5}\). If the primary voltage is \(\frac{5}{4}\) kV, what is the secondary voltage?
\[V_s = \frac{2}{5} \times \frac{5}{4} = \frac{2 \times 5}{5 \times 4} = \frac{10}{20} = \frac{1}{2}\text{ kV} = 500\text{ V}\]A circuit operates at \(\frac{3}{4}\) of its rated current through a resistance whose value is \(\frac{8}{3}\) Ω. Power \(P = I^2 R\), but since current is a fraction of 1 A:
\[P = \left(\frac{3}{4}\right)^2 \times \frac{8}{3} = \frac{9}{16} \times \frac{8}{3} = \frac{72}{48} = \frac{3}{2} = 1.5\text{ W}\]Dividing fractions
Division by a fraction is equivalent to multiplication by its reciprocal (the fraction turned upside down). This is sometimes remembered as "keep, change, flip": keep the first fraction, change ÷ to ×, flip the second fraction.
Why it works: Dividing by \(\frac{a}{b}\) is the same as multiplying by \(\frac{b}{a}\) because \(\frac{a}{b} \times \frac{b}{a} = 1\) — the two fractions cancel each other.
- Keep the first fraction unchanged.
- Change ÷ to ×.
- Flip (invert) the second fraction to its reciprocal.
- Multiply and simplify.
\(P = \frac{V^2}{R}\), so \(R = \frac{V^2}{P}\). If \(V^2 = \frac{9}{4}\text{ V}^2\) and \(P = \frac{3}{8}\text{ W}\):
\[R = \frac{9}{4} \div \frac{3}{8} = \frac{9}{4} \times \frac{8}{3} = \frac{72}{12} = 6\text{ Ω}\]Adding and subtracting mixed fractions
The safest method is to convert each mixed fraction to an improper fraction first, then apply the LCD method. To convert a mixed fraction \(n\frac{a}{b}\) to an improper fraction: multiply the whole number by the denominator and add the numerator: \(\frac{(n \times b) + a}{b}\).
Two cable runs of \(6\frac{3}{4}\) m and \(3\frac{5}{8}\) m must be joined. Total length:
\[6\frac{3}{4} + 3\frac{5}{8} = \frac{27}{4} + \frac{29}{8} = \frac{54}{8}+\frac{29}{8} = \frac{83}{8} = 10\frac{3}{8}\text{ m}\]Multiplying mixed fractions
Always convert to improper fractions before multiplying. Attempting to multiply the integer and fractional parts separately leads to errors.
A motor produces a torque expressed as \(2\frac{1}{2}\text{ Nm}\) at a speed of \(1\frac{1}{5}\) rev/s. Useful output (using \(P = T \times \omega\), with \(\omega = 2\pi n\)):
\[T \times n = \frac{5}{2} \times \frac{6}{5} = \frac{30}{10} = 3\text{ W (approx., ignoring } 2\pi\text{)}\]Decimals and fractions
To convert a fraction to a decimal: divide the numerator by the denominator (long division, or use a calculator). For example \(\frac{22}{7} \approx 3.143\) (the approximation for \(\pi\) used in electrical calculations).
To convert a decimal to a fraction: write the decimal digits over the appropriate power of ten, then simplify. For example \(0.625 = \frac{625}{1000} = \frac{5}{8}\).
| Fraction / expression | Decimal | Use in electrical work |
|---|---|---|
| \(\frac{1}{4}\) | 0.25 | Power factor of 0.25 — a very poor, highly inductive circuit |
| \(\frac{1}{2}\) | 0.5 | 50% efficiency or a 0.5 power factor |
| \(\frac{3}{4}\) | 0.75 | Typical minimum acceptable power factor for a commercial installation |
| \(\frac{22}{7}\) | 3.143 | \(\pi\) — used in \(X_L = 2\pi f L\) and \(X_C = \frac{1}{2\pi fC}\) |
Algebra uses symbols or letters (upper or lower case) to represent unknown or variable quantities, allowing general rules to be expressed as concise formulae. In algebra the multiplication sign is usually omitted — so \(a \times b\) is written as \(ab\), and \(3 \times y\) is written as \(3y\). This shorthand is used throughout electrical theory: Ohm's Law is written \(V = IR\) rather than \(\text{voltage} = \text{current} \times \text{resistance}\). Transposing (rearranging) formulae is a core skill that allows you to find whichever quantity is unknown — R, I, or V — from any two known values.
Fractions and Percentages — Conversion
Being able to move fluently between fractions, decimals, and percentages is important in both calculations and understanding data sheets.
Fraction → Percentage: divide the numerator by the denominator to get a decimal, then multiply by 100.
\[\frac{3}{8} = 0.375 \;\Rightarrow\; 0.375 \times 100 = 37.5\%\]Percentage → Fraction: write the percentage over 100, then cancel to the simplest form.
\[75\% = \frac{75}{100} = \frac{3}{4}\]A motor has a power factor of 0.85. As a percentage: \(0.85 \times 100 = 85\%\). As a fraction: \(0.85 = \frac{85}{100} = \frac{17}{20}\). This means the motor uses \(\frac{17}{20}\) of its apparent power as useful true power.
Find the power dissipated in a 5 Ω resistor when the current is 3 A and the formula is \(P = I^2 R\):
Correct (Order/power first): \(P = 3^2 \times 5 = 9 \times 5 = 45\text{ W}\)
Incorrect (left-to-right): \(P = 3 \times 3 \times 5\) — this actually gives the same answer here, but in a bracketed formula like \(P = (I_1 + I_2)^2 R\) the bracket must be evaluated first.
Addition and Subtraction — Transposition
When rearranging an equation, terms move across the equals sign and change sign: a term that was added on one side becomes subtracted on the other, and vice versa. Think of it as balancing scales — whatever you do to one side, you must do to the other to maintain equality.
If \(V + W = X - Y\), to find X, add Y to both sides:
\[X = V + W + Y\]In a series circuit: \(V_S = V_1 + V_2 + V_3\). To find the voltage across the third component \(V_3\):
\[V_3 = V_S - V_1 - V_2\]With \(V_S = 230\text{ V},\; V_1 = 85\text{ V},\; V_2 = 112\text{ V}\): \(V_3 = 230 - 85 - 112 = 33\text{ V}\)
The formula \(R_2 = R_1 + \Delta R\) relates resistances at two temperatures. To find the change in resistance:
\[\Delta R = R_2 - R_1\]Multiplication and Division — Transposition
When a variable is multiplied by something, divide both sides by that thing to isolate it. When a variable is divided by something, multiply both sides by that thing. A useful mental image: if a quantity is on the top of a fraction on one side, it moves to the bottom on the other side when it crosses the equals sign.
Given \(\frac{WXY}{Z} = \frac{RS}{T}\), to find S (multiply both sides by T, divide both sides by R):
\[S = \frac{WXYT}{ZR}\]Starting from \(V = IR\):
- To find current: \(I = \dfrac{V}{R}\)
- To find resistance: \(R = \dfrac{V}{I}\)
Practical example: a 2.2 kΩ resistor carries 0.05 A. Voltage drop = \(0.05 \times 2200 = 110\text{ V}\).
Simple equations
To solve a simple equation, collect all terms involving the unknown on one side and constants on the other, then divide to find the unknown. Always check your answer by substituting back.
Solve \(5a - 6 = 3a - 8\). Collect a terms on the left, numbers on the right:
\[5a - 3a = -8 + 6 \;\Rightarrow\; 2a = -2 \;\Rightarrow\; a = -1\]Check: \(5(-1)-6 = -11\) and \(3(-1)-8 = -11\) ✓
A circuit has two series resistors. The total is 47 Ω; one resistor is 22 Ω. Find the other:
\[R_1 + 22 = 47 \;\Rightarrow\; R_1 = 47 - 22 = 25\text{ Ω}\]Two parallel branches share 12 A total. One branch carries twice the current of the other: \(I + 2I = 12\):
\[3I = 12 \;\Rightarrow\; I = 4\text{ A}, \quad 2I = 8\text{ A}\]Transposition of Formulae
Many electrical formulae link several quantities. You need to be able to rearrange them freely to find whichever quantity is unknown. The golden rule: perform the same operation to both sides of the equation at each step.
If \(V = 230\text{ V}\) and \(R = 46\text{ Ω}\): \(I = \frac{230}{46} = 5\text{ A}\)
Starting from \(P = IV\), \(P = I^2R\), and \(P = \frac{V^2}{R}\):
| Find | Formula | Example |
|---|---|---|
| \(P\) | \(P = I^2 R\) | \(3^2 \times 5 = 45\text{ W}\) |
| \(I\) | \(I = \sqrt{P/R}\) | \(\sqrt{45/5} = 3\text{ A}\) |
| \(R\) | \(R = P/I^2\) | \(45/9 = 5\text{ Ω}\) |
Formula: \(P = \dfrac{2\pi NT}{60}\), multiply both sides by 60, divide by \(2\pi N\):
\[T = \frac{60P}{2\pi N}\]If \(P = 7500\text{ W}\) and \(N = 1500\text{ rpm}\): \(T = \frac{60 \times 7500}{2\pi \times 1500} = \frac{450000}{9424.8} \approx 47.7\text{ Nm}\)
Given \(Z = \sqrt{R^2 + X_L^2}\), square both sides, subtract \(R^2\), then take the square root:
\[Z^2 = R^2 + X_L^2 \;\Rightarrow\; X_L^2 = Z^2 - R^2 \;\Rightarrow\; X_L = \sqrt{Z^2 - R^2}\]If \(Z = 50\text{ Ω}\) and \(R = 30\text{ Ω}\): \(X_L = \sqrt{2500 - 900} = \sqrt{1600} = 40\text{ Ω}\)
Given \(P = \dfrac{2\pi rN(F_1-F_2)}{60}\), first isolate the bracket:
\[(F_1 - F_2) = \frac{60P}{2\pi rN}\]Then add \(F_2\) to both sides:
\[F_1 = \frac{60P}{2\pi rN} + F_2\]\(R = \dfrac{\rho l}{A}\). To find the maximum length of a 2.5 mm² copper conductor (\(\rho = 17.2 \times 10^{-9}\text{ Ωm}\)) with resistance ≤ 0.86 Ω:
\[l = \frac{R \times A}{\rho} = \frac{0.86 \times 2.5 \times 10^{-6}}{17.2 \times 10^{-9}} = \frac{2.15 \times 10^{-6}}{17.2 \times 10^{-9}} = 125\text{ m}\]Indices (singular: index) are a shorthand way of writing repeated multiplication. The base is the number being multiplied and the index (or exponent) shows how many times. So \(10^3\) means \(10 \times 10 \times 10 = 1\,000\). Indices are indispensable in electrical work because the quantities we deal with span an enormous range — from picofarads (\(10^{-12}\) F) to terawatts (\(10^{12}\) W) — and writing out all the zeros is both cumbersome and error-prone. The SI prefix system (kilo, mega, milli, micro, etc.) is built entirely on powers of ten.
| Power | Value | Prefix |
|---|---|---|
| \(10^{12}\) | 1 000 000 000 000 | T (tera) |
| \(10^9\) | 1 000 000 000 | G (giga) |
| \(10^6\) | 1 000 000 | M (mega) |
| \(10^3\) | 1 000 | k (kilo) |
| \(10^2\) | 100 | h (hecto) |
| \(10^1\) | 10 | d (deka) |
| \(10^0\) | 1 | — |
| \(10^{-1}\) | 0.1 | d (deci) |
| \(10^{-3}\) | 0.001 | m (milli) |
| \(10^{-6}\) | 0.000 001 | µ (micro) |
| \(10^{-9}\) | 0.000 000 001 | n (nano) |
| \(10^{-12}\) | 0.000 000 000 001 | p (pico) |
Squares, Cubes and Square Roots
Square (index 2): a number multiplied by itself once: \(3\times3 = 3^2 = 9\). Used constantly in power calculations: \(P = I^2R\) and \(P = V^2/R\).
Cube (index 3): a number multiplied by itself twice more: \(5\times5\times5 = 5^3 = 125\). Cube calculations appear in volume computations for enclosures and busbar-sizing heat-loss models.
Square root (\(\sqrt{\phantom{x}}\)): the reverse operation of squaring. \(7^2 = 49\) so \(\sqrt{49} = 7\). Square roots arise when finding current from power and resistance (\(I = \sqrt{P/R}\)) or when using the impedance triangle (\(Z = \sqrt{R^2 + X_L^2}\)).
A 10 Ω resistor dissipates 90 W. Find the current:
\[P = I^2 R \;\Rightarrow\; I = \sqrt{\frac{P}{R}} = \sqrt{\frac{90}{10}} = \sqrt{9} = 3\text{ A}\]Laws of Indices
These three laws govern all index calculations. They apply to any base, not just 10, but they are most frequently used with powers of 10 when converting between SI prefixes.
\(3^2 \times 3^5 = 3^{2+5} = 3^7\)
Electrical example: \(10^3 \times 10^{-6} = 10^{3+(-6)} = 10^{-3}\) — multiplying 1 kΩ by 1 µF gives the time constant in milliseconds: \(1 \times 10^3 \times 1 \times 10^{-6} = 1 \times 10^{-3}\text{ s} = 1\text{ ms}\).
\(\dfrac{5^4}{5^2} = 5^{4-2} = 5^2 = 25\)
Electrical example: Converting 1 MW to kW: \(\dfrac{10^6\text{ W}}{10^3\text{ W/kW}} = 10^{6-3} = 10^3\text{ kW} = 1000\text{ kW}\).
\((6^4)^3 = 6^{4\times3} = 6^{12}\)
Electrical example: \((10^3)^2 = 10^6\) — squaring a 1 kΩ value when computing \(V^2/R\) with V in kilovolts.
Negative indices
A negative index means "one over" the positive index: \(a^{-n} = \dfrac{1}{a^n}\). This is the mathematical basis of all "sub-multiple" SI prefixes.
\[\frac{1}{3^4} = 3^{-4} = \frac{1}{81}\]| Negative index | Fraction | SI prefix | Electrical unit example |
|---|---|---|---|
| \(10^{-3}\) | \(\frac{1}{1000}\) | milli (m) | mA — milliampere (RCD trip current) |
| \(10^{-6}\) | \(\frac{1}{1\,000\,000}\) | micro (µ) | µF — microfarad (motor run capacitor) |
| \(10^{-9}\) | \(\frac{1}{10^9}\) | nano (n) | nF — nanofarad (electronic filter capacitor) |
| \(10^{-12}\) | \(\frac{1}{10^{12}}\) | pico (p) | pF — picofarad (stray capacitance in cables) |
Standard Form (Scientific Notation)
Standard form expresses any number as \(A \times 10^n\), where \(1 \leq A < 10\) (exactly one non-zero digit to the left of the decimal point). The exponent \(n\) tells you how many places the decimal point has moved:
- If the original number is large (>1), \(n\) is positive.
- If the original number is small (< 1), \(n\) is negative.
Standard form is essential when entering electrical quantities into calculations — component datasheets, cable resistance tables, and BS 7671 all use it to avoid ambiguous strings of zeros.
| Ordinary number | Standard form | Electrical meaning |
|---|---|---|
| 45 750 000 | \(4.575 \times 10^7\) | 45.75 MW — large generating station output |
| 268 500 | \(2.685 \times 10^5\) | 268.5 kV — EHV transmission voltage |
| 0.000 004 8 | \(4.8 \times 10^{-6}\) | 4.8 µF — motor run capacitor |
| 0.000 000 004 8 | \(4.8 \times 10^{-9}\) | 4.8 nF — electronic filter capacitor |
| 0.000 017 2 | \(1.72 \times 10^{-5}\) | Resistivity of copper in units of Ω·m ×10⁻⁹: \(17.2 \times 10^{-9}\text{ Ωm}\) |
\(X_L = 2\pi f L = 2 \times 3.142 \times 50 \times 0.000\,318\)
Enter L in standard form: \(3.18 \times 10^{-4}\text{ H}\). This avoids misplacing a decimal.
\[X_L = 6.284 \times 50 \times 3.18 \times 10^{-4} = 314.2 \times 3.18 \times 10^{-4} = 0.0999... \approx 0.1\text{ Ω}\]Engineering Notation
Engineering notation is a close relative of standard form, but uses only powers of 10 that are multiples of 3 (…10⁻⁹, 10⁻⁶, 10⁻³, 10⁰, 10³, 10⁶, 10⁹…). This makes it align perfectly with SI prefixes, which is why it is the notation of choice on multimeters, cable datasheets, BS 7671 tables, and component specifications used every day in electrical work.
Standard form: A × 10ⁿ where 1 ≤ A < 10 (any integer n)
Engineering notation: A × 10ⁿ where 1 ≤ A < 1000 and n is a multiple of 3
Because n must be a multiple of 3, A can range from 1 up to (but not including) 1000, rather than being locked to a single leading digit.
Conversion steps
- Write the number in standard form first.
- Adjust the exponent to the nearest lower multiple of 3.
- Compensate by moving the decimal point in A by the same number of places.
- Replace 10ⁿ with its SI prefix if one exists.
| Ordinary number | Standard form | Engineering notation | SI prefix form |
|---|---|---|---|
| 4 700 000 Ω | 4.7 × 10⁶ Ω | 4.7 × 10⁶ Ω | 4.7 MΩ |
| 0.025 A | 2.5 × 10⁻² A | 25 × 10⁻³ A | 25 mA |
| 0.000 047 F | 4.7 × 10⁻⁵ F | 47 × 10⁻⁶ F | 47 µF |
| 0.000 000 220 F | 2.2 × 10⁻⁷ F | 220 × 10⁻⁹ F | 220 nF |
| 33 000 Ω | 3.3 × 10⁴ Ω | 33 × 10³ Ω | 33 kΩ |
| 0.000 001 5 H | 1.5 × 10⁻⁶ H | 1.5 × 10⁻⁶ H | 1.5 µH |
| 400 000 000 Hz | 4 × 10⁸ Hz | 400 × 10⁶ Hz | 400 MHz |
Step 1 — Standard form: 0.025 = 2.5 × 10⁻²
Step 2 — Nearest lower multiple of 3: 10⁻² → use 10⁻³ (i.e. shift exponent down by 1)
Step 3 — Compensate A: move decimal 1 place right → A becomes 25
Result: 25 × 10⁻³ A = 25 mA
Step 1 — Standard form: 4.7 × 10⁻⁵ F
Step 2 — Nearest lower multiple of 3: 10⁻⁵ → use 10⁻⁶ (shift exponent down by 1)
Step 3 — Compensate A: move decimal 1 place right → A becomes 47
Result: 47 × 10⁻⁶ F = 47 µF
Step 1 — Standard form: 4 × 10⁸ Hz
Step 2 — Nearest lower multiple of 3: 10⁸ → use 10⁶ (shift exponent down by 2)
Step 3 — Compensate A: move decimal 2 places right → A becomes 400
Result: 400 × 10⁶ Hz = 400 MHz
- Multimeters display readings in mA, kΩ, µF — all engineering notation.
- Cable resistance tables quote values in mΩ/m or Ω/km.
- BS 7671 Appendix 4 uses mΩ for impedance calculations (Zs, Ze, R1+R2).
- Capacitor markings use µF, nF, pF; inductor markings use mH, µH.
- Supply frequencies are MHz or GHz in communications work.
| SI Prefix | Symbol | Power of 10 | Electrical example |
|---|---|---|---|
| Giga | G | 10⁹ | 1 GΩ = 1 000 000 000 Ω (insulation resistance) |
| Mega | M | 10⁶ | 4.7 MΩ = 4 700 000 Ω (high-value resistor) |
| kilo | k | 10³ | 33 kΩ = 33 000 Ω (pull-up resistor) |
| — | — | 10⁰ | 230 Ω (base unit, no prefix) |
| milli | m | 10⁻³ | 25 mA = 0.025 A (LED current) |
| micro | µ | 10⁻⁶ | 47 µF = 0.000 047 F (motor run capacitor) |
| nano | n | 10⁻⁹ | 220 nF = 0.000 000 220 F (filter capacitor) |
| pico | p | 10⁻¹² | 100 pF = 0.000 000 000 1 F (trimmer capacitor) |
Use of Brackets
If brackets are multiplied by a positive (+) quantity, all signs inside remain unchanged.
If brackets are multiplied by a negative (−) quantity, every sign inside changes (+ becomes −, and − becomes +).
The total resistance of a circuit is \(R_T = R_1 + 2(R_2 - R_3)\). With \(R_1 = 10\), \(R_2 = 8\), \(R_3 = 3\):
\[R_T = 10 + 2(8 - 3) = 10 + 2 \times 5 = 10 + 10 = 20\text{ Ω}\]Power in a belt drive: \(P = \frac{2\pi rN(F_1-F_2)}{60}\). The bracket \((F_1-F_2)\) is the net driving force. Expanding brackets would give \(\frac{2\pi rN F_1 - 2\pi rN F_2}{60}\) — useful for solving when \(F_1\) and \(F_2\) are each expressed as functions of other variables.
Statistics is the branch of mathematics concerned with collecting, organising, analysing and presenting data. It is obtained by two main methods:
- By counting (discrete data) – e.g. number of consumer units sold by an electrical wholesaler each month; number of RCD trips reported in a building over a year.
- By measurement (continuous data) – e.g. the supply voltage recorded at a distribution board every hour; the insulation resistance values of all circuits in a new installation.
In electrical installation and maintenance work, statistics are used to track energy consumption trends, identify fault patterns, plan maintenance schedules, and produce compliance reports. Understanding how to read and construct simple charts ensures you can interpret the data you record and present it clearly to clients or site managers.
Tables, Charts and Diagrams
Raw data becomes much more useful once it is organised into a table or chart. The choice of chart depends on the story the data tells:
- Vertical bar (column) charts – best for comparing discrete categories or showing change over time. The height of each bar represents the value. Used to compare fault call-outs per month, or energy use per floor of a building.
- Horizontal bar charts – the same as vertical bar charts, but drawn sideways. Useful when category labels are long, or when comparing ranked items (e.g. the top ten pieces of equipment by energy consumption).
- Line graphs – best for showing continuous data changing over time, e.g. a 24-hour voltage profile at a distribution board, or insulation resistance trending downward over several months, indicating a developing fault.
Example: Vertical bar chart — Power Tools Stock Check
Example: Horizontal bar chart — Electrical Hand Tools Stock Check
Example: Line graph — Insulation Resistance Trend Over Time
Example — employee statistics for a large road assistance company:
| Type of personnel | Number employed | Percentage | Pie chart angle |
|---|---|---|---|
| Motor engineers | 260 | 65% | 234° |
| Call centre staff | 60 | 15% | 54° |
| Support team | 40 | 10% | 36° |
| Staff canteen staff | 20 | 5% | 18° |
| Office cleaners | 10 | 5% | 18° |
| Total | 400 | 100% | 360° |
Calculating pie chart angles — rationale and method
A full circle = 360°. Each sector's angle is found by expressing the category as a fraction of the total, then multiplying by 360°. The formula is:
\[\text{Sector angle} = \frac{\text{category value}}{\text{total}} \times 360°\]- Motor engineers: \(\frac{260}{400} \times 360 = 0.65 \times 360 = 234°\)
- Call centre staff: \(\frac{60}{400} \times 360 = 0.15 \times 360 = 54°\)
- Support team: \(\frac{40}{400} \times 360 = 0.10 \times 360 = 36°\)
- Staff canteen staff: \(\frac{20}{400} \times 360 = 0.05 \times 360 = 18°\)
- Office cleaners: \(\frac{10}{400} \times 360 = 0.05 \times 360 = 18°\)
- Check: \(234 + 54 + 36 + 18 + 18 = 360°\) ✓
Pie chart — Employee Statistics for a Large Road Assistance Company
Additional worked example — energy consumption pie chart
| End use | kWh | % | Sector angle |
|---|---|---|---|
| Lighting | 2400 | 30% | 108° |
| Heating / AC | 3200 | 40% | 144° |
| IT equipment | 1600 | 20% | 72° |
| Other | 800 | 10% | 36° |
| Total | 8000 | 100% | 360° |
This type of chart would typically accompany an energy audit report, enabling the building manager to see at a glance that heating and air-conditioning represents the largest saving opportunity.
Pythagoras' Theorem
In any right-angled triangle, the square of the hypotenuse (the longest side, always opposite the right angle) equals the sum of the squares of the other two sides. If we label the hypotenuse c and the other two sides a and b:
\[c^2 = a^2 + b^2 \quad\Rightarrow\quad c = \sqrt{a^2 + b^2}\]Rearranged to find either shorter side: \(a = \sqrt{c^2 - b^2}\) and \(b = \sqrt{c^2 - a^2}\).
Right-angled triangle: a = opposite, b = adjacent, c = hypotenuse (AC)
Pythagoras' theorem appears in electrical work wherever right-angled geometry arises. The most important is the impedance triangle: resistance (R), reactance (X) and impedance (Z) form a right-angled triangle where \(Z = \sqrt{R^2 + X^2}\). It is also used to find cable lengths, heights of cable routes, and the geometry of voltage phasors.
Impedance triangle: Z² = R² + X² — applying Pythagoras in AC circuit analysis
Triangle with sides a = 40 cm and b = 9 cm. Find the hypotenuse:
\[c = \sqrt{40^2 + 9^2} = \sqrt{1600+81} = \sqrt{1681} = 41\text{ cm}\]A circuit has resistance R = 30 Ω and inductive reactance \(X_L\) = 40 Ω. Find impedance Z:
\[Z = \sqrt{30^2 + 40^2} = \sqrt{900+1600} = \sqrt{2500} = 50\text{ Ω}\]A load draws 6 kW (true power) and 8 kVAr (reactive power). Find apparent power (kVA):
\[S = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10\text{ kVA}\]Voltage across R is 90 V; voltage across L is 120 V. Total supply voltage:
\[V_S = \sqrt{90^2 + 120^2} = \sqrt{8100 + 14400} = \sqrt{22500} = 150\text{ V}\]Note: you cannot simply add 90 + 120 = 210 V in an ac circuit — the voltages are 90° out of phase, so the triangle rule must be used.
Trigonometry
Trigonometry is the study of the relationship between the angles and sides of triangles. In a right-angled triangle, three ratios link any angle (θ) to the sides:
- The hypotenuse — the longest side, opposite the right angle (90°).
- The opposite side — the side directly across from angle θ.
- The adjacent side — the side next to angle θ (not the hypotenuse).
A memory aid: SOH-CAH-TOA — Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent.
The inverse functions (sin⁻¹, cos⁻¹, tan⁻¹) — also written arcsin, arccos, arctan — work backwards from the ratio to give the angle. Use these when you know two sides and need to find an angle.
Sides: hypotenuse = 5, opposite = 3, adjacent = 4.
\[\sin\theta = \frac{3}{5} = 0.6 \qquad \cos\theta = \frac{4}{5} = 0.8 \qquad \tan\theta = \frac{3}{4} = 0.75\]\(\theta = \sin^{-1}(0.6) = \cos^{-1}(0.8) = \tan^{-1}(0.75) = 36.87°\)
In a circuit, true power P = 6 kW and apparent power S = 10 kVA. The power factor angle θ:
\[\cos\theta = \frac{P}{S} = \frac{6}{10} = 0.6 \;\Rightarrow\; \theta = \cos^{-1}(0.6) = 53.13°\]The power factor is 0.6 lagging — meaning current lags voltage by 53.13°.
R = 30 Ω, \(X_L\) = 40 Ω, Z = 50 Ω. Find the phase angle:
\[\tan\theta = \frac{X_L}{R} = \frac{40}{30} = 1.333 \;\Rightarrow\; \theta = \tan^{-1}(1.333) = 53.13°\]Checking with sin: \(\sin 53.13° = \frac{40}{50} = 0.8\) ✓
A cable must pass diagonally over a water tank. The horizontal distance is 12 m and the cable must rise at an angle of \(\theta = 22.62°\) to clear the tank. Find the cable length (hypotenuse):
\[\cos 22.62° = \frac{\text{adjacent}}{\text{hyp}} \;\Rightarrow\; \text{hyp} = \frac{12}{\cos 22.62°} = \frac{12}{0.9231} \approx 13\text{ m}\]Alternatively using the opposite side (height = 5 m): \(\text{hyp} = \frac{5}{\sin 22.62°} = \frac{5}{0.3847} \approx 13\text{ m}\)
🃏 Chapter 1 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Covers fractions, percentages, algebra, indices, standard form, statistics and trigonometry.
(b) Standard form: 2.2 × 10⁻⁷ F (A must be 1 ≤ A < 10)
(c) Ordinary decimal: 0.000 000 220 F. Note: only standard form has A strictly between 1 and 10; engineering notation allows A up to 999.
📝 Chapter 1 — Multiple Choice Questions
Select your answer and click Submit. You have two attempts per question. First-attempt score shown at the end.
1What is the value of ¾ + ⅙?
2A cable carries a current of 20 A and the allowable volt drop is 5%. The supply voltage is 230 V. What is the maximum permitted volt drop in volts?
3Solve for x: 5x + 3 = 28
4Transpose V = IR to make I the subject.
5Evaluate using BODMAS: 10 + 3 × 2² − (12 ÷ 4)
6Simplify: 2³ × 2⁻¹ ÷ 2²
7Express 4,700,000 Ω in standard form.
8Five daily energy readings (kWh) are: 45, 50, 45, 60, 55. What is the mean daily energy consumption?
9In a pie chart of building energy use, lighting accounts for 90 kWh out of a total 450 kWh. What is the sector angle for lighting?
10A right-angled impedance triangle has R = 8 Ω and XL = 6 Ω. What is the impedance Z and the power factor (cos θ)?
11Using trigonometry, an ac circuit has Z = 25 Ω and R = 20 Ω. What is the phase angle θ?
12The decimal 0.000 001 is equal to which SI prefix?
13Which of the following correctly expresses 0.0082 A in engineering notation?
14A resistor is labelled 4.7 × 10⁴ Ω in standard form. What is this value expressed in engineering notation with its SI prefix?
Sends your Chapter 1 MCQ score summary from your registered email to your tutor.
Standard Units of Measurement and Measuring Instruments
- Identify the base units and derived units of the SI system
- Name symbols and factors for multiples and submultiples of SI units
- Use SI units for measurement of general variables (area, volume, density, etc.)
- Identify and determine values of basic SI units for electrical variables
- Identify appropriate electrical instruments for measurement and calculation of electrical values
The SI system (Système International d'Unités — International System of Units) is the modern, internationally agreed metric system of measurement. It is built on seven independent base units from which all other units are derived. In the UK, SI has largely replaced the imperial system (yards, feet, inches, pounds) in engineering and science, although some older documents still use imperial measurements. As an electrician you must be comfortable working in SI and converting between its multiples and sub-multiples (kilo, milli, micro, etc.).
There are also two supplementary units: the radian (rad) for plane angle and the steradian (sr) for solid angle — these appear in lighting calculations and some AC theory.
Base Units (Table 2.1)
The five base units most relevant to electrical work are shown below. Each is precisely defined by international agreement so that measurements anywhere in the world are consistent and comparable.
| Variable | Unit | Unit symbol | Variable symbol | Electrical relevance |
|---|---|---|---|---|
| Length | metre | m | l | Cable runs, conduit lengths, circuit distances for volt drop calculations |
| Mass | kilogram | kg | m | Weight of switchgear, cable reels, and busbar trunking for structural loading |
| Time | second | s | t | Fault clearance times (overcurrent protection), charge and discharge rates of capacitors |
| Electric current | ampere | A | I | Load current, fuse and breaker ratings, shock risk assessment (lethal at ~30 mA) |
| Temperature* | kelvin | K | T | Conductor operating temperature affects resistance and current-carrying capacity |
*The degree Celsius (°C) is used for everyday temperature. The relationship is: \(\text{K} = \text{°C} + 273.15\), so 0°C = 273.15 K (absolute zero is 0 K = −273.15°C). For temperature conversion between Celsius and Fahrenheit: \(\text{°C} = \frac{5}{9}(\text{°F}-32)\) and \(\text{°F} = \frac{9}{5}\text{°C}+32\).
A cable is rated at a conductor temperature of 70°C. The ambient temperature is 35°C (rather than the standard 30°C reference). Convert both to kelvin to check the temperature rise:
\[70\text{°C} = 343.15\text{ K}; \quad 35\text{°C} = 308.15\text{ K}; \quad \Delta T = 343.15\text{ K} - 308.15\text{ K} = 35\text{ K}\]The 5°C increase in ambient above the 30°C standard reference requires a derating of the cable's current-carrying capacity (correction factor from BS 7671 Table 4B1).
Derived Units (Table 2.2)
| Variable | Unit | Unit symbol | Variable symbol |
|---|---|---|---|
| Acceleration | metre/second² | m/s² | a |
| Area (cross-sectional) | square metre | m² | A |
| Capacitance | farad | F | C |
| Charge (quantity of electricity) | coulomb | C | Q |
| Density | kilogram/cubic metre | kg/m³ | ρ |
| Electromotive force (emf) | volt | V | V |
| Electrical energy | joule / kilowatt hour | J / kWh | W |
| Force | newton | N | F |
| Frequency | hertz | Hz | f |
| Inductance | henry | H | L |
| Potential difference | volt | V | V |
| Power | watt | W | P |
| Resistance | ohm | Ω | R |
| Resistivity | ohm metre | Ωm | ρ |
| Speed (velocity) | metre/sec | m/s | v |
| Volume | cubic metre | m³ | V |
| Magnetic flux | weber | Wb | Φ |
| Magnetic flux density | tesla | T | B |
Multiples and Submultiples (Table 2.3)
| Prefix | Symbol | Value |
|---|---|---|
| tera | T | \(10^{12}\) = 1 000 000 000 000 |
| giga | G | \(10^9\) = 1 000 000 000 |
| mega | M | \(10^6\) = 1 000 000 |
| kilo | k | \(10^3\) = 1 000 |
| deci | d | \(10^{-1}\) = 0.1 |
| centi | c | \(10^{-2}\) = 0.01 |
| milli | m | \(10^{-3}\) = 0.001 |
| micro | µ | \(10^{-6}\) = 0.000 001 |
| nano | n | \(10^{-9}\) = 0.000 000 001 |
| pico | p | \(10^{-12}\) = 0.000 000 000 001 |
Common electrical conversions: 1 MW = \(10^6\) W; 1 kW = \(10^3\) W; 1 mA = \(10^{-3}\) A; 1 µF = \(10^{-6}\) F.
Speed, Velocity and Acceleration
Speed is a scalar quantity — it has magnitude only and no direction. Velocity is a vector quantity — it has both magnitude and direction. In practice, a motor that drives a conveyor at "5 m/s" quotes its speed; but if you say the conveyor moves "5 m/s to the right", that is velocity. Both are measured in metres per second (m/s).
Acceleration is the rate of change of velocity: \(a = \frac{\Delta v}{t}\), measured in m/s². It matters in electrical work when selecting motor starters — a direct-on-line (DOL) start produces a sudden high current because the motor accelerates rapidly, while a soft starter reduces the initial acceleration and current.
A car goes from rest to 30 m/s in 3 seconds:
\[a = \frac{30 - 0}{3} = 10\text{ m/s}^2\]An electrician pushes a cable drum trolley from rest to 2 m/s in 4 seconds across a workshop floor. Calculate the acceleration.
\[a = \frac{\Delta v}{t} = \frac{2 - 0}{4} = 0.5\text{ m/s}^2\]The trolley accelerates at 0.5 m/s². The larger the mass of cable on the trolley, the greater the force required to produce this acceleration — linking directly to Newton's second law (F = ma).
An electric lift motor decelerates a 1000 kg lift from 3 m/s to rest in 2 s. Deceleration (negative acceleration):
\[a = \frac{0 - 3}{2} = -1.5\text{ m/s}^2\]The braking force required: \(F = m \times a = 1000 \times 1.5 = 1500\text{ N}\)
Area
Area measures the amount of two-dimensional space a shape occupies. The SI unit is the square metre (m²), but in electrical installation work you frequently use mm² (for conductor cross-sectional area) and cm² (for small panel spaces). Key formulae:
Circle: \(A = \pi r^2 = \dfrac{\pi d^2}{4}\)
Triangle: \(A = \frac{1}{2} \times \text{base} \times \text{height}\)
Rectangle: A = l × b
Circle: A = πr²
Triangle: A = ½ × b × h
Area calculations are used in electrical work to: determine the cross-sectional area (csa) of conductors (which governs current-carrying capacity and resistance), calculate the space factor of conduits and trunking, and size containment systems.
A single-core cable has a conductor diameter of 6 mm. Calculate the cross-sectional area (csa) of the conductor in mm².
Use the formula: \(A = \dfrac{\pi d^2}{4}\) — give your answer to 2 decimal places.
▶ Show worked answer
The cross-sectional area of the conductor is 28.27 mm².
A steel trunking has internal dimensions of 70 mm × 50 mm. Calculate: (a) the internal cross-sectional area, and (b) the maximum usable cable fill area, given that BS 7671 limits cable fill to 45% of the internal csa for single-layer installations.
▶ Show worked answer
(a) Internal csa:
\[\text{Internal csa} = 70 \times 50 = 3500\text{ mm}^2\](b) Maximum usable cable fill area (45%):
\[\text{Max fill} = 3500 \times 0.45 = 1575\text{ mm}^2\]The maximum usable area for cables is 1575 mm². This is the total csa of all cables that can be installed in the trunking without exceeding the BS 7671 fill ratio.
Three 4 mm² single-core cables are to be drawn into a circular conduit of internal diameter 20 mm. Each cable has an overall diameter of 6.9 mm. Check whether the conduit is adequate, given that BS 7671 requires the total cable csa to not exceed 40% of the conduit bore for three or more cables.
▶ Show worked answer
Step 1 — Conduit internal csa and 40% limit:
\[\text{Conduit csa} = \frac{\pi \times 20^2}{4} = 314.2\text{ mm}^2 \qquad 40\% \text{ limit} = 314.2 \times 0.4 = 125.7\text{ mm}^2\]Step 2 — Total cable csa:
\[\text{Each cable} = \frac{\pi \times 6.9^2}{4} = 37.4\text{ mm}^2 \qquad \text{Three cables} = 3 \times 37.4 = 112.2\text{ mm}^2\]Step 3 — Compare:
\[112.2\text{ mm}^2 < 125.7\text{ mm}^2 \checkmark\]The total cable area of 112.2 mm² is less than the 40% limit of 125.7 mm² — the conduit is adequate.
Volume
Volume is the amount of three-dimensional space a body occupies. SI unit: cubic metre (m³); also cm³ and litres (1 litre = 0.001 m³ = 1 000 cm³) for liquid volumes in battery rooms and cooling systems.
Cylinder: \(V = \dfrac{\pi d^2 \times l}{4} = \pi r^2 l\)
Volume shapes: Cuboid V = l × b × d | Cylinder V = πr² × l
Diameter 0.5 m, height 2 m:
\[V = \frac{3.142 \times 0.5^2 \times 2}{4} = \frac{3.142 \times 0.25 \times 2}{4} = 0.393\text{ m}^3\]The volume of liquid this tank holds = \(0.393 \times 1000 = 393\text{ litres}\) — useful when considering splash risk for cable routing decisions.
A switch room is 4 m × 3 m × 2.5 m high. Volume:
\[V = 4 \times 3 \times 2.5 = 30\text{ m}^3\]Ventilation engineers need this figure to calculate the number of air changes per hour required to keep transformer and switchgear temperatures within safe limits.
Density and Mass
Mass is the quantity of matter in a body. SI unit: kilogram (kg). Do not confuse mass with weight — weight is a force (measured in newtons, N) and depends on gravitational acceleration (g = 9.81 m/s²): \(W = m \times g\).
Density (symbol ρ) is the mass per unit volume: \(\rho = m/V\). SI unit: kg/m³. Density is important when calculating the mass of copper or aluminium conductors, busbar sections, or transformer cores.
A rectangular copper busbar is 100 mm × 10 mm × 2 m long. Density of copper = 8900 kg/m³. Volume:
\[V = 0.1 \times 0.01 \times 2 = 0.002\text{ m}^3\] \[m = \rho V = 8900 \times 0.002 = 17.8\text{ kg}\]This mass figure is used to check the structural loading of the busbar supports.
| Material | Density (kg/m³) | Electrical use |
|---|---|---|
| Copper | 8900 | Conductors, busbars, windings |
| Aluminium | 2700 | Overhead lines, large cable conductors |
| Iron / steel | 7870 | Motor cores, transformer cores, conduit |
| PVC insulation | ~1380 | Cable sheath and insulation |
Resistance (symbol R)
Resistance is the property of a material that opposes the flow of electric current, converting electrical energy into heat in the process. Every conductor — including the cables in an installation — has resistance. While conductors are chosen to minimise resistance, that resistance still causes voltage drop and power loss. The SI unit of resistance is the ohm (Ω), defined as one volt per ampere: \(1\text{ Ω} = 1\text{ V/A}\).
Four factors affect the resistance of a conductor:
- Length (l) — resistance is directly proportional to length. Double the length → double the resistance. This is why BS 7671 requires a volt drop calculation: a long cable run has higher resistance and therefore a larger volt drop.
- Cross-sectional area (A) — resistance is inversely proportional to csa. Double the csa → half the resistance. Choosing a larger cable size reduces both resistance and volt drop.
- Material (resistivity ρ) — different materials have inherently different resistivities. Copper has low resistivity and is preferred for fixed wiring; aluminium is used for large distribution cables because it is lighter and cheaper despite higher resistivity.
- Temperature (T) — for most metallic conductors, resistance increases with temperature. This matters when calculating fault loop impedance at operating temperature (BS 7671 uses a multiplier of 1.2 for copper at 70°C compared to 20°C).
| Material | Resistivity at 20°C (Ω·m) | Use |
|---|---|---|
| Silver | \(15.9 \times 10^{-9}\) | Special high-conductivity contacts |
| Copper | \(17.2 \times 10^{-9}\) | Fixed wiring conductors, busbars |
| Aluminium | \(28.4 \times 10^{-9}\) | Overhead lines, large LV/MV cables |
| Nichrome | \(1110 \times 10^{-9}\) | Heating elements, resistors |
Find the resistance of a 50 m run of 2.5 mm² copper conductor (\(\rho = 17.2 \times 10^{-9}\text{ Ω·m}\), \(A = 2.5 \times 10^{-6}\text{ m}^2\)):
\[R = \frac{\rho l}{A} = \frac{17.2 \times 10^{-9} \times 50}{2.5 \times 10^{-6}} = \frac{860 \times 10^{-9}}{2.5 \times 10^{-6}} = \frac{860}{2500} = 0.344\text{ Ω}\]The volt drop at 16 A (the maximum current for a 2.5 mm² circuit): \(V_d = IR = 16 \times 0.344 = 5.5\text{ V}\). For a 230 V circuit, this is \(\frac{5.5}{230} \times 100 = 2.39\%\) — within the 3% BS 7671 lighting limit.
The same 50 m run but using 1.5 mm² cable:
\[R = \frac{17.2 \times 10^{-9} \times 50}{1.5 \times 10^{-6}} = 0.573\text{ Ω}\]Volt drop at 13 A: \(13 \times 0.573 = 7.45\text{ V}\) = 3.24% — exceeds the 3% lighting limit. The smaller cable creates a compliance issue.
Resistance and temperature
For most metallic conductors (copper, aluminium, nichrome), resistance increases with temperature — they have a positive temperature coefficient (PTC). Semiconductor materials and carbon, by contrast, have a negative temperature coefficient (NTC) — their resistance falls as temperature rises.
The formula relating resistance at two temperatures is:
\[\frac{R_1}{R_2} = \frac{1 + \alpha t_1}{1 + \alpha t_2}\]where α = temperature coefficient of resistance (the increase in resistance per degree Celsius for a material with 1 Ω at 0°C), \(t_1\) and \(t_2\) are temperatures in °C.
A copper conductor has resistance 0.5 Ω at 20°C. Find its resistance at 70°C (maximum operating temperature for 70°C PVC insulated cable). For copper, α = 0.004/°C.
\[R_{70} = R_{20} \times \frac{1 + \alpha \times 70}{1 + \alpha \times 20} = 0.5 \times \frac{1 + (0.004 \times 70)}{1 + (0.004 \times 20)} = 0.5 \times \frac{1.28}{1.08} = 0.593\text{ Ω}\]The resistance has increased by about 18.5% — consistent with the BS 7671 multiplier of approximately 1.2 used in fault loop impedance calculations.
Voltage (symbol V)
Electromotive force (emf) is the energy per unit charge supplied by a source (battery, generator, transformer) — the "driving force" that pushes electrons around a circuit. Potential difference (pd) is the energy per unit charge transferred between two points in a circuit. Both are measured in volts (V), defined as one joule per coulomb: \(1\text{ V} = 1\text{ J/C}\).
Emf is the open-circuit voltage; pd is measured across a component under load. In most practical calculations the terms are used interchangeably as "voltage".
| System | Nominal voltage | Application |
|---|---|---|
| Single-phase (L–N) | 230 V ac (±10%) | Standard socket outlets, lighting |
| Three-phase (L–L) | 400 V ac | Motor feeds, industrial equipment |
| ELV / SELV | ≤ 50 V ac / 120 V dc | Garden/outdoor lighting, data systems |
| High voltage | 11 kV / 33 kV | Distribution substations |
Current (symbol I)
Electric current is the rate of flow of electric charge through a conductor. The SI unit is the ampere (A). One coulomb contains \(6.3 \times 10^{18}\) electrons; one ampere is defined as one coulomb passing a point per second:
\[I = \frac{Q}{t} \quad\text{or}\quad Q = It\]Current causes heating in conductors (\(P = I^2 R\)) and creates magnetic fields in coils. The current rating of a cable is set by the maximum temperature its insulation can withstand continuously.
A 13 A circuit runs for 2 hours. Total charge:
\[Q = It = 13 \times (2 \times 3600) = 93\,600\text{ C}\]A 1.5 mm² cable rated at 16 A has resistance 0.012 Ω/m. At rated current, heat generated per metre = \(16^2 \times 0.012 = 3.07\text{ W/m}\). At 20 A (overload): \(20^2 \times 0.012 = 4.8\text{ W/m}\) — a 57% increase in heat generation, which will raise conductor temperature above the 70°C insulation rating if sustained.
Power (symbol P)
Power is the rate of energy conversion. SI unit: watt (W) = 1 J/s. For dc and purely resistive ac circuits:
\[P = \frac{W}{t} = V \times I = I^2 R = \frac{V^2}{R}\]A 46 Ω element connected to 230 V:
\[P = \frac{V^2}{R} = \frac{230^2}{46} = \frac{52900}{46} = 1150\text{ W} = 1.15\text{ kW}\]A 2.5 mm² cable (R = 0.344 Ω) carries 16 A. Heat dissipated in the cable:
\[P_{\text{loss}} = I^2 R = 16^2 \times 0.344 = 88.1\text{ W}\]This heat must be allowed to dissipate — if the cable is in thermal insulation, its rating must be derated (BS 7671 Table 4C1).
Energy (symbol W)
Energy is the capacity to do work. SI unit: joule (J). For practical electrical metering, the kilowatt-hour (kWh) is used:
\[W\text{(J)} = P\text{(W)} \times t\text{(s)} \qquad W\text{(kWh)} = P\text{(kW)} \times t\text{(h)}\]Conversion: \(1\text{ kWh} = 3.6 \times 10^6\text{ J} = 3.6\text{ MJ}\).
A 2 kW heater runs for 4 hours:
\[W = 2 \times 4 = 8\text{ kWh} = 28.8\text{ MJ}\]At 28 p/kWh: cost = \(8 \times 28 = 224\text{p} = £2.24\).
A premises uses 1800 kWh/month. At 0.233 kg CO₂/kWh:
\[\text{CO}_2 = 1800 \times 0.233 = 419.4\text{ kg/month}\]Energy meters record this consumption for billing and carbon-reporting purposes; the modern smart meter transmits readings automatically to the supplier.
Frequency and Periodic Time
An alternating supply reverses direction continuously. The number of complete cycles per second is the frequency (f), measured in hertz (Hz). The time for one complete cycle is the periodic time (T):
\[f = \frac{1}{T} \qquad T = \frac{1}{f}\]The UK standard is 50 Hz. Frequency directly affects reactance: \(X_L = 2\pi fL\) and \(X_C = 1/(2\pi fC)\) — so equipment rated for 50 Hz may malfunction at 60 Hz (USA) and vice versa.
Within each 20 ms cycle the voltage reaches a peak of approximately \(230 \times \sqrt{2} = 325\text{ V}\) in each direction.
The very high frequency allows much smaller transformers and capacitors — this is why laptop chargers and LED drivers are compact despite handling significant power.
Inductance and Inductive Reactance (XL)
An inductor is a coil of wire. Connected to a dc supply it acts as a resistor. On ac it develops inductance (L), measured in henrys (H). The opposition to ac is the inductive reactance XL, measured in ohms.
\[X_L = 2\pi f L\]Capacitance (symbol C)
A capacitor stores electric charge. It consists of two conductive plates separated by a dielectric insulator. Capacitance (C) is measured in farads (F), but practical values use µF, nF or pF:
\[1\,\mu\text{F} = 10^{-6}\text{ F} \qquad 1\,\text{nF} = 10^{-9}\text{ F} \qquad 1\,\text{pF} = 10^{-12}\text{ F}\] \[C = \frac{Q}{V} \quad\Rightarrow\quad Q = CV\]Capacitance depends on: plate area (larger → greater C), plate separation (smaller → greater C), and the type of dielectric material.
Dielectric Constants (Table 2.4)
| Material | Dielectric constant |
|---|---|
| Air | 1.0 |
| Glass | 7.6 |
| Mica | 7.5 |
| Mylar | 3.0 |
| Paper | 2.5 |
| Porcelain | 6.3 |
| Quartz | 5.0 |
| Aluminium oxide | 10.0 |
| Tantalum oxide | 11.0 |
Capacitive Reactance (XC)
The opposition a capacitor offers to ac is the capacitive reactance XC, measured in ohms:
\[X_C = \frac{1}{2\pi f C} \qquad\text{(or if C is in \mu F:)}\quad X_C = \frac{10^6}{2\pi f C}\]Impedance (symbol Z)
Impedance is the total opposition to ac current flow in a circuit — the combined effect of resistance (R), inductive reactance (XL) and capacitive reactance (XC). Unit: ohm (Ω).
It is found using the impedance triangle and Pythagoras' theorem:
R-C circuit: \(Z = \sqrt{R^2 + X_C^2}\)
R-L-C circuit (XL > XC): \(Z = \sqrt{R^2 + (X_L - X_C)^2}\)
R-L-C circuit (XC > XL): \(Z = \sqrt{R^2 + (X_C - X_L)^2}\)
Power Factor
In a dc circuit, true power = V × I. In an ac circuit, V × I gives apparent power (VA). The ratio of true power to apparent power is the power factor:
\[\text{Power factor} = \cos\theta = \frac{\text{true power (kW)}}{\text{apparent power (kVA)}} = \frac{R}{Z}\]Power factor can be:
- Unity (1) — true power = apparent power (purely resistive circuit)
- Lagging — inductive circuit, current lags voltage
- Leading — capacitive circuit, current leads voltage
A good power factor is close to 1 (e.g. 0.9); a poor power factor is close to 0 (e.g. 0.4).
Power in AC Circuits
Three types of power exist in an ac circuit:
- True (actual) power — dissipated in resistance, measured in watts (W) or kW
- Reactive power — wattless power in the reactive components, measured in VAr or kVAr
- Apparent power — combined true and reactive power, measured in VA or kVA
Measuring Resistance
The instrument used to measure resistance is an ohmmeter. Modern multifunction testers and dedicated installation testers include an ohmmeter function. In an ohmmeter, the meter's own battery drives a small current through the unknown resistance; the resulting voltage drop is processed and displayed as an ohms reading. Self-ranging digital instruments automatically select the correct measurement range.
Before use: check the battery condition (low battery gives inaccurate readings), then short-circuit the test leads and confirm the display reads 0 Ω (or null/zero the meter on instruments with a zero-adjust). Good-quality leads with low resistance are essential.
During an electrical installation condition report (EICR), a 500 V dc insulation resistance (IR) test is performed. The ohmmeter sends 500 V dc across the insulation. A healthy circuit shows IR > 1 MΩ (BS 7671 minimum). A reading of 0.8 MΩ would indicate a fault requiring investigation.
Measuring Voltage
Voltage is measured with a voltmeter connected in parallel (across) the component or supply terminals — never in series. A voltmeter must have a very high internal resistance (typically > 10 MΩ for a digital multimeter) so that it draws negligible current and does not disturb the circuit under test.
Voltage readings are taken with the circuit live. Key safety requirements (from the Electricity at Work Regulations 1989 and HSE Guidance Note GS38):
- Test leads must be rated for the voltage being measured (CAT rating — see below).
- If the voltage exceeds 50 V ac or 120 V dc, test probes must comply with GS38: finger-barrier probes, fused leads, exposed metal tip no longer than 4 mm.
- Test instruments must be appropriate for the Measurement Category (CAT) of the circuit: CAT III for distribution boards; CAT IV for supply intake and metering equipment.
Before working on a circuit: (1) switch off and lock off, (2) use a voltage indicator or multimeter to confirm 0 V between L–N, L–E, and N–E. The prove-safe routine is: prove tester on known source → test circuit → prove tester on known source again. This confirms the tester itself is working before and after the test.
Measuring Current
An ammeter is connected in series with the load so that all the load current flows through it. Because a series connection requires the circuit to be broken, the supply must be switched off when inserting or removing an ammeter.
A clamp-on ammeter (tong tester / clamp meter) avoids circuit interruption — it measures the magnetic field around a single conductor using a current transformer (CT) built into the jaw. Opening the jaw, placing it around a single insulated conductor, and closing it gives an immediate current reading. Clamp meters are ideal for measuring load currents in service without disturbing the circuit.
To measure the current drawn by a single-phase motor: open the clamp jaw, place it around only the line (live) conductor of the supply cable, close the jaw. With the motor running at full load, the clamp displays — say — 8.5 A. Compare this with the nameplate full-load current (FLC) to check for overloading.
Analogue Instruments
Analogue instruments produce a mechanical deflection proportional to the measured quantity, displayed by a pointer on a calibrated scale. Two main types are used in electrical work:
- Moving coil — a rectangular coil is pivoted between the poles of a permanent horseshoe magnet. When current flows through the coil, the interaction between the coil's magnetic field and the permanent magnet produces a torque, deflecting the pointer against a hair-spring. The scale is linear (uniform) — equally spaced divisions throughout. Moving coil instruments can only measure dc directly; a full-wave rectifier must be added to measure ac (and the scale must be recalibrated for rms values). They are sensitive and accurate but fragile.
- Moving iron — two soft-iron vanes are placed inside a fixed coil. When current flows, both vanes are magnetised with the same polarity and repel each other, deflecting a pointer. Because the force depends on the square of the current, the scale is non-linear (cramped at low values). Moving iron instruments measure both dc and true rms ac directly without a rectifier, making them useful for current and voltage measurement in both types of circuit.
Analogue instruments have largely been replaced by digital types for installation testing, but moving-iron panel ammeters and voltmeters are still found on older switchboards and control panels.
Digital Instruments
Digital instruments convert the measured quantity into a numerical value displayed on an LCD or LED display. They work by measuring a voltage (or a voltage developed across an internal shunt or burden resistor for current), converting it using an analogue-to-digital converter (ADC), and displaying the result.
Advantages of digital over analogue:
- Higher accuracy and resolution (typically ±0.5% or better vs ±2–3% for analogue)
- No parallax reading error (reading varies with viewing angle on analogue)
- Auto-ranging eliminates range selection errors
- Input impedance typically 10 MΩ — minimal loading effect on the circuit
- Data-hold, min/max capture, and data-logging functions available
Extending the Range of Ammeters and Voltmeters
Moving coil movements are delicate and designed for a specific full-scale deflection current (typically microamps to milliamps). External components extend their range:
Ammeter shunt: a precision low-resistance resistor connected in parallel with the meter movement. Most of the current bypasses the movement through the shunt; only a small known fraction passes through the meter. By choosing the shunt resistance, any current range can be accommodated. The meter reading is then scaled accordingly.
Voltmeter multiplier: a precision high-resistance resistor connected in series with the meter movement. The multiplier drops most of the applied voltage, leaving only a small proportion across the meter movement. Again, choosing the multiplier resistance sets the voltage range.
A moving coil meter has full-scale deflection at 1 mA with an internal resistance of 100 Ω. Find the multiplier needed to extend the range to 100 V:
\[V_{\text{meter}} = I_{\text{fsd}} \times R_m = 0.001 \times 100 = 0.1\text{ V}\] \[R_{\text{mult}} = \frac{V_{\text{range}} - V_{\text{meter}}}{I_{\text{fsd}}} = \frac{100 - 0.1}{0.001} = 99900\text{ Ω} \approx 100\text{ kΩ}\]Ammeter and Voltmeter Method for Impedance
In an ac circuit where a direct impedance meter is unavailable, impedance Z can be measured by connecting an ammeter in series and a voltmeter in parallel with the component, then applying Ohm's Law:
\[Z = \frac{V}{I} = \frac{\text{Voltmeter reading}}{\text{Ammeter reading}}\]A single-phase motor is supplied at 230 V ac. The clamp ammeter reads 4.6 A. Impedance of the motor winding:
\[Z = \frac{230}{4.6} = 50\text{ Ω}\]Compare with the dc resistance measured by an ohmmeter (say 30 Ω): the difference indicates inductive reactance \(X_L = \sqrt{Z^2 - R^2} = \sqrt{2500 - 900} = 40\text{ Ω}\).
Part 10 — Measurement of Power
Power in a dc circuit: measure V and I with a voltmeter and ammeter, then \(P = V \times I\).
True power in an ac circuit is measured with an electrodynamic (dynamometer) wattmeter. It has a current coil (in series, like an ammeter) and a voltage coil (in parallel, like a voltmeter).
\[\text{Power factor} = \frac{\text{true power (wattmeter reading)}}{\text{voltage} \times \text{current}}\]Measurement of Energy
An energy meter (kilowatt-hour meter) measures power consumed over time. The older type uses an aluminium disc that spins in proportion to energy consumed; modern meters use electronic pulses.
Measuring Frequency
Digital multi-meters and clamp multi-meters have functions for measuring frequency and are widely used.
Power Factor Measurement
Power factor can be measured by:
- A power factor meter (connected like a wattmeter)
- Using a wattmeter, voltmeter and ammeter together: \(\text{PF} = \dfrac{\text{W}}{V \times I}\)
🃏 Chapter 2 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Covers SI units, temperature, speed/velocity/acceleration, area, volume, density, resistance, electrical quantities and measuring instruments.
📝 Chapter 2 — Multiple Choice Questions
Select your answer and click Submit. You have two attempts per question. First-attempt score shown at the end.
1What are the seven SI base units? Which one directly measures electric current?
2Convert 85°C to Kelvin.
3Velocity is measured in which unit?
4The cross-sectional area of a 75 mm × 75 mm trunking is:
5A room measures 8 m × 5 m × 3 m. What is its volume?
6A copper busbar is 1 m long with a 40 mm × 5 mm cross-section. Density of copper = 8900 kg/m³. What is its mass?
7Conductor 1: diameter 4 mm, length 2 m. Conductor 2: same material, diameter 2 mm. What length of conductor 2 gives the same resistance?
8What is the difference between EMF and potential difference (pd)?
9The UK mains frequency is 50 Hz. What is the periodic time of one complete cycle?
10Resistance measurements with an ohmmeter must be taken with the supply disconnected, and the ohmmeter uses its own internal supply. Which statement is correct?
11In an AC circuit a wattmeter indicates the:
12A test instrument is rated CAT III. Which location is it suitable for?
Sends your Chapter 2 MCQ score summary from your registered email to your tutor.
Mechanical Science
- Specify what is meant by mass, force and weight
- Explain the basic mechanical principles of levers, pulleys and gears
- Describe the main principles and interrelationships of force, work, energy, power and efficiency
- Calculate values of electrical energy, power and efficiency
Mass
Mass is the quantity of matter contained in a body. It is measured in kilograms (kg) and is a scalar quantity — it has magnitude only and no direction. Crucially, mass does not change regardless of where an object is: a motor with a mass of 50 kg has a mass of 50 kg on earth, on the moon, or in space. Mass is found by weighing an object against calibrated standard masses on a balance.
In electrical installation work, knowing the mass of equipment is essential for structural calculations — cable trays, busbars, switchgear panels, and transformer cores must all be supported by structures rated for the load they carry. A cable drum weighing 800 kg will require lifting equipment rated to at least that mass; a transformer weighing 2 tonnes must be positioned on a floor slab that can bear its weight.
Force
Force is any interaction that tends to change the motion of an object. The SI unit is the newton (N), defined as the force required to give a mass of 1 kg an acceleration of 1 m/s²: \(1\text{ N} = 1\text{ kg·m/s}^2\).
Newton's Second Law of Motion states that the net force acting on a body equals its mass multiplied by its acceleration:
\[F = m \times a\]where F = force (N), m = mass (kg), a = acceleration (m/s²).
- If a force acts on a stationary object in the direction of intended motion, it accelerates the object.
- If it acts against the direction of motion, it decelerates (brakes) the object.
- Gravity is a special case of force: the earth's gravitational field gives every mass a downward acceleration of g = 9.81 m/s².
A cable drum of mass 400 kg has a net horizontal pulling force of 50 N applied to it (friction has already been overcome). What is its acceleration?
\[a = \frac{F}{m} = \frac{50}{400} = 0.125 \text{ m/s}^2\]This is a slow acceleration — confirming that large masses need sustained forces to move appreciably.
A length of steel trunking has a mass of 400 kg. What minimum vertical force is required to lift it?
\[F = m \times g = 400 \times 9.81 = 3924 \text{ N} \approx 3.9 \text{ kN}\]This is the minimum force to just hold the trunking stationary. Any additional force will accelerate it upward. Lifting equipment (chain blocks, tirfors, cranes) must be rated above this value.
A 150 kg electric motor mounted on a trolley is moving at 0.5 m/s and must be brought to rest in 2 seconds. Required braking force:
\[a = \frac{\Delta v}{t} = \frac{0.5}{2} = 0.25 \text{ m/s}^2\] \[F = m \times a = 150 \times 0.25 = 37.5 \text{ N}\]Weight
Weight is the gravitational force acting on a mass. Since gravity on earth accelerates all masses at g = 9.81 m/s², weight is simply:
\[W = m \times g = m \times 9.81 \text{ N}\]Inertia is the tendency of a body to resist changes to its state of rest or uniform motion. A heavy transformer has high inertia — it takes a large force to start it moving, and a large braking force to stop it. This is directly relevant to specifying lifting equipment and restraints for heavy electrical plant.
| Equipment | Typical mass (kg) | Approximate weight (N) | Implication |
|---|---|---|---|
| Steel conduit (3 m length, 25 mm) | ~2 | ~19.6 | Manual handling acceptable |
| Full cable drum (70 mm² Cu, 100 m) | ~120 | ~1177 | Fork-lift or drum stand required |
| 400 A distribution board | ~60 | ~589 | Two-person lift; wall fixings rated accordingly |
| 11 kV/415 V transformer (100 kVA) | ~500 | ~4905 | Crane or heavy lifting required |
Pressure
Pressure is the force applied per unit area of a surface. When a force F acts on an area A, the pressure P is:
\[P = \frac{F}{A}\]SI unit: pascal (Pa), where \(1\text{ Pa} = 1\text{ N/m}^2\). A smaller contact area concentrates the same force into a higher pressure — relevant when calculating foundation loading for heavy switchgear (the contact pressure on the floor must not exceed the slab's rated bearing capacity), or when specifying anti-vibration pads under motor bases.
A 500 kg transformer sits on a base plate of 0.6 m × 0.4 m. Weight = \(500 \times 9.81 = 4905\text{ N}\). Base area = \(0.6 \times 0.4 = 0.24\text{ m}^2\). Floor pressure:
\[P = \frac{F}{A} = \frac{4905}{0.24} = 20\,437.5 \text{ Pa} \approx 20.4 \text{ kPa}\]Typical domestic floor slabs can support around 1.5–2 kN/m², while industrial concrete floors are typically rated at 5–10 kN/m². This transformer exerts ~20 kN/m² — a specialist floor slab or spreader plate is needed.
Vectors and Phasor Diagrams
A scalar quantity has magnitude only (e.g. mass, speed, energy). A vector quantity has both magnitude and direction (e.g. force, velocity, displacement). Vectors are represented graphically as arrows: the length of the arrow is proportional to the magnitude, and the arrow points in the direction of action.
In electrical engineering, voltage and current in ac circuits also have magnitude and direction (phase angle), so they are treated as vectors — specifically called phasors. Phasor diagrams are vector diagrams applied to ac quantities, and the same rules of vector addition apply to both force diagrams and phasor diagrams.
- Same direction: resultant = arithmetic sum. E.g. two forces of 6 N and 8 N both pushing right: resultant = 14 N to the right.
- Opposite directions: resultant = arithmetic difference. E.g. 8 N right and 6 N left: resultant = 2 N to the right.
- Equal and opposite: resultant = zero — body is in equilibrium (no net force, no acceleration).
- At an angle: use the parallelogram of forces (draw both vectors from the same point; complete the parallelogram; the diagonal is the resultant). For 90°, Pythagoras applies directly.
A cable is pulled horizontally with 8 N and simultaneously lifted with 4 N (vertical). The resultant tension in the cable:
\[\text{Resultant} = \sqrt{8^2 + 4^2} = \sqrt{64+16} = \sqrt{80} = 8.94 \text{ N}\]The angle of pull above horizontal: \(\theta = \tan^{-1}(4/8) = 26.6°\)
In a series RL circuit, the voltage across the resistor is 90 V and across the inductor is 120 V. These voltages are 90° out of phase, so the supply voltage is found using the same vector rule:
\[V_S = \sqrt{90^2 + 120^2} = \sqrt{8100+14400} = \sqrt{22500} = 150 \text{ V}\]This is identical in method to finding the resultant of two forces at 90° — demonstrating the direct link between mechanical vectors and electrical phasors.
Levers
A lever is a rigid bar that pivots about a fixed point called the fulcrum. By varying the positions of the load, the effort, and the fulcrum relative to each other, a small effort can move a much larger load — this is mechanical advantage (MA). The principle is used extensively on site: crowbars and conduit benders are levers, and even the spindle of a switch can be analysed as a lever.
There are three orders (classes) of lever, distinguished by the relative positions of the fulcrum (F), load (L), and effort (E):
| Order | Arrangement | MA | Everyday example | Electrical/site example |
|---|---|---|---|---|
| First | F between L and E | Can be >1, =1, or <1 depending on distances | Crowbar, seesaw, scissors | Claw hammer removing a nail from timber; conduit bending bar |
| Second | L between F and E | Always >1 (effort < load) | Wheelbarrow, nutcracker | Cable drum trolley; hand truck for heavy switchgear |
| Third | E between F and L | Always <1 (effort > load) | Tweezers, fishing rod | Forearm holding a heavy cable — speed and range of movement traded for force |
Moments
The moment of a force (or torque) is the turning effect produced by a force acting at a perpendicular distance from a pivot. It is measured in newton metres (Nm):
\[M = F \times d\]where F = force (N) and d = perpendicular distance from the fulcrum (m).
The principle of moments states that when a body is in equilibrium (not rotating), the sum of all clockwise moments equals the sum of all anticlockwise moments about any pivot:
\[F_1 d_1 = F_2 d_2 \quad (\text{for a simple two-force lever})\]This principle is directly analogous to Kirchhoff's Voltage Law: just as voltages around a loop sum to zero, moments in equilibrium sum to zero.
A crowbar (first-order lever) has its fulcrum 200 mm from the load end. The operator pushes down at a point 1 metre from the fulcrum. Load = 25 kg.
\[F_1 d_1 = F_2 d_2 \;\Rightarrow\; E \times 1 = (25 \times 9.81) \times 0.2\] \[E = \frac{245.25 \times 0.2}{1} = 49.05 \text{ N}\]Mechanical advantage: \(\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{245.25}{49.05} = 5\). The lever multiplies the operator's force by a factor of 5.
An installer uses a 600 mm spanner to tighten a gland nut. They apply a force of 40 N at the end of the spanner. The torque applied to the nut:
\[M = F \times d = 40 \times 0.6 = 24 \text{ Nm}\]Torque values like this appear in manufacturer's specifications for cable glands, terminal screws, and busbar joints — over-tightening can strip threads; under-tightening can cause loose connections and overheating.
A wall-mounted distribution board (mass 40 kg) is fixed at two points: a top bracket 0.3 m from the board's centre of mass, and a bottom bracket 0.5 m below the centre of mass. Find the reaction forces at each fixing (treat as a simply supported beam):
Taking moments about the bottom bracket: \(R_{\text{top}} \times 0.8 = (40 \times 9.81) \times 0.5\)
\[R_{\text{top}} = \frac{392.4 \times 0.5}{0.8} = 245.25 \text{ N}\] \[R_{\text{bottom}} = 392.4 - 245.25 = 147.15 \text{ N}\]Both fixing points must be bolted to studs or use cavity fixings rated above these values.
Pulleys
A pulley is a wheel with a grooved rim mounted on a shaft, designed to carry a rope, belt, or chain. Pulleys are used to change the direction of a force and to gain mechanical advantage when lifting heavy loads. They are common on construction sites for raising cable drums, switchgear, and conduit runs to upper floors.
| Type | Description | MA | Effect |
|---|---|---|---|
| Fixed (Class 1) | Axle is fixed to a rigid support. Rope runs over the wheel. | 1 | Changes the direction of pull only — you pull down to lift a load up. No force multiplication. |
| Movable (Class 2) | Axle moves with the load. Both sides of the rope support the load. | 2 | Halves the effort needed — both rope sections carry half the load weight. |
| Compound | Combination of fixed and movable pulleys (block and tackle). | n (number of rope sections) | Effort = Load ÷ n. Each additional movable pulley doubles the MA. |
Load = \(160 \times 9.81 = 1569.6\text{ N}\). Effort required:
- Single fixed pulley (MA = 1): effort = 1569.6 N — no advantage, just redirects the force.
- One movable + one fixed (MA = 2): effort = 784.8 N — one person can manage.
- Four-pulley compound (MA = 4): effort = 392.4 N — easily within one person's capability.
- Eight-pulley compound (MA = 8): effort = 196.2 N — suitable for very heavy equipment like transformers.
In any pulley system, the rope at the effort end must move n times as far as the load rises. If you use a 4-pulley system to raise a load by 1 m, you must pull 4 m of rope. This ratio is the velocity ratio (VR):
\[\text{VR} = \frac{\text{distance moved by effort}}{\text{distance moved by load}} = n\]In a real (non-ideal) pulley system, friction reduces the mechanical advantage below the velocity ratio. Efficiency:
\[\eta = \frac{\text{MA}}{\text{VR}} \times 100\%\]If a 4-pulley system (VR = 4) actually delivers MA = 3.2 due to friction: \(\eta = (3.2/4) \times 100 = 80\%\)
Part 3 — Belt Drives and Gears
Belt Drives
Belt drives transmit rotational motion and torque between two shafts that are not directly connected. A belt runs over a pulley on each shaft; the friction between the belt and the pulleys transfers force. Belt drives are found on older motor-driven equipment such as lathes, fans, and pumps.
- Flat belt: simple but prone to slipping under high loads; largely replaced by vee belts.
- Vee belt: wedge-shaped cross-section sits in a matching vee groove on the pulley, dramatically increasing friction and grip. Much less slip than flat belts. Most common type on industrial motors.
- Toothed (synchronous) belt: has teeth that engage with matching grooves on the pulley — no slip at all; used where precise speed ratio is essential (e.g. camshafts, servo drives).
When two pulleys are connected by a belt and driven by a motor, they rotate in the same direction. The speed ratio depends on the pulley diameters:
\[\frac{N_1}{N_2} = \frac{D_2}{D_1}\]where N₁, N₂ = shaft speeds (rpm) and D₁, D₂ = pulley diameters.
A motor pulley of diameter 100 mm drives a fan pulley of diameter 300 mm at 1500 rpm. Fan speed:
\[N_2 = N_1 \times \frac{D_1}{D_2} = 1500 \times \frac{100}{300} = 500 \text{ rpm}\]The smaller motor pulley spins faster, driving the larger fan pulley at one-third the speed — but with three times the torque (ignoring losses).
Gears
A gear is a rotating machine part with precisely machined teeth that mesh with another toothed component to transmit torque and motion without slipping. Gear drives are used throughout electrical machines — motors, generators, gearboxes, and actuators all use gears internally.
| Gear type | Description | Application |
|---|---|---|
| Spur gear | Teeth parallel to shaft axis; simplest type | Gearboxes, clocks, most standard drives |
| Helical gear | Teeth at an angle to shaft axis; quieter, stronger than spur | Car gearboxes, high-speed drives |
| Worm gear | Screw-like worm meshes with a worm wheel; very high ratio possible | Winches, lifts, speed reducers on motors |
| Bevel gear | Conical shape; transmits rotation between shafts at an angle | Right-angle drives, differential gears |
| Rack and pinion | Converts rotational to linear motion | Steering systems, CNC machines |
Gears are used for four fundamental reasons:
- To increase or decrease shaft speed (speed reduction is most common — motors typically run at 1500 rpm but machines may need 50 rpm)
- To reverse the direction of rotation (two meshed external gears always rotate in opposite directions)
- To change the axis of rotation (bevel or worm gears for right-angle drives)
- To synchronise two shafts at a precise speed ratio
The gear ratio is determined by the number of teeth on each gear (not their diameter — this is why gears are more precise than belt drives):
\[\text{Gear ratio} = \frac{T_2}{T_1} = \frac{N_1}{N_2}\]where T₁, T₂ = number of teeth; N₁, N₂ = shaft speeds.
A motor gear (driver) has 20 teeth and rotates at 1440 rpm. The driven gear has 80 teeth. Output speed:
\[\frac{N_1}{N_2} = \frac{T_2}{T_1} \;\Rightarrow\; N_2 = N_1 \times \frac{T_1}{T_2} = 1440 \times \frac{20}{80} = 360 \text{ rpm}\]Gear ratio = 4:1 (speed reduction). The output torque is increased by the same ratio (less friction losses): \(T_{\text{out}} \approx 4 \times T_{\text{in}}\).
A worm gear reducer has a ratio of 40:1. The motor runs at 1500 rpm. Output shaft speed of the hoist drum:
\[N_2 = \frac{1500}{40} = 37.5 \text{ rpm}\]A worm gear is inherently self-locking — the load cannot back-drive the worm, providing a natural mechanical brake. This is why worm gears are used in lifts and hoists as a safety feature.
| Feature | Belt drive (vee) | Gear drive (spur) |
|---|---|---|
| Slip | Up to ~3% under overload | Zero — exact ratio guaranteed |
| Speed ratio accuracy | Depends on pulley diameter (±mm) | Exact (tooth count) |
| Maintenance | Belt replacement every few years | Lubrication; longer service life |
| Overload protection | Belt slips, protecting motor | No slip — overload damages gears or motor |
| Noise | Quieter | Can be noisy at high speed |
Physical Quantities Summary (Table 3.1)
Before working through the calculations in this section, it is useful to see all the mechanical quantities together. Note how each is defined in terms of the others — they form an interconnected chain: Force acts through Distance to do Work; Work done in Time is Power; Power sustained over Time is Energy.
| Variable | Symbol | Unit | Formula | Electrical equivalent |
|---|---|---|---|---|
| Force | F | newton (N) | \(F = ma\) | EMF / voltage (drives current just as force drives motion) |
| Mass | m | kilogram (kg) | — | Inductance (resists change in current, like inertia resists change in motion) |
| Weight | W | newton (N) | \(W = mg\) | Static load on a circuit |
| Work | W | joule (J) | \(W = Fd\) | Electrical energy (\(W = Pt\)) |
| Power | P | watt (W) | \(P = W/t\) | Electrical power (\(P = VI\)) |
| Energy | E | joule (J) or kWh | \(E = Pt\) | Electrical energy (billed in kWh) |
| Efficiency | η | % (dimensionless) | \(\eta = P_{\text{out}}/P_{\text{in}}\times100\) | Same formula for motors, transformers, generators |
Work
Work is done when a force moves its point of application in the direction of the force. If you push on a wall and it does not move, no mechanical work is done — force without displacement = zero work. The SI unit is the joule (J):
\[1 \text{ J} = 1 \text{ N} \times 1 \text{ m} = 1 \text{ Nm}\] \[W = F \times d\]where F = force (N) and d = distance moved in the direction of the force (m).
Work and electrical energy share the same unit (joule) because work is a form of energy transfer. Whenever an electric motor lifts a load, it converts electrical energy into mechanical work.
A cable drum of mass 10 kg is lifted 6 m vertically from floor level to a cable tray. Work done against gravity:
\[W = F \times d = (10 \times 9.81) \times 6 = 98.1 \times 6 = 588.6 \text{ J}\]A cable is pulled horizontally through a 30 m conduit. The pulling tension (friction force) is 80 N. Work done:
\[W = F \times d = 80 \times 30 = 2400 \text{ J}\]This work is entirely converted to heat (friction). In practice, a pulling lubricant reduces the friction force and therefore the work needed.
A passenger lift (total mass 800 kg including occupants) rises 24 m (8 floors × 3 m each). Work done:
\[W = F \times d = (800 \times 9.81) \times 24 = 7848 \times 24 = 188\,352 \text{ J} \approx 188.4 \text{ kJ}\]Energy
Energy is the capacity to do work. It cannot be created or destroyed — only converted from one form to another (the law of conservation of energy). Energy comes in many forms; in electrical installation work you deal with all of these:
| Form of energy | Example in electrical work | Conversion |
|---|---|---|
| Electrical | Current flowing in a circuit | → Heat, light, mechanical |
| Chemical | Battery / accumulator | → Electrical (discharge); Electrical → Chemical (charging) |
| Mechanical (KE) | Rotating motor shaft | → Electrical (generator) |
| Mechanical (PE) | Raised counterweight in a lift | → KE when released (regenerative braking) |
| Heat (thermal) | Resistive heating element | Electrical → Heat |
| Light | LED, fluorescent lamp | Electrical → Light (+ some heat) |
The SI unit of energy is the joule (J). For practical billing and metering, the kilowatt-hour is used: \(1\text{ kWh} = 3\,600\,000\text{ J} = 3.6\text{ MJ}\).
A steel support bracket of mass 6 kg is lifted 12 m:
\[W = m \times g \times d = 6 \times 9.81 \times 12 = 706.32 \text{ J}\]A 250 V dc generator delivers 10 A for 2 minutes (120 s):
\[P = V \times I = 250 \times 10 = 2500 \text{ W}\] \[\text{Energy} = P \times t = 2500 \times 120 = 300\,000 \text{ J} = 300 \text{ kJ}\]In kWh: \(300\,000 / 3\,600\,000 = 0.0833\text{ kWh}\) — a small amount, as expected for only 2 minutes of operation.
Kinetic Energy (KE)
Kinetic energy is the energy possessed by a body due to its motion. Since KE depends on the square of velocity, doubling speed quadruples the kinetic energy — this is why high-speed impacts are so much more destructive than low-speed ones, and why regenerative braking in lifts and electric vehicles recovers significant energy.
\[\text{KE} = \frac{1}{2}mv^2\]A flywheel of mass 500 kg has a peripheral velocity of 20 m/s:
\[\text{KE} = 0.5 \times 500 \times 20^2 = 0.5 \times 500 \times 400 = 100\,000 \text{ J (100 kJ)}\]This stored kinetic energy can be used to ride through brief supply interruptions (uninterruptible power supply using a flywheel).
An electric delivery van of mass 800 kg travels at 25 m/s. It must stop in 40 m. Find the braking force:
\[\text{KE} = 0.5 \times 800 \times 25^2 = 250\,000 \text{ J}\] \[F = \frac{\text{KE}}{d} = \frac{250\,000}{40} = 6250 \text{ N} = 6.25 \text{ kN}\]In a regenerative braking system, most of this energy (minus losses) is fed back into the battery rather than wasted as heat.
A cable drum rolling at 2 m/s (mass 120 kg): \(\text{KE} = 0.5 \times 120 \times 4 = 240\text{ J}\).
At 4 m/s (double): \(\text{KE} = 0.5 \times 120 \times 16 = 960\text{ J}\) — four times the energy. This illustrates why slow controlled movement of heavy equipment is a core manual handling principle.
Potential Energy (PE)
Potential energy is stored energy arising from an object's position or condition. Two types are relevant in electrical engineering:
- Gravitational PE (GPE): energy stored by raising a mass against gravity. Released as the mass falls (or as a lift descends). Regenerative lift systems use this energy to help pull the counterweight up when the lift goes down.
- Elastic PE (EPE): energy stored in a deformed spring or elastic material. Relevant in circuit breaker mechanisms (the spring stores energy to snap contacts open rapidly on a fault) and in contactor return springs.
where m = mass (kg), g = 9.81 m/s², h = height (m).
A switchgear panel of mass 50 kg is raised 5 m to a mezzanine floor. GPE stored:
\[\text{GPE} = 50 \times 9.81 \times 5 = 2452.5 \text{ J}\]If the panel were to fall, this energy would be converted to KE — giving it a velocity on impact of \(v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 5} = 9.9\text{ m/s}\). This highlights why securing heavy equipment during installation is a critical safety requirement.
An empty lift car (mass 400 kg) descends 15 m. Ignoring friction, the GPE released:
\[\text{GPE} = 400 \times 9.81 \times 15 = 58\,860 \text{ J}\]A regenerative drive system converts this to electrical energy fed back to the building supply — reducing net energy consumption. At 80% efficiency, the energy recovered is \(58\,860 \times 0.8 = 47\,088\text{ J}\).
Power
Power is the rate of doing work — how quickly energy is transferred or converted. A more powerful machine does the same work in less time. The SI unit is the watt (W) where \(1\text{ W} = 1\text{ J/s}\):
\[P = \frac{W}{t} \qquad\text{(mechanical)}\] \[P = V \times I \qquad\text{(electrical)}\]These are linked: a motor's mechanical output power equals its electrical input power multiplied by its efficiency.
A hoist raises 200 kg at a velocity of 4 m/s. Power required (ignoring friction):
\[F = 200 \times 9.81 = 1962 \text{ N}\] \[P = F \times v = 1962 \times 4 = 7848 \text{ W} \approx 7.85 \text{ kW}\]Note: \(P = F \times v\) is a convenient shortcut when velocity is given instead of time.
A hoist lifts 20 kg through 15 m in 40 s:
\[W = F \times d = (20 \times 9.81) \times 15 = 2943 \text{ J}\] \[P = \frac{W}{t} = \frac{2943}{40} = 73.6 \text{ W}\]This matches the self-assessment question answer of 73.58 W.
A pump raises 600 litres of water per minute through 10 m. Pump efficiency = 50%.
Mass of water per second: \(600/60 = 10 \text{ kg/s}\) (1 litre ≈ 1 kg).
\[F = 10 \times 9.81 = 98.1 \text{ N/s} \;\Rightarrow\; \text{Useful power} = 98.1 \times 10 = 981 \text{ W}\] \[\text{Motor input power} = \frac{981}{0.50} = 1962 \text{ W} \approx 2 \text{ kW}\]The motor must be rated at least 2 kW. In practice it would be specified at the next standard frame size above this value.
Efficiency (η)
No real machine is 100% efficient — there are always losses due to friction, heat, vibration, and (in electrical machines) copper losses (\(I^2R\)) and iron losses. Efficiency η (the Greek letter eta) compares what you get out with what you put in:
\[\eta = \frac{\text{useful output}}{\text{total input}} \times 100\%\]In terms of power: \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%\). In terms of energy: \(\eta = \frac{E_{\text{out}}}{E_{\text{in}}} \times 100\%\).
The losses are: \(\text{Losses} = P_{\text{in}} - P_{\text{out}}\) and efficiency can also be written as \(\eta = 1 - \frac{\text{losses}}{P_{\text{in}}}\).
A hoist lifts 250 kg through 200 m. The system efficiency is 30% (very poor — significant friction losses in old equipment). How many kWh of electrical energy are consumed?
\[\text{Useful work} = 250 \times 9.81 \times 200 = 490\,500 \text{ J}\] \[\text{Total energy input} = \frac{490\,500}{0.30} = 1\,635\,000 \text{ J}\] \[\text{Energy (kWh)} = \frac{1\,635\,000}{3\,600\,000} = 0.454 \text{ kWh}\]At 80% efficiency (modern equipment): input = \(490\,500 / 0.80 = 613\,125\text{ J} = 0.170\text{ kWh}\) — less than 40% of the energy consumed by the 30%-efficient system.
A single-phase motor draws 15 A at 230 V and delivers 2880 W of mechanical output:
\[P_{\text{in}} = V \times I = 230 \times 15 = 3450 \text{ W}\] \[\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100 = \frac{2880}{3450} \times 100 = 83.5\%\] \[\text{Losses} = 3450 - 2880 = 570 \text{ W (heat and friction)}\]A 100 kVA transformer has iron losses of 500 W and copper losses at full load of 1200 W. Full-load efficiency (assuming unity power factor):
\[P_{\text{in}} = 100\,000 + 500 + 1200 = 101\,700 \text{ W}\] \[\eta = \frac{100\,000}{101\,700} \times 100 = 98.3\%\]Transformers are typically the most efficient electrical machines — modern distribution transformers achieve 98–99.5% efficiency. Compare this with a typical induction motor (85–95%) or incandescent lamp (~5% — 95% of input is wasted as heat).
An electric motor (η = 90%) drives a pump (η = 75%) via a belt drive (η = 95%). Overall system efficiency:
\[\eta_{\text{overall}} = 0.90 \times 0.95 \times 0.75 = 0.641 = 64.1\%\]In a chain of machines, multiply the individual efficiencies. Each inefficiency compounds the others — this is why direct-coupled motor-pump sets (eliminating the belt drive) improve overall system efficiency.
🃏 Chapter 3 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Covers mass, force, weight, pressure, vectors, levers, moments, pulleys, gears, work, energy, power and efficiency.
📝 Chapter 3 — Multiple Choice Questions
Select your answer and click Submit. Two attempts per question. First-attempt score shown at the end.
1A bundle of conduit has a mass of 200 kg. What force is required to lift it?
2A transformer of mass 800 kg has a base area of 0.8 m². What pressure does it exert on the floor?
3Which of the following is a vector quantity?
4A lever has a load of 300 N at 0.2 m from the pivot. What effort at 0.6 m from the pivot maintains equilibrium?
5The effort required to lift a 40 kg load using a four-pulley system is:
6A compound pulley system has:
7A driving gear has 30 teeth and meshes with a driven gear of 90 teeth. If the driver rotates at 1500 rpm, what is the driven gear speed?
8How much kinetic energy is possessed by a 2 kg mass moving at 4.5 m/s?
9A cable drum of mass 50 kg is lifted 3 m. How much gravitational potential energy is stored?
10A hoist lifts a mass of 20 kg through 15 m in 40 s. What is the output power?
11A motor delivers 5 kW of mechanical output from a 7 kW electrical input. What is the efficiency?
12A motor (efficiency 85%) drives a gearbox (efficiency 92%). What is the overall system efficiency?
Sends your Chapter 3 MCQ score summary from your registered email to your tutor.
Electrical Science
- Describe the principles of electron flow and current flow
- Distinguish between conductors and insulators; state types and properties of electrical cables
- Calculate electrical variables related to Ohm's Law
- Explain and calculate current, voltage and resistance in series and parallel dc circuits
- Calculate values of power in series and parallel dc circuits
- Describe voltage drop, chemical and thermal effects of electrical current
Atomic Structure
All matter is made up of atoms — the smallest unit of a chemical element that retains the element's properties. Every atom has two main parts:
- The central nucleus — contains positively charged protons and neutral neutrons. The nucleus is very dense and carries virtually all the atom's mass.
- Orbiting electrons — negatively charged particles that circle the nucleus in defined energy levels called shells (or orbits). Each shell can hold a maximum number of electrons (inner shell: 2; second shell: 8; third shell: 18; etc.).
In a neutral atom, the number of protons equals the number of electrons, so the atom has no overall charge. The atomic number is the number of protons (and therefore electrons in the neutral state). For example, copper has atomic number 29.
The Copper Atom — why copper conducts so well
The copper atom has 29 electrons arranged in four shells: 2 + 8 + 18 + 1. The outermost (valence) shell contains just one electron, held very weakly because it is far from the nucleus and shielded by the inner shells. This single outer electron can break free from its parent atom with very little energy — it becomes a free electron.
In a copper conductor at room temperature, vast numbers of free electrons drift randomly between atoms. This is what makes copper such an excellent conductor: a large supply of free charge carriers is always available. Silver has a similar structure and slightly higher conductivity, but copper is used instead because it is far cheaper and almost as good.
| Element | Atomic number | Outer shell electrons | Conductivity (relative) | Use |
|---|---|---|---|---|
| Silver (Ag) | 47 | 1 | Best conductor | High-end contacts, plating |
| Copper (Cu) | 29 | 1 | Excellent | Cable conductors, busbars, windings |
| Aluminium (Al) | 13 | 3 | Good | Overhead lines, large cables |
| Iron (Fe) | 26 | 2 | Poor (relative) | Cores, structural, conduit |
| Carbon (C) | 6 | 4 | Poor | Brushes in motors and generators |
Electron and Conventional Current Flow
When a potential difference (voltage) is applied across a conductor — for example, by connecting a battery — the electric field pushes free electrons towards the positive terminal. This movement of free electrons is electron flow, which travels from the negative terminal to the positive terminal of the supply.
Historically, before the electron was discovered, scientists assumed current flowed from positive to negative (from high potential to low). This is conventional current flow and is still used in circuit diagrams, Ohm's Law, and most electrical calculations. The two directions are simply the opposite of each other:
- Electron flow: negative (−) → positive (+)
- Conventional current flow: positive (+) → negative (−)
Both conventions give the same results in calculations — the choice is purely historical convention. All circuit diagrams and formulae in BS 7671 and standard electrical theory use conventional current flow.
A 1.5 V AA cell is connected to a 5 Ω bulb. Conventional current flows from the + terminal of the cell, through the bulb (converting electrical energy to light and heat), and back to the − terminal. The current is:
\[I = \frac{V}{R} = \frac{1.5}{5} = 0.3 \text{ A}\]Meanwhile, free electrons flow in the opposite direction — from − to + inside the cell and from the − terminal, through the bulb, to the + terminal externally. The rate of electron flow is still 0.3 A (since current is defined as charge per second regardless of carrier sign).
One ampere is defined as one coulomb of charge passing a point per second. Since one electron carries a charge of \(1.6 \times 10^{-19}\) C, the number of electrons passing per second in a 1 A current is:
\[n = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18} \text{ electrons per second}\]Even a modest 1 A circuit involves over six billion billion electron movements per second — illustrating why current flows appear instantaneous even though individual electrons drift quite slowly (~mm/s).
Conductors
A good electrical conductor is a material that allows electric current to flow through it with minimal opposition. Conductors have low resistivity because their atoms have loosely bound outer electrons (free electrons) that can drift easily under an applied voltage. The lower the resistivity, the less energy is lost as heat in the conductor, and the thinner (and cheaper) the conductor needed for a given current.
Common conductors used in electrical installations, in order of conductivity:
| Material | Resistivity at 20°C (Ω·m) | Key property | Electrical use |
|---|---|---|---|
| Silver | \(15.9 \times 10^{-9}\) | Best conductor, expensive | Plated contacts, specialist connectors |
| Copper | \(17.2 \times 10^{-9}\) | Excellent conductor, ductile, solderable | All fixed wiring, busbars, motor windings |
| Aluminium | \(28.4 \times 10^{-9}\) | Good conductor, very light, low cost | Overhead lines, large distribution cables |
| Brass | \(~70 \times 10^{-9}\) | Moderate conductor, hard-wearing | Terminals, plug pins, lamp holders |
| Carbon | \(~35 \times 10^{-6}\) | Poor conductor but self-lubricating | Motor and generator brushes |
Insulators
Insulators are materials whose electrons are tightly bound to their parent atoms — no free electrons, so no current can flow under normal voltages. Insulators are used to separate live conductors from each other and from earth, preventing shock, short circuits, and fires.
Key insulating materials used in electrical work:
| Material | Max. service temp. | Key property | Application |
|---|---|---|---|
| PVC (polyvinyl chloride) | 70°C conductor | Flexible, cheap, self-extinguishing | General cable insulation and sheath |
| XLPE (cross-linked polyethylene) | 90°C conductor | Better heat resistance than PVC, tougher | Armoured cables, underground MV cables |
| Silicon rubber | 150–180°C | Remains flexible at very high and low temps | FP200, oven wiring, aerospace |
| Magnesium oxide (MgO) | 250°C+ continuous; 1000°C short-term | Inorganic — will not burn | MIMS cable insulation |
| Mica | 500°C+ | Excellent high-temp insulator, brittle | Heater elements, switchgear arc barriers |
| Glass / ceramic | Very high | Rigid, chemical-resistant | Overhead line insulators, fuse carriers |
| Air | — | Free, self-healing dielectric | Switchgear arc gaps, overhead line clearances |
A 2.5 mm² PVC-insulated cable has a maximum conductor temperature of 70°C. If the cable carries too much current, its temperature rises above 70°C — the PVC softens, can crack, and eventually carbonises. Once carbonised, PVC becomes a conductor, which can cause a short circuit or fire. This is why BS 7671 specifies current ratings that keep the conductor temperature within the insulation's limit, and why derating factors are applied when cables are grouped or enclosed in insulation.
Conductor Materials — Copper vs Aluminium
The two main conductor materials for fixed wiring in the UK are copper and aluminium. Understanding their differences helps explain why each is used in particular applications:
| Property | Copper | Aluminium |
|---|---|---|
| Resistivity | \(17.2 \times 10^{-9}\) Ω·m | \(28.4 \times 10^{-9}\) Ω·m (65% higher) |
| Density | 8900 kg/m³ | 2700 kg/m³ (70% lighter) |
| Tensile strength | High | Lower — needs larger csa for same current |
| Termination | Easy — solders, crimps, clamps directly | Needs special lugs; oxidises rapidly (oxide is an insulator) |
| Cost | Higher | Lower per kg |
| Typical use | All domestic/commercial fixed wiring | Overhead lines, large LV/MV distribution cables ≥ 16 mm² |
Annealed copper is copper that has been heat-treated (annealed) to make it soft and ductile. This process removes the work-hardening from the wire-drawing manufacturing process, making the conductor flexible enough to be bent repeatedly without cracking — essential for flexible cables and multicore conductors.
Cable Insulation Types
Cable insulation polymers are classified by how they behave when heated:
- Thermoplastic (e.g. PVC) — softens reversibly when heated and re-hardens on cooling. Can be recycled. Relatively low maximum temperature. Most common cable insulation for general wiring.
- Thermosetting (e.g. XLPE, EPR, silicon rubber) — when first manufactured, a chemical cross-linking process permanently bonds the polymer chains. The result does not soften on reheating. Better thermal performance, better chemical resistance, and higher current ratings than thermoplastic at the same conductor size.
Types of Electrical Cables
A cable consists of one or more conductors, each insulated individually, with overall mechanical protection. The selection of cable type depends on: the installation method, the environment (damp, corrosive, flammable), the required fire performance, the operating temperature, and the voltage level.
| Cable Type | Reference | Application | Key Properties and Rationale |
|---|---|---|---|
| PVC/PVC twin & multicore flat | 6242Y / 6243Y | Domestic & light commercial fixed wiring | Max conductor temp 70°C. The flat profile sits neatly under plaster. The grey or white outer sheath is thermoplastic PVC. The bare CPC (green/yellow in modern cables) relies on the outer sheath for insulation — it must not be installed where it could be damaged mechanically. Not suitable for damp locations without additional protection. |
| PVC-insulated singles | 6491X | Commercial & industrial wiring in conduit or trunking | Individual cores colour-coded to BS 7671. No mechanical protection of their own — must always be enclosed in conduit, trunking, or ducting. Allows flexible circuit design (any number of cores in any combination). Very cost-effective where containment already exists. |
| SWA armoured (XLPE or PVC insulated) | BS 5467 / BS 6346 | Industrial, underground, between buildings | Steel wire armour (SWA) provides mechanical protection against crushing and rodent damage. The armour itself can serve as the circuit protective conductor (CPC) but must be correctly terminated with armoured cable glands. XLPE-insulated versions have a 90°C conductor rating — higher current capacity than PVC equivalents of the same conductor size. The outer PVC sheath protects against moisture and corrosion. |
| FP200 (fire-resistant, silicon rubber) | FP200 / FP400 | Fire alarms, emergency lighting, escape routes | Silicon rubber insulation withstands 150°C in normal operation. In a fire, the silicon rubber decomposes to a white silica ash that remains as a solid insulator — maintaining circuit integrity. An aluminium tape screen provides additional physical protection and electromagnetic screening. BS 5839 and BS 5266 specify fire-resistant cable for fire alarm and emergency lighting circuits respectively. |
| MIMS (mineral-insulated, metal-sheathed) | BS EN 60702 / BS 6207 | Fire alarm circuits, petro-chemical, food industry, hazardous areas | Copper conductors compressed inside a copper sheath with magnesium oxide (MgO) powder insulation. Virtually indestructible and can operate at extremely high temperatures (250°C continuous, 1000°C short-term). The copper sheath is the CPC. Drawback: MgO is hygroscopic — absorbs moisture — so terminations must always use a brass pot and seal to exclude moisture, otherwise insulation resistance collapses. Requires special termination skills. |
| LSF / LSHF cables | Various | Public buildings, transport, schools, hospitals | LSF (Low Smoke and Fume): limits the volume of smoke produced in a fire — important for evacuation visibility. LSHF / LSOH (Low Smoke, Zero Halogen): no chlorine, fluorine, or bromine in the polymer — when burned, produces no hydrogen chloride (HCl) gas, which is toxic and corrosive. Standard PVC releases HCl when burning, which is dangerous in confined spaces. Specified on all contracts where the BS 7671 Special Installations for public buildings or fire safety applies. |
| Fibre optic cable | IEC 60793 / BS EN 60793 | Data networks, telecoms, CCTV, building management | A glass core (typically 9 µm single-mode or 50/62.5 µm multi-mode) surrounded by a glass cladding (125 µm outer diameter) transmits light signals using total internal reflection. No electrical current flows — completely immune to electromagnetic interference (EMI), no shock risk, no spark hazard. Ideal for data connections between buildings (no earth potential difference issue). Signal loss is far lower than copper for long runs. Requires specialist splicing and termination tools. |
Cable Voltage Rating
Cables are rated by their ability to withstand voltage. The rating is expressed as U₀/U where:
- U₀ = nominal voltage between any conductor and earth (or the metallic sheath)
- U = nominal voltage between any two phase conductors
For example, a 600/1000 V cable (the most common for LV fixed wiring) is suitable for use on the UK 230/400 V system with substantial margin. This margin accommodates voltage surges and confirms the cable's insulation is tested to these higher values during manufacture.
A separate outbuilding (garage with workshop) needs a 32 A single-phase supply run 25 m underground from the house. The appropriate cable selection would be:
- Type: 6 mm² two-core + CPC SWA armoured cable (e.g. BS 5467 XLPE/SWA/PVC)
- Rationale: Armoured for mechanical protection underground; XLPE insulation for 90°C rating giving higher current capacity; SWA provides CPC path; 600/1000 V rating appropriate for 230 V supply; buried at least 600 mm deep (BS 7671 Appendix 4) or in suitable duct.
Resistance and Resistivity
Every conductor has resistance. Even a short length of copper cable has some resistance — small, but never zero. This resistance causes voltage drop (energy lost as heat in the cable rather than delivered to the load) and contributes to the earth fault loop impedance that determines whether a protective device will operate fast enough to clear a fault. The formula linking resistance to the physical properties of a conductor is:
\[R = \frac{\rho l}{A}\]where ρ = resistivity (Ω·m), l = length (m), A = cross-sectional area (m²). This formula is transposed in practice to find the required conductor size (A) for a given resistance limit, or to check the resistance of an existing conductor.
A 20 Ω lamp is fed by two 5 m conductors of 1 mm² copper (ρ = 17.2 × 10⁻⁹ Ω·m).
Resistance of one conductor: \(R = \frac{17.2 \times 10^{-9} \times 5}{1 \times 10^{-6}} = 0.086\text{ Ω}\)
Both conductors (line + neutral): \(2 \times 0.086 = 0.172\text{ Ω}\)
\[R_\text{circuit} = 0.172 + 20 = 20.172 \; \Omega\]Voltage drop across conductors at rated current \((I = 230/20.172 = 11.4\text{ A})\): \(V_d = 11.4 \times 0.172 = 1.96\text{ V}\) — only 0.85%, well within the 3% limit.
A 40 m circuit (80 m total cable length) must have a total resistance ≤ 0.5 Ω for fault protection purposes. What minimum copper conductor csa is required?
\[A = \frac{\rho l}{R} = \frac{17.2 \times 10^{-9} \times 80}{0.5} = \frac{1376 \times 10^{-9}}{0.5} = 2.75 \times 10^{-6} \text{ m}^2 = 2.75 \text{ mm}^2\]Select the next standard size up: 4 mm² copper conductor.
Ohm's Law
Georg Simon Ohm (1789–1854) discovered that for a conductor at constant temperature, the current through it is directly proportional to the voltage across it and inversely proportional to its resistance. This relationship, now called Ohm's Law, is the single most important formula in electrical work:
\[V = I \times R \qquad I = \frac{V}{R} \qquad R = \frac{V}{I}\]The three forms are used in different situations: V = IR when you know current and resistance; I = V/R when you know voltage and resistance (most common — finding circuit current); R = V/I when measuring voltage and current to find an unknown resistance.
A 200 V dc supply feeds a 10 Ω resistive load. Current:
\[I = \frac{V}{R} = \frac{200}{10} = 20 \text{ A}\]A voltmeter reads 229.5 V across a load; an ammeter reads 9.5 A. The load resistance:
\[R = \frac{V}{I} = \frac{229.5}{9.5} = 24.2 \; \Omega\]A 230 V circuit has a total earth fault loop impedance of 1.15 Ω. Under a line-to-earth fault:
\[I_{\text{fault}} = \frac{V}{Z} = \frac{230}{1.15} = 200 \text{ A}\]A 32 A BS 88 fuse must disconnect within 5 s at 200 A — checking the time-current characteristic confirms this is well within its operating range. If Z were higher (say 3 Ω), fault current would be only 77 A and the fuse might not operate in time.
Series Circuits
In a series circuit, components are connected end-to-end so there is only one path for current. The same current flows through every component. Series connections are used for: current-limiting resistors, voltage dividers, pilot lamps, emergency stop buttons (where any open contact breaks the circuit), and fuses (a fuse is always in series with the load it protects).
Rules for series circuits:
- Current is the same through all components: \(I_T = I_1 = I_2 = I_3\)
- Total resistance is the sum: \(R_T = R_1 + R_2 + R_3 + \ldots\)
- Voltage divides in proportion to resistance: \(V_n = I \times R_n\)
- Sum of voltage drops = supply voltage: \(V_S = V_1 + V_2 + V_3 + \ldots\) (Kirchhoff's Voltage Law)
12 V supply, R₁ = 1 Ω, R₂ = 3 Ω in series:
\[R_T = 1 + 3 = 4 \; \Omega \qquad I_T = \frac{12}{4} = 3 \text{ A}\] \[V_1 = 3 \times 1 = 3 \text{ V} \qquad V_2 = 3 \times 3 = 9 \text{ V} \qquad \text{Check: } 3+9=12\text{ V} ✓\]A conveyor has three emergency stop (E-stop) push-buttons wired in series with a 24 V relay coil (resistance 400 Ω). All buttons must be closed (contacts made) for the relay to energise. Current through the relay when all buttons closed:
\[I = \frac{24}{400} = 0.06 \text{ A} = 60 \text{ mA}\]If any one E-stop button is pressed, the circuit is broken, current falls to zero, the relay de-energises, and the conveyor stops — the series wiring provides a fail-safe control circuit.
Parallel Circuits
In a parallel circuit, components are connected between the same two nodes — each has the same voltage across it, but carries its own independent current. Parallel connections are used for virtually all building wiring: each socket outlet, light, and appliance connects in parallel so it operates at the full supply voltage independently of others.
Rules for parallel circuits:
- Voltage is the same across all branches: \(V_T = V_1 = V_2 = V_3\)
- Total current is the sum of branch currents: \(I_T = I_1 + I_2 + I_3 + \ldots\) (Kirchhoff's Current Law)
- Total resistance is always less than the smallest branch resistance
- Two resistors: \(R_T = \dfrac{R_1 \times R_2}{R_1 + R_2}\) (product over sum — only for two resistors)
- Three or more: \(\dfrac{1}{R_T} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \ldots\)
Two 8 Ω resistors in parallel (product over sum):
\[R_T = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4 \; \Omega\]Note: two equal resistors in parallel always give half the individual resistance. Three equal resistors give one third, and so on.
24 V supply, R₁ = 6 Ω, R₂ = 3 Ω, R₃ = 8 Ω in parallel:
\[\frac{1}{R_T} = \frac{1}{6}+\frac{1}{3}+\frac{1}{8} = \frac{4+8+3}{24} = \frac{15}{24} \;\Rightarrow\; R_T = 1.6 \; \Omega\] \[I_T = \frac{24}{1.6} = 15 \text{ A}\] \[I_1 = \frac{24}{6} = 4\text{ A}, \quad I_2 = \frac{24}{3} = 8\text{ A}, \quad I_3 = \frac{24}{8} = 3\text{ A}\]Check (KCL): \(4 + 8 + 3 = 15\text{ A} = I_T\) ✓. The lowest resistance branch (R₂ = 3 Ω) draws the most current.
A 230 V ring final circuit feeds four socket outlets with loads: a 2 kW kettle (R = 26.45 Ω), a 1 kW microwave (R = 52.9 Ω), a 0.5 kW phone charger (R = 105.8 Ω), and a 0.1 kW lamp (R = 529 Ω) — all in parallel at 230 V. Total current:
\[I_T = \frac{2000}{230} + \frac{1000}{230} + \frac{500}{230} + \frac{100}{230} = 8.7 + 4.35 + 2.17 + 0.43 = 15.65 \text{ A}\]All four appliances operate at 230 V regardless of each other. Switching off the kettle doesn't affect the microwave — as it would in a series circuit. This independence of parallel branches is why all final circuits are wired in parallel.
Combined Series-Parallel Circuits
Real circuits often combine series and parallel sections — for example, a supply cable (series resistance) feeding several parallel loads, or a distribution board with a series protective device feeding parallel final circuits. The method is always: simplify the parallel section first, then treat the result as a series circuit.
Step-by-step method:
- Identify and calculate the resistance of the parallel section (RP).
- Add RP to any series resistances to get RT.
- Calculate total current: \(I_T = V_S / R_T\).
- Calculate voltage drop across series component(s): \(V_S = I_T \times R_\text{series}\).
- Voltage across parallel section: \(V_P = V_S - V_\text{series drops}\).
- Calculate individual branch currents: \(I_n = V_P / R_n\).
R₁ = 2.2 Ω in series with three parallel resistors R₂ = 7 Ω, R₃ = 14 Ω, R₄ = 21 Ω. Total current IT = 2 A.
\[\frac{1}{R_P} = \frac{1}{7}+\frac{1}{14}+\frac{1}{21} = \frac{6+3+2}{42} = \frac{11}{42} \;\Rightarrow\; R_P = 3.82 \; \Omega\] \[R_T = 2.2 + 3.82 = 6.02 \; \Omega \qquad V_S = 2 \times 6.02 = 12 \text{ V}\] \[V_{R1} = 2 \times 2.2 = 4.4 \text{ V} \qquad V_P = 12 - 4.4 = 7.6 \text{ V}\] \[I_2 = \frac{7.6}{7} = 1.09\text{ A},\quad I_3 = \frac{7.6}{14} = 0.54\text{ A},\quad I_4 = \frac{7.6}{21} = 0.36\text{ A}\]Check (KCL): \(1.09 + 0.54 + 0.36 = 1.99 \approx 2\text{ A} = I_T\) ✓ (rounding)
A 230 V supply feeds a sub-distribution board via 25 m of 6 mm² copper cable (R_cable ≈ 0.072 Ω each conductor, total = 0.144 Ω). The board has three parallel loads: 10 Ω, 15 Ω, 30 Ω.
\[\frac{1}{R_P} = \frac{1}{10}+\frac{1}{15}+\frac{1}{30} = \frac{3+2+1}{30} = \frac{6}{30} \;\Rightarrow\; R_P = 5 \; \Omega\] \[R_T = 0.144 + 5 = 5.144 \; \Omega \qquad I_T = \frac{230}{5.144} = 44.7 \text{ A}\] \[V_\text{cable drop} = 44.7 \times 0.144 = 6.44 \text{ V} \qquad V_\text{board} = 230 - 6.44 = 223.6 \text{ V}\]Volt drop percentage: \((6.44/230) \times 100 = 2.8\%\) — within the 5% BS 7671 limit for power circuits.
Power Formulae
Power (P) is the rate at which electrical energy is converted into another form (heat, light, mechanical work). The three power formulae are derived from \(P = VI\) and Ohm's Law \((V = IR)\), giving three equivalent expressions depending on which two quantities are known:
Choosing the right form saves steps: if you know V and R but not I, use \(P = V^2/R\) directly rather than calculating I first. If you know I and R but not V, use \(P = I^2R\).
100 Ω resistor on 24 V dc supply:
\[I = \frac{24}{100} = 0.24 \text{ A}\] \[P = VI = 24 \times 0.24 = 5.76 \text{ W} \quad P = I^2R = 0.0576 \times 100 = 5.76 \text{ W} \quad P = \frac{V^2}{R} = \frac{576}{100} = 5.76 \text{ W}\]A 2.5 mm² copper cable run of 50 m (total resistance 0.344 Ω) carries 16 A:
\[P_\text{cable} = I^2 R = 16^2 \times 0.344 = 256 \times 0.344 = 88.1 \text{ W}\]This heat is wasted energy — it adds to the building's cooling load and raises conductor temperature. Using 4 mm² cable (R ≈ 0.215 Ω) reduces losses to \(256 \times 0.215 = 55\text{ W}\) — a saving of 33 W per circuit, potentially significant in large installations.
A 470 Ω resistor is connected across a 12 V dc supply. The power dissipated:
\[P = \frac{V^2}{R} = \frac{144}{470} = 0.306 \text{ W}\]Standard resistors are rated at 0.125 W, 0.25 W, 0.5 W, 1 W, 2 W, etc. A 0.5 W rated resistor would run hot; a 1 W resistor should be specified to allow a safety margin (typically select the next standard rating that is at least twice the calculated dissipation).
Power in Parallel and Series DC Circuits
The total power dissipated in any circuit (series or parallel) equals the sum of the power dissipated by each component, which also equals the power delivered by the supply:
\[P_T = P_1 + P_2 + P_3 + \ldots = V_S \times I_T\]This is the electrical equivalent of energy conservation — all power in equals all power dissipated.
Two 12 V, 2.4 Ω lamps connected in parallel across a 12 V supply. Each lamp is at its rated voltage:
\[P_\text{each} = \frac{V^2}{R} = \frac{12^2}{2.4} = \frac{144}{2.4} = 60 \text{ W} \qquad P_T = 60+60 = 120 \text{ W}\]Total current from supply: \(I_T = P_T/V_S = 120/12 = 10\text{ A}\). Each lamp operates at full brightness.
The two 2.4 Ω lamps connected in series across the same 12 V supply:
\[R_T = 2.4+2.4 = 4.8 \; \Omega \qquad I = \frac{12}{4.8} = 2.5 \text{ A}\] \[P_\text{each} = I^2 R = 2.5^2 \times 2.4 = 6.25 \times 2.4 = 15 \text{ W} \qquad P_T = 15+15 = 30 \text{ W}\]Each lamp now dissipates only 15 W instead of 60 W — one quarter of its rated power. The lamps appear very dim. This shows why lamps and appliances must be connected in parallel, not in series — series connection reduces voltage and power to each load.
Self-assessment check: The same principle applies to two 12 V, 50 W lamps in series on 12 V. \(R_\text{each} = V^2/P = 144/50 = 2.88\text{ Ω}\). \(I = 12/(2 \times 2.88) = 2.08\text{ A}\). \(P_T = 12 \times 2.08 = 25\text{ W}\) — confirming the self-assessment answer of 25 W.
A 230 V final distribution board has the following parallel loads: 6 × 60 W downlights, a 2 kW water heater, and a 0.75 kW fan. Total power and total current drawn:
\[P_T = (6 \times 60) + 2000 + 750 = 360 + 2000 + 750 = 3110 \text{ W}\] \[I_T = \frac{P_T}{V_S} = \frac{3110}{230} = 13.5 \text{ A}\]This confirms a 16 A circuit breaker is adequate (13.5 A < 16 A), with some spare capacity for additional loads.
Voltage Drop in Electrical Circuits
Every conductor has resistance. When current flows through that resistance, a voltage is dropped across it (Ohm's Law: \(V = IR\)). This means the voltage arriving at the load is less than the supply voltage — the difference is the voltage drop. Voltage drop is not a fault; it is an inevitable consequence of having resistance in the circuit. However, if it is excessive, equipment performance suffers:
- Lamps are dimmer and have reduced life
- Motor starting torque is reduced (torque ∝ V²) — motors may fail to start or stall under load
- Contactors and relays may fail to pick up or may drop out
- Electronic equipment may malfunction or shut down
- Fluorescent and discharge lamps may fail to strike
In a two-conductor circuit (line + neutral/return), both conductors contribute resistance, so the total conductor resistance is twice the resistance of one conductor.
12 V supply, line and neutral conductors each 0.5 Ω, lamp load 24 Ω:
\[R_T = 0.5+0.5+24 = 25 \; \Omega \qquad I = \frac{12}{25} = 0.48 \text{ A}\] \[V_d = I \times R_\text{conductors} = 0.48 \times (0.5+0.5) = 0.48 \text{ V}\] \[V_\text{lamp} = 12 - 0.48 = 11.52 \text{ V} \qquad \text{Drop} = \frac{0.48}{12} \times 100 = 4\%\]Impact on lamp power: \(P = V^2/R\). Without drop: \(12^2/24 = 6\text{ W}\). With drop: \(11.52^2/24 = 5.53\text{ W}\) — the lamp operates at only 92% of its rated power.
A 2.5 mm² copper twin-and-earth cable runs 25 m to a lighting circuit carrying 10 A. The resistance per metre for 2.5 mm² copper is approximately 0.0073 Ω/m per conductor (from BS 7671 Appendix 4). Total resistance of both conductors:
\[R = 2 \times 0.0073 \times 25 = 0.365 \; \Omega\] \[V_d = I \times R = 10 \times 0.365 = 3.65 \text{ V}\] \[\%V_d = \frac{3.65}{230} \times 100 = 1.59\%\]Maximum permitted for lighting (BS 7671 Section 525): 3% = 6.9 V. The circuit complies (1.59% < 3%) ✓.
A 6 mm² copper SWA cable runs 40 m to a 16 A socket outlet circuit. Resistance per metre for 6 mm² copper ≈ 0.00305 Ω/m per conductor. Total volt drop:
\[V_d = I \times 2 \times R_m \times L = 16 \times 2 \times 0.00305 \times 40 = 16 \times 0.244 = 3.9 \text{ V}\] \[\%V_d = \frac{3.9}{230} \times 100 = 1.70\%\]Maximum permitted for power circuits: 5% = 11.5 V. Complies (1.70% < 5%) ✓. The cable has been sized primarily for current-carrying capacity; volt drop is not the limiting factor here.
Two conductors each drop 0.6 V. Total conductor drop = \(2 \times 0.6 = 1.2\text{ V}\). Voltage at load:
\[V_\text{load} = 24 - 1.2 = 22.8 \text{ V}\]This confirms the self-assessment answer of 22.8 V ✓.
- Lighting circuits: maximum 3% of nominal supply voltage (6.9 V on 230 V single-phase)
- Other (power) circuits: maximum 5% (11.5 V on 230 V single-phase)
Three Main Effects of Current Flow
When electric current flows through a conductor or circuit, it always produces one or more of three physical effects. Recognising which effect is dominant in a given application is fundamental to understanding how electrical equipment works and how faults manifest themselves.
| Effect | Useful applications | Unwanted consequences |
|---|---|---|
| Heating (thermal) | Immersion heaters, fan heaters, electric cookers, filament lamps, welding | Cable I²R losses, overheating of joints, fire risk from overloaded circuits |
| Chemical | Battery charging, electroplating, electrolytic capacitors | Electrolytic corrosion of buried metalwork, stray current corrosion of pipes |
| Magnetic | Motors, generators, transformers, relays, contactors, solenoids | EMI affecting sensitive electronics, circulating currents in metal enclosures |
Heating (Thermal) Effect
When current flows through any resistance, electrical energy is converted to heat — this is Joule's Law. The power dissipated (and therefore the rate of heat production) is:
\[P = I^2 R\]This effect is exploited in all resistive heating applications. The heating element in a storage heater, an electric oven, or an immersion heater is simply a resistive conductor designed to dissipate large amounts of power safely. In contrast, the heating effect is a loss and safety hazard in cables, joints, and switchgear — where \(I^2R\) losses must be minimised through correct cable sizing and tight terminations.
A 230 V fan heater takes 5 A. Power consumed:
\[P = V \times I = 230 \times 5 = 1150 \text{ W} = 1.15 \text{ kW}\]A terminal connection has developed a resistance of 0.5 Ω due to oxidation and poor contact pressure. A 16 A circuit flows through it. Power dissipated at the terminal:
\[P = I^2 R = 16^2 \times 0.5 = 256 \times 0.5 = 128 \text{ W}\]128 W concentrated at a small terminal will rapidly raise the temperature above the melting point of PVC insulation (and could ignite surrounding materials). This illustrates why correct terminal torque values (specified by the manufacturer) are a fire prevention requirement, and why thermal imaging surveys are used during periodic inspection to find hot joints before they cause fires.
A 13 A BS 1362 cartridge fuse uses the thermal effect deliberately: the fuse element has a small cross-section and melts (opens the circuit) when excessive current heats it above its melting point. At 13 A the element reaches equilibrium temperature and survives. At 26 A (200%) it melts within seconds. At fault currents of hundreds of amps it melts almost instantaneously — protecting the circuit from damage.
Chemical Effect
The chemical effect of current arises from the interaction between electricity and ionic (charged particle) solutions or compounds. It is the basis of battery technology and electroplating.
Batteries and Cells
A cell converts chemical energy to electrical energy through a chemical reaction at two electrodes immersed in an electrolyte. Cells are classified as:
| Type | Description | Examples | Electrical use |
|---|---|---|---|
| Primary cell | Single-use — the chemical reaction is not reversible. Once the chemicals are exhausted, the cell is discarded. | Alkaline (AA, AAA, PP3), lithium coin cells (CR2032) | Smoke detectors, remote controls, emergency lighting test buttons, multimeter batteries |
| Secondary cell (accumulator) | Rechargeable — passing an external current through the cell in reverse reverses the chemical reaction and restores the stored energy. | Lead-acid (car/van batteries, UPS), NiMH, Li-ion | UPS (uninterruptible power supply) systems, electric vehicles, emergency lighting central battery systems |
A 12 V, 7 Ah lead-acid battery in an uninterruptible power supply (UPS) is kept on trickle charge during normal operation. During a mains failure, it delivers 12 V to the inverter. At a discharge current of 3.5 A, the theoretical discharge time is:
\[t = \frac{\text{Capacity (Ah)}}{I} = \frac{7}{3.5} = 2 \text{ hours}\]In practice, the usable capacity is less (typically 80% of rated) due to internal resistance and temperature effects — around 1.6 hours of backup. This calculation is used when sizing UPS battery banks for server rooms and fire alarm panels.
Electrolysis (Electroplating)
When a direct current is passed through an electrolyte (a liquid containing ions — dissolved salts or acids), positive ions (cations) migrate to the negative electrode (cathode) and negative ions (anions) migrate to the positive electrode (anode). At each electrode, a chemical reaction deposits or dissolves material.
In electroplating, the object to be plated is the cathode; the plating metal forms the anode; both are immersed in a solution containing the plating metal's ions. The flow of current causes metal to dissolve from the anode and deposit as a thin layer on the cathode.
Copper plating on a steel busbar improves conductivity at the surface and prevents oxidation. The steel busbar (cathode) and a copper plate (anode) are placed in copper sulphate solution. Passing a dc current deposits copper atoms from the anode onto the busbar surface. The amount deposited is proportional to current × time (Faraday's Law of Electrolysis).
Stray direct currents from dc traction systems (underground railways, trams) can flow through buried metalwork (gas pipes, water mains, cable sheaths). Where the current leaves the metal to return through the earth to the supply, electrochemical corrosion rapidly destroys the metal — a pit can form in a gas pipe within months. Cathodic protection systems counter this by deliberately making the pipe the cathode (not the anode) of the electrolytic cell. This is why BS 7671 requires careful attention to earthing near dc traction systems.
- Thermal: \(P = I^2R\) — heat at every resistance. Useful in heaters; dangerous in loose connections. The basis of fuse and circuit breaker operation.
- Chemical: drives batteries and electroplating; causes corrosion where unwanted. Always involves dc or pulsating dc (ac cannot sustain directional ion migration).
- Magnetic: produced by every current-carrying conductor — covered in Chapter 5. The basis of motors, generators, transformers, and contactors.
🃏 Chapter 4 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Covers atomic structure, conductors, insulators, cables, Ohm's Law, series/parallel circuits, power and volt drop.
📝 Chapter 4 — Multiple Choice Questions
Select your answer and click Submit. Two attempts per question. First-attempt score shown at the end.
1How many electrons does a copper atom have in its neutral state?
2Conventional current flows from _____ to _____ in an external circuit.
3Which cable insulating material has the highest temperature rating?
4A conductor has resistivity ρ = 1.72 × 10⁻⁸ Ωm, length 100 m, CSA 2.5 mm². What is its resistance?
5A 110 V supply feeds two 22 Ω resistors in series. What is the circuit current?
6Two 12 V 50 W lamps are connected in series across a 12 V supply. What is the total power dissipated?
7Three resistors of 6 Ω, 12 Ω and 4 Ω are connected in parallel. What is the total resistance?
8A 230 V circuit has a measured load current of 10 A. What is the power consumed?
9A 24 V supply circuit has a diode with a 0.6 V forward volt drop. What is the voltage across the load?
10BS 7671 limits volt drop on a lighting circuit to 3% of the 230 V nominal supply. What is the maximum permitted volt drop?
11Which of the following correctly describes the thermal effect of electric current?
12What is the chemical effect of electric current called when current passes through an electrolyte causing material to deposit or dissolve?
Sends your Chapter 4 MCQ score summary from your registered email to your tutor.
Magnetism and Electricity
- Describe the magnetic effects of electric currents and sketch magnetic flux patterns
- State and apply units, symbols and variables for magnetic flux and magnetic flux density
- Determine the force on a current-carrying conductor in a magnetic field
- Determine the emf induced in a conductor moving through a magnetic field
- Explain how emf may be produced by self induction and mutual induction
- Describe the basic principles of generating an ac supply and explain RMS and average values of a sine wave
Magnetic Field Patterns
Every magnet — permanent or electromagnet — affects the space around it. Any ferrous (iron-containing) material or other magnet placed in this space experiences a force. That region of influence is the magnetic field. Because a magnetic field is invisible, we represent it with imaginary lines of magnetic flux — a model that makes it possible to predict the direction and strength of magnetic forces.
Lines of flux obey five fundamental rules that govern their behaviour in every application — from the field of a bar magnet to the air gap of a motor:
| Rule | Explanation | Practical consequence |
|---|---|---|
| Always form complete closed loops | Flux leaves the north pole, travels through surrounding space, re-enters at the south pole, and completes the loop inside the magnet | There is no magnetic equivalent of an open circuit — flux always returns to its source |
| Never cross one another | Each line represents a unique direction at that point; if two crossed they would indicate two directions simultaneously, which is impossible | Allows us to identify exactly which way the force acts at any point |
| Have a definite direction | Conventional direction: N → S outside the magnet; S → N inside | Determines which way a motor conductor is pushed (Fleming's left-hand rule) |
| Try to contract (shorten) | Lines behave like stretched elastic — they try to become as short as possible | Creates an attractive force between unlike poles; pulls iron cores into solenoids |
| Repel sideways neighbours | Parallel flux lines with the same direction push apart — they cannot occupy the same space | Creates forces between parallel current-carrying conductors |
The closer together the flux lines, the stronger (more dense) the field. This is why flux density is highest inside the magnetic material (where lines are most concentrated) and weakest far from the magnet. Unlike poles attract because flux lines link north to south, contracting to pull them together. Like poles repel because their flux lines, running in the same direction side-by-side, push each other apart.
Concentric Magnetic Fields Around a Current-Carrying Conductor
Electricity and magnetism are inseparable: every electric current produces a magnetic field. When direct current flows through a straight conductor, it produces a magnetic field in the form of concentric circles centred on the conductor. This is the electromagnetic effect that makes motors, transformers, relays, and contactors possible.
Two rules determine the direction of the magnetic field relative to the current direction:
- Right-hand grasp rule: Wrap the right hand around the conductor with the thumb pointing in the direction of conventional current flow. The fingers curl in the direction of the magnetic field around the conductor.
- Maxwell's Corkscrew Rule: Imagine driving a right-handed screw in the direction of conventional current flow. The direction in which the screw turns is the direction of the magnetic field around the conductor.
- ⊗ (cross) — current flowing away from you (into the page) — like the tail feathers of an arrow
- ⊙ (dot) — current flowing towards you (out of the page) — like the point of an arrow
A single-core cable carries conventional current flowing away from you (⊗). Applying the right-hand grasp rule: thumb points away (into page), fingers curl clockwise. Therefore the magnetic field around the conductor runs clockwise when viewed from behind. This field pattern surrounds the cable in every direction — this is why cables generate EMI that can affect nearby sensitive instrumentation and why cables are twisted or run in pairs to cancel their fields.
Right-Hand Grasp Rule Applied to a Solenoid (Coil / Electromagnet)
A solenoid is a coil of wire wound in a helix. When current flows through it, each turn produces a concentric field; the fields from all turns reinforce each other inside the coil to produce a strong, approximately uniform field along the axis — very similar to that of a bar magnet.
Applying the right-hand grasp rule to the whole coil: wrap the right hand around the solenoid with the fingers pointing in the direction of conventional current flow through the turns. The thumb points towards the north pole.
Reversing the current direction reverses the polarity (north becomes south and vice versa). This is exploited in:
- Relays and contactors: energising the coil creates an electromagnet that attracts the armature, closing contacts
- Solenoid valves: the plunger is attracted into the coil when energised, opening or closing a fluid path
- Electric bells and buzzers: the solenoid attracts a striker repeatedly while current flows
- Motor field windings: coils create the magnetic field that interacts with the rotor conductors
A 230 V ac contactor coil is wound with conventional current flowing left-to-right through the top turns when viewed from the front. Using the right-hand rule, the thumb points to the left — the left end of the coil is the north pole, which attracts the iron armature downward to close the contacts. If the supply connections are reversed, the current direction reverses but (since it is ac) the polarity alternates anyway — ac contactors work on the principle of magnetic attraction regardless of polarity.
Force Between Parallel Current-Carrying Conductors
Two parallel conductors each carrying current both produce their own magnetic fields. Where the fields overlap, they interact and produce a force between the conductors. This effect is significant in busbars, cable trays with multiple cables, and motor windings carrying high fault currents:
| Current directions | Field interaction | Resultant force | Practical example |
|---|---|---|---|
| Same direction (both →) | Fields cancel between conductors; strengthen outside — conductors pulled into the weaker region between | Attraction | Busbars carrying the same phase current are attracted — supports must resist this force to prevent contact |
| Opposite directions (→ and ←) | Fields add between conductors; weaken outside | Repulsion | During a three-phase fault, adjacent busbar phases carry opposing surge currents and are violently repelled — busbar clamps and supports must withstand thousands of newtons |
During a fault, two 33 kV busbars 100 mm apart each carry 20 kA peak fault current. The force per metre between them is approximately:
\[F = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{4\pi \times 10^{-7} \times 20000 \times 20000}{2\pi \times 0.1} = \frac{4 \times 10^{-7} \times 4 \times 10^8}{0.2} = 800 \text{ N/m}\]At 800 N per metre of busbar length, substantial clamps and insulator supports are essential to prevent busbar deflection and contact during faults — this is why short-circuit withstand ratings are specified for all busbar systems.
Magnetic Flux (symbol Φ, unit weber Wb)
Magnetic flux (Φ) is the total amount of magnetic field passing through a given surface — it represents the "quantity" of magnetism. The unit is the weber (Wb) (pronounced "vayber"), named after German physicist Wilhelm Eduard Weber. One weber is a large unit; in practice motor air gaps and transformer cores carry flux values measured in milliwebers (mWb) or microwebers (µWb).
Magnetic flux is analogous to electric current in a circuit: just as current is the quantity of charge flowing, flux is the quantity of magnetic field threading through a surface. The path through which flux flows is called a magnetic circuit — a concept used to design transformer cores, motor cores, and relay cores.
Magnetic Flux Density (symbol B, unit tesla T)
Magnetic flux density (B) is the flux per unit area — it measures how concentrated (intense) the field is at a particular location. A large flux spread over a large area gives a low flux density; the same flux concentrated into a small area gives a high flux density. The unit is the tesla (T):
\[B = \frac{\Phi}{A} \qquad 1\text{ T} = 1\text{ Wb/m}^2\]Flux density directly determines the force on a current-carrying conductor (F = BIl) and the emf induced in a moving conductor (E = Blv). Higher flux density means stronger motor torque and greater generator output for the same conductor current and speed.
| Application | Typical flux density (T) | Notes |
|---|---|---|
| Earth's magnetic field | ~0.00005 | Too weak for practical machines |
| Permanent magnet (ferrite) | 0.2–0.4 | Used in small dc motors, loudspeakers |
| Permanent magnet (rare earth NdFeB) | 1.0–1.4 | Modern servo and brushless motors |
| Transformer core (silicon steel) | 1.2–1.7 | Design limit before saturation |
| Large power motor air gap | 0.8–1.2 | Typical industrial induction motor |
| MRI scanner (medical) | 1.5–3.0 | Superconducting electromagnet |
A transformer core has a cross-sectional area of 0.01 m² and carries a flux of 12 mWb. Flux density:
\[B = \frac{\Phi}{A} = \frac{0.012}{0.01} = 1.2\text{ T}\]This is near the saturation point for silicon steel — increasing the primary voltage further would saturate the core, dramatically increasing magnetising current and losses.
A motor air gap has a flux density of 0.9 T across a pole face area of 0.05 m². Total flux:
\[\Phi = B \times A = 0.9 \times 0.05 = 0.045\text{ Wb} = 45\text{ mWb}\]Magnetomotive Force (MMF) and Magnetic Field Strength (H)
The force that drives flux around a magnetic circuit is the magnetomotive force (MMF), measured in ampere-turns (At). For a coil of N turns carrying current I:
\[\text{MMF} = N \times I \quad (\text{ampere-turns, At})\]The magnetic field strength H is the MMF per unit length of the magnetic circuit:
\[H = \frac{N \times I}{l} \quad (\text{A/m})\]where l = mean length of the magnetic circuit (m). H is the cause; B is the effect — the relationship between them is determined by the core material's permeability.
Magnetic Field Strength of a Solenoid
The magnetic field strength and therefore the flux density inside a solenoid depends on three factors:
| Factor | Effect on field strength | How to increase field strength |
|---|---|---|
| Current (I) | Directly proportional — double I → double MMF and H | Increase supply current (limited by coil rating) |
| Number of turns (N) | Directly proportional — double N → double MMF | Wind more turns (increases coil inductance and resistance) |
| Core material | Ferromagnetic materials (iron, silicon steel, ferrite) concentrate flux far more than air — permeability can be hundreds or thousands of times that of air | Insert an iron or silicon-steel core — most effective method |
A relay coil has 500 turns and is energised by 48 V dc. The coil resistance is 240 Ω. MMF:
\[I = \frac{V}{R} = \frac{48}{240} = 0.2\text{ A}\] \[\text{MMF} = N \times I = 500 \times 0.2 = 100\text{ At}\]The 100 At drives flux through the relay's magnetic circuit (core + air gap). If the relay requires 80 At to operate (pick up), this coil has a 25% margin — sufficient to allow for reduced supply voltage and increased air gap due to worn contacts.
The interaction between a magnetic field and a current-carrying conductor is the fundamental operating principle of virtually all electric motors. When a conductor carrying current is placed in an external magnetic field, its own field distorts the main field: on one side the two fields oppose and weaken each other; on the other side they reinforce and strengthen. The conductor is pushed from the strong-field side towards the weak-field side — this is the motor effect or electromagnetic force.
Two conditions must both be present for a force to be produced:
- The conductor must carry current
- The conductor must be in a magnetic field
Reversing either the current direction or the magnetic field direction reverses the force direction. Reversing both simultaneously leaves the force direction unchanged — this is why ac motors that reverse both field and current together still develop torque in the same direction for each half-cycle.
Fleming's Left-Hand (Motor) Rule
Fleming's left-hand rule gives the direction of the force (motion) when the field and current directions are known. Hold the left hand with the first finger, second finger, and thumb all mutually at right angles — like three sides of a corner of a box:
| Digit | Represents | Memory aid |
|---|---|---|
| First finger | Direction of the magnetic Field (north → south) | First finger = Field |
| seCond finger | Direction of Conventional Current flow | Current starts with C |
| Thumb | Direction of Motion (force / thrust) | Thumb = Thrust = Motion |
Important: The left-hand rule applies to motors (electrical energy converted to mechanical motion). The right-hand rule (generator rule — Part 4) applies to generators (mechanical motion converted to electrical energy). Use the correct hand for the correct machine.
A dc motor has a horizontal magnetic field pointing left to right (north on left, south on right). The armature conductor carries conventional current flowing away from you (into the page, ⊗). Applying the left-hand rule: first finger points right (field N→S), second finger points away from you (current direction). The thumb therefore points upward — the conductor experiences an upward force, rotating the armature clockwise when viewed from the commutator end.
Calculating the Force: F = BIl
The magnitude of the electromagnetic force on a current-carrying conductor is given by:
where F = force (N), B = flux density (T), I = current (A), l = active length of conductor within the field (m).
This formula applies when the conductor is at exactly 90° to the magnetic field — giving the maximum possible force. At any other angle θ, the force is reduced: \(F = BIl \sin\theta\). At 0° (conductor parallel to field) the force is zero. Motors are designed so their conductors always sweep through the field at right angles to maximise torque.
A 0.3 m conductor at right angles to a 1.5 T field carries 30 A:
\[F = B \times I \times l = 1.5 \times 30 \times 0.3 = 13.5\text{ N}\]A dc motor armature has 20 conductors, each 0.15 m long in a field of 0.8 T, carrying 5 A. Total force on all conductors:
\[F_\text{total} = 20 \times B \times I \times l = 20 \times 0.8 \times 5 \times 0.15 = 12\text{ N}\]If the armature radius is 0.05 m, the torque produced is:
\[T = F \times r = 12 \times 0.05 = 0.6\text{ Nm}\]This simple calculation shows how increasing B (stronger magnet) or I (more current) directly increases motor torque — the basis of motor speed and torque control.
A busbar conductor 0.5 m long carries a fault current of 10 kA in a field of 0.05 T created by the adjacent conductor. Force on the busbar:
\[F = 0.05 \times 10000 \times 0.5 = 250\text{ N}\]250 N on a 0.5 m busbar section is a very large force — equivalent to hanging a 25 kg weight from that point. Busbar clamps and supports must be rated for the short-circuit withstand forces specified in IEC 61439.
A relay armature requires a minimum force of 0.05 N to close. The relay magnet has a flux density of 0.4 T across a 0.02 m conductor. What minimum current is needed?
\[I = \frac{F}{B \times l} = \frac{0.05}{0.4 \times 0.02} = \frac{0.05}{0.008} = 6.25\text{ A}\]Electromagnetic Induction — Faraday's Law
Michael Faraday discovered in 1831 that an emf is induced in a conductor whenever there is a change in the magnetic flux linking that conductor. This is Faraday's Law of Electromagnetic Induction and is the single most important principle underlying generators, transformers, inductors, and the back-emf of motors.
There are two ways to change the flux linking a conductor:
- Move the conductor through a stationary magnetic field (generator principle)
- Change the flux through a stationary conductor or coil (transformer / inductor principle)
In both cases the magnitude of the induced emf is proportional to the rate of change of flux linkage.
EMF Induced by a Moving Conductor: E = Blv
When a straight conductor of length l moves with velocity v at right angles through a uniform magnetic field of flux density B, the induced emf is:
\[E = B \times l \times v\]where E = induced emf (V), B = flux density (T), l = active length of conductor in the field (m), v = velocity perpendicular to both conductor and field (m/s).
This formula gives maximum emf only when the conductor moves at exactly 90° to the field. As the angle decreases towards 0° (motion parallel to field), the emf drops to zero — \(E = Blv\sin\theta\) in the general case. This explains why a generator's output varies sinusoidally as the coil rotates.
A 0.2 m conductor moves at 12 m/s at right angles through a 0.9 T field:
\[E = 0.9 \times 0.2 \times 12 = 2.16\text{ V}\]The same generator needs to produce 4.32 V. The flux density and conductor length are fixed. What velocity is needed?
\[v = \frac{E}{B \times l} = \frac{4.32}{0.9 \times 0.2} = \frac{4.32}{0.18} = 24\text{ m/s}\]Doubling the speed doubles the emf — this is why generator speed control directly governs output voltage and frequency.
A dc motor armature is also a generator. As it rotates, its conductors move through the field and induce an emf that opposes the supply voltage (back-emf, E_b). At full speed a 230 V motor might have E_b = 215 V, leaving only 15 V to drive current through the armature resistance. At start-up E_b = 0 so the full 230 V drives current through a small resistance — producing a very high starting current (hence the need for motor starters on large dc motors).
Fleming's Right-Hand (Generator) Rule
The right-hand rule gives the direction of the induced current (and emf) when both the field and the conductor's motion are known. Hold the right hand with the first finger, thumb, and second finger mutually at right angles:
| Digit | Represents | Memory aid |
|---|---|---|
| First finger | Direction of the magnetic Field (N → S) | First = Field |
| Thumb | Direction of Motion (velocity of conductor) | Thumb = Thrust/motion |
| seCond finger | Direction of induced Current (and emf) | Current = Comes out at second finger |
Contrast with left-hand rule: Use left hand for motors (current in → motion out). Use right hand for generators (motion in → current out). Both hands have Field on first finger and Current on second finger — only the thumb differs (motor: thrust; generator: motion input).
A generator conductor moves upward (thumb up) in a field pointing left to right (first finger right). Applying the right-hand rule: second finger points away from you (into the page ⊗). The induced conventional current flows into the page in this conductor — which, continuing around the external circuit, flows from the generator terminal to the load.
EMF Induced by Rate of Change of Flux: Faraday's Law for Coils
For a coil with N turns, if the flux through it changes from Φ₁ to Φ₂ in time t, the induced emf is:
\[E = \frac{N(\Phi_2 - \Phi_1)}{t} = N \frac{\Delta\Phi}{\Delta t}\]This is the formula used for transformer design and for calculating the back-emf of inductors switching on or off. A faster flux change, more turns, or a larger flux change all produce a higher induced emf.
A 600-turn coil experiences flux rising from 60 mWb to 120 mWb in 100 ms:
\[E = \frac{600 \times (0.12 - 0.06)}{0.1} = \frac{600 \times 0.06}{0.1} = 360\text{ V}\]A relay coil of 500 turns and 0.2 H carries 0.5 A. The circuit is broken by a switch in 2 ms. The flux collapses from its full value to zero. Induced emf:
\[E = L \frac{\Delta I}{\Delta t} = 0.2 \times \frac{0.5}{0.002} = 0.2 \times 250 = 50\text{ V}\]A 50 V spike on a 24 V control circuit can damage the switching transistor or PLC output module. This is why a freewheeling diode (flyback diode) is connected across every relay or solenoid coil in electronic control circuits — it provides a path for the collapsing flux current and limits the voltage spike to about 0.7 V.
A coil must induce 240 V when the flux changes from 0 to 80 mWb in 50 ms. How many turns are needed?
\[N = \frac{E \times t}{\Delta\Phi} = \frac{240 \times 0.05}{0.08} = \frac{12}{0.08} = 150\text{ turns}\]Self-Inductance and Lenz's Law
When current through a coil changes, the changing flux induces an emf in the same coil — this is self-inductance. The induced emf always acts in a direction that opposes the change that produced it — stated by Lenz's Law:
Consequences of Lenz's Law:
- When current is switched on in an inductive coil, the self-induced emf opposes the rise — current builds up gradually (exponentially), not instantly.
- When current is switched off, the collapsing flux induces a large emf that tries to maintain the current — causing a voltage spike (see Example 2 above).
- A motor back-emf opposes the supply — the motor draws less current as it accelerates.
- Transformer reflected impedance — a load on the secondary is "reflected" back to oppose changes in the primary current.
The unit of self-inductance is the henry (H): a coil has an inductance of 1 H if a current changing at 1 A/s produces a self-induced emf of 1 V:
\[E_L = L \frac{\Delta I}{\Delta t}\]Practical coils (inductors, chokes) have inductances measured in mH or µH. Large transformer and motor windings have inductances of several henries.
Mutual Inductance — The Transformer Principle
Mutual inductance occurs when two coils are magnetically linked — a change in current in one coil (primary) induces an emf in the other (secondary). The magnitude of the mutual inductance M (measured in henries) determines how effectively flux from one coil links the other:
\[E_2 = M \frac{\Delta I_1}{\Delta t}\]The transformer exploits mutual inductance deliberately: the primary and secondary windings share a common iron core that guides virtually all the flux from primary to secondary (near-perfect coupling). An alternating current in the primary creates a continuously changing flux, which continuously changes flux linkage in the secondary, inducing a continuous alternating emf. This is why transformers only work on alternating current — a steady dc current produces constant flux, which induces zero emf.
A transformer has a mutual inductance M = 0.5 H. The primary current changes at a rate of 20 A/s. The emf induced in the secondary:
\[E_2 = M \times \frac{\Delta I_1}{\Delta t} = 0.5 \times 20 = 10\text{ V}\]For comparison, if a dc current of 20 A flows steadily in the primary (no change), ΔI/Δt = 0, so E₂ = 0 — confirming transformers do not work on dc.
Single-Phase AC Generator (Alternator)
All large-scale electricity generation exploits electromagnetic induction: a conductor moving through a magnetic field has an emf induced in it. In an alternator, a coil rotates continuously in a stationary magnetic field (or, in large power station alternators, a rotating electromagnet spins inside stationary coils). The result is a continuously changing induced emf — an alternating current supply.
A simple single-phase alternator consists of a single rectangular coil mounted on a shaft between the poles of a permanent magnet. The coil connects to the external circuit through slip rings (conducting rings mounted on the shaft) and carbon brushes (stationary contacts pressing on the rings). This arrangement allows the rotating coil to maintain a continuous electrical connection to the external circuit.
As the coil rotates through one complete revolution (360°):
| Position (rotation angle) | Coil orientation relative to field | Rate of flux cutting | Induced emf |
|---|---|---|---|
| 0° (start) | Coil parallel to field — conductors move parallel to flux lines | Zero | Zero |
| 90° | Coil perpendicular to field — conductors cut flux at maximum rate | Maximum | Maximum positive (+V_max) |
| 180° | Coil parallel again (opposite side) | Zero | Zero |
| 270° | Coil perpendicular again — but conductors now move in opposite direction | Maximum (reverse) | Maximum negative (−V_max) |
| 360° | Back to start | Zero | Zero (one full cycle complete) |
The emf at any angle θ is: \(e = V_\text{max} \sin\theta\) — a sinusoidal waveform. This is why the UK supply waveform is called a sine wave. The continuously reversing direction of the emf is what makes it alternating — it is not constant in direction like dc.
The UK supply completes 50 cycles per second (frequency f = 50 Hz), meaning the generator rotates at 50 rev/s (3000 rpm for a 2-pole machine). The periodic time is T = 1/50 = 20 ms.
Three-Phase Generators
All power stations and large alternators generate three-phase ac. Instead of one coil, there are three coils placed 120° apart on the stator. As the rotor magnet sweeps past each coil, it induces a sine wave in each — but because the coils are 120° apart, the three sine waves are 120° out of phase with each other.
The three phases are conventionally labelled L1, L2, L3 (or A, B, C) with conductor colours brown, black, and grey in modern BS 7671 notation. Each reaches its peak 1/3 of a cycle (6.67 ms at 50 Hz) after the previous one.
Advantages of three-phase over single-phase:
- More efficient transmission — three conductors carry the power of six single-phase conductors
- The three phases balance each other — at any instant, their sum is zero (no neutral current in a balanced load)
- Three-phase induction motors are self-starting and have smooth continuous torque
- The rotating magnetic field produced by three-phase windings directly drives induction motors
In a balanced three-phase system: line voltage (between phases) = \(\sqrt{3} \times\) phase voltage (phase to neutral).
UK supply: phase voltage = 230 V rms. Line voltage = \(230 \times \sqrt{3} = 230 \times 1.732 = 398.4 \approx 400\text{ V}\).
This is why UK industrial supplies are described as 230/400 V — 230 V phase-to-neutral, 400 V line-to-line.
Sine Wave Characteristics and Terminology
The key characteristics of a sinusoidal ac waveform are:
| Term | Symbol | Definition | UK supply example |
|---|---|---|---|
| Peak (maximum) value | V_max or V_p | Highest instantaneous value of the waveform | 325 V |
| Peak-to-peak value | V_pp | Total excursion from +peak to −peak = 2 × V_max | 650 V |
| RMS value | V_rms | Equivalent dc value for equal power dissipation = 0.707 × V_max | 230 V |
| Average value | V_av | Average of the waveform over one half-cycle = 0.637 × V_max | 207 V |
| Frequency | f | Number of complete cycles per second (Hz) | 50 Hz |
| Periodic time | T | Time for one complete cycle = 1/f | 20 ms |
| Form factor | — | V_rms / V_av = 0.707/0.637 = 1.11 | 1.11 (for pure sine wave) |
RMS (Root Mean Square) Value — the Most Important AC Value
The RMS value is the most important and most commonly used value of an ac waveform because it represents the equivalent dc voltage (or current) that would produce the same heating effect in a resistor. When we say the UK supply is 230 V, we mean 230 V RMS — this is what voltmeters measure, what ratings on appliances refer to, and what appears in calculations.
The name "root mean square" comes from its mathematical derivation: take the instantaneous values of the waveform, square them (making them all positive), find the mean (average), then take the square root. For a pure sine wave this gives:
\[V_\text{rms} = \frac{V_\text{max}}{\sqrt{2}} = V_\text{max} \times 0.707\] \[V_\text{max} = V_\text{rms} \times \sqrt{2} = V_\text{rms} \times 1.414\]The 0.707 factor only applies to pure sine waves. For non-sinusoidal waveforms (square waves, triangular waves, distorted supplies) the RMS value must be calculated differently — this is why true-RMS meters are needed to accurately measure current in circuits with non-linear loads such as variable speed drives and switched-mode power supplies.
The 230 V rms supply actually reaches ±325 V at its peaks. This is why electrical equipment is rated for at least 300 V minimum, and why 1000 V test voltages are used in insulation resistance tests — the insulation must withstand more than just the nominal supply voltage.
This confirms the self-assessment answer of 282 V ✓.
An oscilloscope shows the UK supply has a peak-to-peak voltage of 650 V. Verify the RMS value:
\[V_\text{max} = \frac{650}{2} = 325\text{ V} \qquad V_\text{rms} = 325 \times 0.707 = 229.8 \approx 230\text{ V}\] ✓Average Value
The average value of a complete sine wave cycle is zero — the positive half and negative half cancel exactly. For this reason, the average value is always stated for one half-cycle only:
\[V_\text{av} = V_\text{max} \times 0.637 \qquad I_\text{av} = I_\text{max} \times 0.637\]The average value is used in rectifier circuits (converting ac to dc) to calculate the mean dc output voltage. It is rarely used in power calculations — RMS is used for those.
This confirms the self-assessment answer of 16 A ✓.
A 10 Ω resistor is connected to a 230 V rms supply. Power dissipated using RMS value:
\[P = \frac{V_\text{rms}^2}{R} = \frac{230^2}{10} = \frac{52900}{10} = 5290\text{ W}\]If the incorrect average value (207 V) were used: \(P = 207^2/10 = 4285\text{ W}\) — significantly wrong. RMS must be used because power is proportional to V² and the squaring makes the relationship non-linear.
- \(V_\text{rms} = 0.707 \times V_\text{max}\) — use for all power and current calculations
- \(V_\text{max} = 1.414 \times V_\text{rms}\) — peak voltage reached by the supply
- \(V_\text{av} = 0.637 \times V_\text{max}\) — average over one half-cycle only; used in rectifier output calculations
- UK supply: 230 V rms → 325 V peak → 207 V average
- All voltmeters and ammeters read RMS values unless stated otherwise
🃏 Chapter 5 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Covers magnetic flux, electromagnets, motor force, electromagnetic induction, generators and AC waveforms.
📝 Chapter 5 — Multiple Choice Questions
Select your answer and click Submit. Two attempts per question. First-attempt score shown at the end.
1The unit of magnetic flux density is the:
2A solenoid coil has 500 turns and carries a current of 2 A. What is the magnetomotive force (MMF)?
3When applying Fleming's left-hand rule, the thumb indicates:
4A conductor 0.4 m long carries 5 A in a field of 0.8 T perpendicular to the conductor. What force acts on it?
5State Lenz's Law. What does it tell us about the direction of an induced EMF?
6A conductor of length 0.5 m moves at 4 m/s through a field of 1.5 T perpendicular to both. What EMF is induced?
7The output from a rotating coil of a single-phase alternator is taken by:
8The RMS voltage of a sinusoidal supply is 200 V. What is the peak voltage?
9The maximum value of a sinusoidal current is 25 A. What is its average value?
10A coil of 200 turns has its flux change by 0.05 Wb in 0.1 s. What is the induced EMF?
11State three advantages of three-phase AC generation over single-phase. Which of the following is NOT an advantage?
12The UK mains supply has a frequency of 50 Hz. What is the angular frequency ω in radians per second?
Sends your Chapter 5 MCQ score summary from your registered email to your tutor.
Supply and Distribution Systems
- Describe how electricity is generated, transmitted and distributed
- Specify features and characteristics of generation, transmission and distribution systems
- Explain other sources of electrical generation (renewables, batteries)
- Describe the main characteristics of single and three-phase supplies
- Describe operating principles, applications and limitations of transformers
- Determine primary and secondary voltages, currents and kVA rating of transformers
Power Stations
All large-scale electricity generation relies on the same fundamental principle from Chapter 5 (electromagnetic induction): rotating the shaft of an alternator (ac generator) causes conductors to cut through a magnetic field, inducing an alternating emf. The primary engineering challenge is not the alternator itself — it is supplying the mechanical energy to spin the shaft. Different power station types solve this problem in different ways, each with its own efficiency, cost profile, and environmental impact.
Understanding the generation mix is relevant to electricians specifying renewable microgeneration systems, advising clients on energy storage, and interpreting supply interruptions during major grid events.
Steam-Cycle Power Stations (Thermal Generation)
The dominant generation principle worldwide: heat a fluid (almost always water) to produce high-pressure steam; the steam expands through a turbine, spinning the alternator shaft. The turbine exhaust steam is condensed back to water by cooling towers or river/sea water, and the cycle repeats. The energy source determines the type of station:
Fossil-Fuel Power Stations (Coal, Oil, Gas)
Fossil fuels are burned in a boiler to heat water to steam. Coal was historically the most widely used fuel in the UK — a 2,000 MW station consumes around 5 million tonnes per year, requiring direct rail or waterway links. Stations need vast cooling water — hence their siting on rivers or the coast. Coal stations have the highest carbon dioxide emissions per kWh of any generation type.
Oil-fired stations are flexible (quick to start) but expensive to run — now mostly used only as peaking plant. Natural gas fired boiler stations have lower emissions than coal and faster start-up.
Typical thermal efficiency of a coal/gas boiler steam station: 33–40% — the rest of the energy input is rejected as heat to the cooling water and atmosphere.
Nuclear Power Stations
Use nuclear fission — splitting uranium or plutonium atoms — to generate heat. The heat produces steam through a heat exchanger, which drives turbines exactly as in a fossil-fuel station. About 4.5 tonnes of uranium fuel per week sustains a 1,000 MW station (compared with ~100,000 tonnes of coal for the equivalent). Advantages: very low carbon emissions during operation, high energy density of fuel. Disadvantages: high capital cost, radioactive waste management, and very long decommissioning timescales. Similar cooling water requirements to coal stations.
Gas Turbine Power Stations (Open Cycle)
Natural gas is burned in a combustion chamber; the hot exhaust gases expand directly through a gas turbine (similar to a jet engine), spinning the alternator. No steam cycle is involved. Advantages: very fast start-up (minutes rather than hours), compact, lower capital cost. Disadvantage: lower efficiency (~30%) since exhaust heat is wasted. Used primarily as peaking plant — brought online rapidly during periods of peak demand or following unexpected generation loss.
Combined Cycle Gas Turbine (CCGT)
The most efficient large-scale fossil-fuel generation technology currently in widespread use. A gas turbine generates electricity (first cycle); its hot exhaust gases (~550°C) pass through a heat recovery steam generator (HRSG) which produces steam; the steam drives a separate steam turbine to generate additional electricity (second cycle). Overall electrical efficiency: 55–60% — significantly better than either cycle alone. CCGT stations now provide the majority of gas-fired generation in the UK.
A 400 MW CCGT station (58% efficient) and an equivalent coal station (36% efficient) both deliver 400 MW of electrical output. Fuel energy input needed:
\[\text{CCGT input} = \frac{400}{0.58} = 690\text{ MW (thermal)} \qquad \text{Coal input} = \frac{400}{0.36} = 1111\text{ MW (thermal)}\]The CCGT needs 38% less fuel energy for the same electrical output — directly reducing both fuel cost and CO₂ emissions per kWh.
Hydro-Electric Power Stations
Water stored in a high-altitude reservoir flows through penstocks (large pipes) down to turbines at the base of the dam. The potential energy of the water (GPE = mgh) is converted to kinetic energy at the turbine, then to electrical energy in the alternator. Advantages: essentially zero fuel cost, very fast response (seconds), no direct emissions, very long station life (50+ years), high efficiency (~90%). Disadvantages: geographically limited (needs mountains and reliable rainfall), large reservoirs displace communities and ecosystems, often far from load centres requiring long transmission lines.
A small hydro scheme has a flow rate of 5 m³/s through a head (height) of 40 m. Water density ≈ 1000 kg/m³. Power available (at 85% efficiency):
\[P = \rho \times g \times h \times Q \times \eta = 1000 \times 9.81 \times 40 \times 5 \times 0.85 = 1\,668\,300\text{ W} \approx 1.67\text{ MW}\]Pumped Storage
A specialised form of hydro generation that acts as large-scale electrical energy storage. During periods of low demand (overnight), surplus grid electricity pumps water uphill from a lower to an upper reservoir. During peak demand, the water flows back down through turbines to generate electricity. The UK's Dinorwig station (Wales) can deliver 1,800 MW within 16 seconds — making it essential for grid frequency stabilisation. Round-trip efficiency is typically 70–80%.
Why High-Voltage Transmission?
The output voltage of a modern power station generator is typically 11–25 kV ac. Before transmission, this is stepped up to 400 kV or 275 kV by a grid transformer. The reason is purely economic and physical: power loss in a transmission line depends on current (P_loss = I²R), not voltage. By stepping voltage up, current is reduced proportionally (P = VI, so doubling V halves I), and since losses vary with I², halving the current reduces losses to one quarter.
A 100 MW power transfer over a line with total resistance 10 Ω. Compare losses at 11 kV vs 400 kV:
At 11 kV: \(I = P/V = 100 \times 10^6 / 11000 = 9091\text{ A}\) → \(P_{\text{loss}} = I^2 R = 9091^2 \times 10 = 826\text{ MW}\) (impossibly high — line would melt)
At 400 kV: \(I = 100 \times 10^6 / 400000 = 250\text{ A}\) → \(P_{\text{loss}} = 250^2 \times 10 = 625\text{ kW}\) — only 0.625% of the transmitted power
This calculation shows definitively why high-voltage transmission is essential — the losses at distribution voltage would be catastrophic.
Transmission Switching Stations (Grid Substations)
At the point where the generator connects to the transmission network, a grid (switching) substation contains:
- Grid transformer — steps generator voltage (25 kV) up to transmission voltage (400 kV or 275 kV)
- Circuit breakers — provide overcurrent and fault protection; operate in milliseconds using SF₆ gas or vacuum as the arc-quenching medium at EHV levels
- Disconnectors / Isolators — manually operated switches used only when the circuit breaker has already opened the circuit; provide a visible air gap for safe working
- Busbars — solid copper or aluminium conductors that act as the common connection point between multiple circuits at the same voltage
- Instrument transformers (CTs and VTs) — step down current and voltage to safe levels (5 A and 110 V respectively) for metering and protection relays
Overhead Transmission Lines
The most economical method of bulk power transmission. Conductors are bare aluminium (ACSR — Aluminium Conductor Steel Reinforced) stranded cables suspended from lattice steel towers (pylons). The air gap between conductors and between conductors and earth provides insulation (typically 4–6 m clearance at 400 kV). String insulators attach the conductors to the tower crossarms and provide the required creepage distance to prevent flashover.
Overhead lines are exposed to lightning strikes — surge diverters and earthing of tower tops (via the earth wire running above the phase conductors) protect against direct strikes. The characteristic "buzzing" or "crackling" heard near pylons is corona discharge — partial ionisation of air around the conductor surface at high electric field intensity.
Underground Cables at Transmission Voltages
Underground cables are used where overhead lines are impractical (urban areas, Areas of Outstanding Natural Beauty, under water). At 400 kV, underground cables are oil-filled or use cross-linked polyethylene (XLPE) insulation with lead or aluminium sheaths. Capital cost is 5–10× that of equivalent overhead lines, and they require forced cooling for long runs. Despite this, the UK has more underground 400 kV cable than any other country — a consequence of planning restrictions on pylon construction.
HVDC Interconnectors
Long-distance undersea links (such as the cross-channel interconnector with France and links with Norway and Belgium) use High Voltage Direct Current (HVDC). AC has reactive power losses over long cable runs that make it impractical beyond about 50 km underground. DC has no such reactive losses. HVDC converter stations at each end convert ac to dc (rectification) and dc back to ac (inversion) using thyristor or IGBT valve stacks. HVDC also allows power to be imported and exported between grids running at different frequencies or phases.
The Transmission System — Voltage Hierarchy
The UK grid operates at several voltage levels, each step serving a specific purpose:
| Voltage Level | System | Purpose | Typical equipment served |
|---|---|---|---|
| 25 kV ac | Generator terminals | Generator output before stepping up | Power station alternator |
| 400 kV / 275 kV | National Transmission System (NTS) | Bulk long-distance power transfer between regions | Major power stations, large-scale wind farms, interconnectors |
| 132 kV | Standard grid / Transmission Distribution Interface | Regional bulk supply; feeds primary substations | Primary substations in major cities and industrial areas |
| 33 kV | Sub-transmission / primary distribution | Distribution to large users and 11 kV networks | Large industrial sites, primary substations |
| 11 kV | Secondary distribution network | Local distribution to commercial, light industrial | 11 kV/LV substations throughout towns |
| 400/230 V | Low Voltage (LV) network | Consumer supply | Houses, offices, small factories |
A coal-fired power station in Yorkshire generates at 25 kV. The route to a domestic socket in London:
- 25 kV → 400 kV: Grid transformer at the power station substation steps voltage up for transmission
- 400 kV transmission: Overhead lines on pylons carry power to a Grid Supply Point (GSP) substation near London
- 400 kV → 132 kV: GSP transformer steps down for the regional distribution network
- 132 kV → 33 kV: Primary substation transformer for an area of London
- 33 kV → 11 kV: Secondary substation feeds local distribution cables
- 11 kV → 400/230 V: Street transformer (pad-mounted or pole-mounted) feeds the LV network
- 230 V to socket: Final connection via the service fuse and consumer unit
The entire journey may span 300+ km, with 6 voltage transformation stages, all achieved automatically with no moving parts except the alternator.
The distribution network delivers electricity from the high-voltage transmission system to individual consumers. It splits into three main sectors: industrial (large factories, process plants), commercial and domestic (offices, shops, housing), and rural (farms, isolated buildings). Wherever possible, distribution cables are laid underground — more expensive initially but avoids storm damage and is visually less intrusive. In rural areas, overhead lines on wooden poles remain cost-effective.
European voltage standardisation (Harmonisation Document HD 60038) specifies:
- Nominal LV supply: 400/230 V, +10%, −6% — so the supply voltage at the consumer's terminals may legitimately range from 216.2 V to 253 V on the single-phase
- Frequency: must not deviate more than ±1% from 50 Hz over a 24-hour period (i.e. 49.5–50.5 Hz); momentary deviations up to ±2% are permitted
These tolerances explain why equipment must be designed to work across a range, and why voltage measurements on a BS 7671 inspection must be interpreted against the ±10%/−6% limits rather than expecting exactly 230 V.
HV Primary Substation (132 kV / 33 kV)
Fed from the transmission system at 132 kV or 33 kV, the primary substation contains high-voltage switchgear, protection relays, and transformers that step down to 33 kV or 11 kV for the distribution network. Primary substations supply 11 kV distribution substations on a ring main system: each 11 kV substation is connected to two feeders, each coming from a different direction around the ring. If one section of cable faults and is isolated, the other direction still feeds the substation — providing security of supply (N-1 redundancy).
Automatic circuit breakers and distance protection relays provide graded overcurrent protection: faults are isolated by the nearest upstream device, leaving the rest of the network intact.
HV/LV Distribution Substation (11 kV → 400/230 V)
The 11 kV/400 V transformer substation is the final voltage transformation before the consumer. There are tens of thousands of these across the UK — typically in small brick buildings, underground vaults, or on poles. The standard transformer configuration is:
- Primary winding: connected in delta at 11 kV — no neutral connection required at this voltage
- Secondary winding: connected in star (wye), with the star point (neutral point) solidly earthed and connected to the neutral conductor
The star secondary gives two voltages simultaneously:
- 400 V between any two line conductors (line-to-line voltage, V_L = √3 × V_P)
- 230 V between any line conductor and neutral (phase-to-neutral voltage)
This confirms that the UK 230/400 V supply is a three-phase star system — not two separate supplies.
Typical supply configurations provided from the LV network:
| Voltage and configuration | Conductors | Typical application |
|---|---|---|
| 230 V single-phase | Line + Neutral + CPC | Domestic housing, small shops |
| 400 V three-phase four-wire | 3 Lines + Neutral + CPC | Industrial, commercial, agricultural — motors, large loads |
| 400 V three-phase three-wire | 3 Lines + CPC (no neutral) | Balanced three-phase motor circuits |
| 230 V single-phase (from 400 V supply) | 1 Line + Neutral + CPC | Commercial premises fed from three-phase intake |
Earthing Systems (TN-S, TN-C-S, TT)
BS 7671 and IEC 60364 classify earthing systems by how the supply neutral and exposed conductive parts are earthed. The system type determines the earth-fault loop path and the protection method required:
| System | Description | Earth-fault return path | Typical UK use |
|---|---|---|---|
| TN-S | Separate neutral (N) and protective earth (PE) conductors throughout; both connected to the earthed star point at the transformer | Via the PE conductor — low impedance | Older urban areas with separate sheath cables |
| TN-C-S (PME) | Combined neutral and earth (PEN) conductor in the supply cable; split into separate N and PE at the consumer's earth block | Via the PEN conductor — very low impedance | Most modern UK domestic and commercial supplies |
| TT | Supply transformer earthed via its own electrode; consumer equipment earthed via a separate electrode at the premises | Via the mass of earth — higher impedance | Rural areas, areas where PME is unavailable |
Earth-Fault Loop Impedance (ZS)
When a live conductor makes contact with exposed metalwork, a fault current flows around the earth-fault loop. For the protective device to disconnect quickly enough to prevent electric shock (BS 7671 regulation 411 specifies maximum disconnection times of 0.4 s for socket outlets and 5 s for fixed equipment on 230 V supplies), the loop impedance must be sufficiently low to allow a high enough fault current.
The fault current is: \(I_f = V_0 / Z_S\) where V₀ = nominal voltage to earth and Z_S = total loop impedance.
The complete earth-fault loop path in a TN-C-S system (starting at the point of fault and tracing the path of fault current) comprises:
- The circuit protective conductor (CPC) from the fault back to the consumer unit
- The main protective bonding conductors and main earthing terminal
- The supply company's PEN conductor (combined neutral/earth) back to the transformer
- The secondary winding of the distribution transformer
- The line conductor from the transformer to the fault
For a TT system, the fault current must also pass through two earth electrodes (consumer and supply transformer) and the mass of earth, which has much higher impedance. This is why TT systems almost always require an RCD as the primary protection device — the earth electrode resistance is too high for overcurrent devices to operate in time.
A 230 V TN-C-S socket outlet circuit has a measured Z_S of 0.8 Ω. The fault current under a line-to-earth fault:
\[I_f = \frac{V_0}{Z_S} = \frac{230}{0.8} = 287.5\text{ A}\]A 32 A Type B MCB must disconnect at 5× rated current (160 A) within 0.1 s. At 287.5 A (9× rated), it will trip instantaneously. Maximum permitted Z_S for a 32 A Type B MCB on 230 V (from BS 7671 Table 41.3): \(Z_S \leq 230/(5 \times 32) = 1.44\text{ Ω}\). The measured 0.8 Ω complies ✓.
Solar Photovoltaic (PV) Systems
Photovoltaic cells convert light energy directly into electrical energy using the photovoltaic effect: photons of light striking a semiconductor p-n junction dislodge electrons, creating a small dc voltage (typically 0.5–0.6 V per cell). Single cells produce very little power, so they are connected in series and parallel strings to form a solar module, and multiple modules form an array. A typical domestic roof-top installation (3–4 kWp) might have 10–12 modules.
Because PV generates dc, an inverter is required to convert to ac for use by the building and export to the grid. Grid-tied inverters must synchronise with the mains frequency and comply with Engineering Recommendation G98/G99 (connection requirements for the Distribution Network Operator).
Key electrical installation requirements for PV (BS 7671 Section 712):
- DC wiring must be run in separate containment (dc faults produce sustained arcs unlike ac)
- Isolation provisions at the inverter, array, and point of connection
- Labelling to warn of dual source of supply
- Fire fighter's switch for arrays on buildings
A 4 kWp array in the UK generates approximately 850–950 kWh per kWp per year. Annual generation estimate:
\[E = 4 \times 900 = 3600\text{ kWh/year}\]A typical UK household uses 3,500–4,000 kWh/year — so a well-sited 4 kWp array can generate approximately the household's annual consumption, though not necessarily when the consumption occurs (hence battery storage systems).
Solar Thermal Heating
Solar thermal panels use sunlight to heat a fluid (water-glycol mix) in roof-mounted collectors. The heated fluid passes through a heat exchanger in a hot water cylinder, preheating domestic hot water. Unlike PV, solar thermal does not generate electricity — it reduces the energy demand on the boiler or immersion heater. Flat-plate and evacuated-tube collectors are the two main types; evacuated tubes perform better in low sunlight conditions.
Wind Generation
Wind turbines convert the kinetic energy of moving air into rotational mechanical energy (via rotor blades and a gearbox), then to electrical energy in a generator. The power available from a wind turbine increases with the cube of wind speed — doubling wind speed gives eight times the power:
\[P = \frac{1}{2} \rho A v^3 \times C_p\]where ρ = air density (≈1.225 kg/m³), A = rotor swept area (m²), v = wind speed (m/s), C_p = power coefficient (max theoretical 0.593, practical 0.35–0.45).
- Micro wind (2–8 kW): Small turbines mounted on buildings or masts at domestic/commercial premises. Generation is variable and must be managed with an inverter and grid connection. Often less cost-effective than roof-mounted PV in the UK due to turbulence around buildings.
- Macro wind (onshore, 2–5 MW per turbine): Large wind farms on elevated sites with consistent wind. Feed directly into the distribution or transmission network.
- Offshore wind (5–15 MW per turbine): Larger turbines in stronger, more consistent winds. The UK has the world's largest installed offshore wind capacity. Output is collected at offshore substations and transmitted onshore via HVDC or HVAC cables.
A 3 MW offshore turbine has a rotor diameter of 90 m (swept area ≈ 6362 m²). At 12 m/s wind speed (rated wind speed), C_p ≈ 0.40:
\[P = 0.5 \times 1.225 \times 6362 \times 12^3 \times 0.40 = 0.5 \times 1.225 \times 6362 \times 1728 \times 0.40 \approx 2.69\text{ MW}\]At 8 m/s (⅔ of rated speed), power = \(2.69 \times (8/12)^3 = 2.69 \times 0.296 \approx 0.80\text{ MW}\) — showing how strongly wind speed affects output.
Wave Energy
Incoming ocean waves cause water to rise and fall in a chamber (Oscillating Water Column, or OWC); as water rises, air above it is compressed through a turbine; as water falls, air is drawn back through the same turbine. Wells turbines (which operate in either direction of airflow) are used. Wave energy is predictable hours ahead (unlike wind) and the UK coastline has significant wave resource. Commercial-scale deployment is still in development.
Micro-Hydro Generation
Systems up to about 100 kW use run-of-river flows — no large dam required. A weir diverts water into a pipe (penstock) leading to a turbine. Used for remote farms and small communities where grid connection is impractical. Output power depends on head and flow rate (same formula as hydro above). Micro-hydro systems have very high availability (running >8,000 hours/year) compared to wind (~3,000 hours) or solar (~1,200 hours in the UK).
Ground Source Heat Pumps (GSHP)
Below about 1 m depth, ground temperature remains stable at approximately 8–12°C year-round in the UK. A GSHP extracts this low-grade heat using a refrigeration cycle: a fluid circulating in buried pipework (ground loop) absorbs heat from the ground; the heat pump compresses the refrigerant to raise its temperature significantly; the high-grade heat is transferred to the building's heating system.
The key metric is the Coefficient of Performance (COP): useful heat output ÷ electrical energy input. A COP of 3.5 means 3.5 kWh of heat is delivered for every 1 kWh of electricity consumed — effectively amplifying the electrical input by the COP factor. This is why GSHPs are considered a low-carbon heating technology even though they consume electricity.
A house needs 12,000 kWh/year of heat. GSHP with COP = 3.5 vs gas boiler (90% efficient):
\[\text{GSHP electricity} = \frac{12000}{3.5} = 3429\text{ kWh/year} \qquad \text{Gas} = \frac{12000}{0.9} = 13333\text{ kWh/year}\]At UK electricity 28 p/kWh and gas 7 p/kWh: GSHP cost = £960/yr vs gas = £933/yr — similar cost but with significantly lower carbon emissions (depending on grid carbon intensity).
Combined Heat and Power (CHP)
A conventional power station wastes roughly 60–65% of its fuel energy as heat in cooling towers. A CHP plant (also called cogeneration) captures this waste heat and uses it for space heating, hot water, or industrial processes, raising overall system efficiency from ~35% to 70–85%.
- Large CHP (hospitals, universities, industrial sites): gas engine or gas turbine drives an alternator; cooling water and exhaust heat supply the building's heating system. Typical: 100 kW–10 MW electrical output.
- Micro CHP (domestic): a small gas engine (Stirling or reciprocating) produces ~1 kW of electricity while heating the house. The unit runs when heat is required, with surplus electricity exported to the grid under a feed-in arrangement.
A 15 kWe/26 kWth CHP unit runs 8,000 hours/year. Annual electrical generation: \(15 \times 8000 = 120\,000\text{ kWh}\). Saving at 28p/kWh (displacing grid electricity): £33,600/year. Additional saving from displaced gas for heat — total payback typically 3–5 years on systems of this scale.
Batteries and Cells
Batteries are the dominant source of portable dc power and increasingly important for stationary energy storage (grid-scale battery storage, building-level storage, UPS systems). A cell is the basic electrochemical unit; a battery is one or more cells connected together.
Each cell consists of two electrodes (positive and negative plates) separated by an electrolyte. A chemical reaction at each electrode drives conventional current from negative to positive internally (from positive to negative externally), establishing the cell's emf.
| Cell type | Positive electrode | Negative electrode | Electrolyte | Cell voltage | Rechargeable? |
|---|---|---|---|---|---|
| Voltaic (simple) | Copper | Zinc | Dilute sulphuric acid | 0.5–1.0 V | No |
| Leclanché (wet) | Carbon | Zinc | Ammonium chloride | 1.4 V | No |
| Zinc-carbon (dry) | Carbon | Zinc | Ammonium chloride paste | 1.5 V | No (primary) |
| Alkaline (dry) | Manganese dioxide | Zinc powder | Potassium hydroxide | 1.5 V | No (primary) |
| Lead-acid | Lead dioxide (PbO₂) | Lead (Pb) | Dilute sulphuric acid | 2.0 V/cell | Yes (secondary) |
| Nickel-metal hydride (NiMH) | Nickel hydroxide | Metal hydride | Potassium hydroxide | 1.2 V/cell | Yes (secondary) |
| Lithium-ion (Li-ion) | Lithium cobalt oxide | Graphite | Lithium salt in solvent | 3.6–3.7 V/cell | Yes (secondary) |
Cells in Series
When cells are connected positive-to-negative in a chain, their individual voltages add together — the total emf is the sum of all cell emfs, while the capacity (Ah) remains the same as one cell. This is used to build higher-voltage battery packs:
- 9 V PP3 battery = six 1.5 V alkaline cells in series
- 12 V car battery = six 2.0 V lead-acid cells in series
- 400 V EV traction battery = approximately 108 × 3.7 V Li-ion cells in series
Cells in Parallel
When cells are connected positive-to-positive and negative-to-negative, the voltage is the same as a single cell but the capacity (Ah) adds together — each cell shares the load current. Used where long operating time (high capacity) is required without increasing voltage:
A fire alarm panel needs a 12 V, 21 Ah battery backup. Six 2 V lead-acid cells in series = 12 V. But each cell has only 7 Ah. Connecting three such series strings in parallel: total capacity = \(3 \times 7 = 21\text{ Ah}\) at 12 V.
Primary vs Secondary Cells
| Type | Description | When to use | Electrical applications |
|---|---|---|---|
| Primary | Single-use — chemical reaction not reversible. Cell is discarded when exhausted. | Low-drain, infrequent-use devices; where recharging is impractical | Smoke detector batteries, multimeter batteries, emergency lighting test buttons, remote controls |
| Secondary | Rechargeable — passing a reverse current restores the chemicals. Hundreds to thousands of charge/discharge cycles possible. | Frequent use or standby applications requiring repeated cycling | UPS batteries (lead-acid), fire alarm central battery systems (sealed lead-acid), EVs (Li-ion), portable tools (Li-ion) |
Battery Capacity and Sizing
Battery capacity is measured in ampere-hours (Ah): a battery rated at 100 Ah can theoretically deliver 10 A for 10 hours, or 1 A for 100 hours. In practice, the usable capacity depends on the discharge rate — higher discharge rates reduce effective capacity (Peukert's Law). Small batteries (phones, tools) are rated in milliampere-hours (mAh).
A server room UPS must maintain 5 kVA (unity power factor = 5 kW) for 30 minutes from a 48 V dc battery bank. Current required:
\[I = \frac{P}{V} = \frac{5000}{48} = 104.2\text{ A}\] \[\text{Capacity} = I \times t = 104.2 \times 0.5\text{ h} = 52.1\text{ Ah}\]Allowing for 80% usable capacity (to avoid deep discharge): \(52.1 / 0.8 = 65\text{ Ah}\). Specify 65 Ah minimum rated battery bank at 48 V.
BS 5266 requires emergency lighting to operate for at least 3 hours. A maintained luminaire (lamp always on) with a 10 W LED is powered from a 3.6 V Li-ion cell. Current draw: \(I = 10/3.6 = 2.78\text{ A}\). Required capacity: \(2.78 \times 3 = 8.33\text{ Ah}\) minimum.
Purpose and Importance of Transformers
A transformer converts an ac supply at one voltage to an ac supply at a different voltage (higher or lower), with no moving parts and near-perfect efficiency. The transformer is perhaps the most important single piece of electrical equipment in modern society: without it, long-distance power transmission at economical cost would be impossible, and all the voltage levels from 400 kV down to 230 V would require separate generation at each level.
Transformers also provide electrical isolation between primary and secondary — the secondary circuit has no direct metallic connection to the primary. This is exploited for safety (shaver sockets, medical equipment supplies, SELV circuits) and for eliminating unwanted earth currents between systems.
Key properties:
- Only works on ac (dc produces no changing flux, so no induced emf)
- Efficiency typically 95–99.5% (losses are real but small)
- No moving parts — very reliable, long service life (25–40 years for power transformers)
- Output voltage and current are both determined solely by the turns ratio
Construction
Transformer construction varies with power rating and application:
| Type | Construction | Cooling | Application |
|---|---|---|---|
| Small signal/audio | Laminated silicon-steel E-I or toroidal core; bobbin-wound coils | Natural air convection | Electronic equipment, audio amplifiers |
| Small power (up to ~5 kVA) | Laminated silicon-steel core; varnish-impregnated windings | Natural air cooling (AN) | Control transformers, bell transformers, ELV supplies |
| Medium distribution (5–2500 kVA) | Wound core or stacked lamination core; windings in oil-filled tank | Oil natural (ONAN) or oil forced (ONAF) | 11 kV/LV substations, industrial supply transformers |
| Large power (>2500 kVA) | Three-phase core; windings in oil tank with radiator banks | Oil with forced air cooling (OFAF); oil pumped through external coolers | Grid substations, power stations |
| Dry-type (cast resin) | Resin-encapsulated windings; no oil | Forced air or natural convection | Indoor installations (buildings, wind turbines), fire-risk environments |
All transformer cores are made from laminated silicon steel — thin sheets (typically 0.3–0.5 mm) insulated from each other. Lamination reduces eddy current losses; silicon steel has high permeability and low hysteresis loss. Toroidal cores (ring-shaped) are more efficient than stacked E-I lamination cores and are used in premium quality equipment.
Operating Principle
The transformer exploits mutual induction (Chapter 5, Part 4): alternating current in the primary winding creates an alternating magnetic flux in the core; this continuously changing flux threads through the secondary winding, inducing an alternating emf. Both windings are on the same core, so they share the same flux.
Since flux is common to both windings, the volts per turn is the same for both windings (a fundamental consequence of Faraday's Law). This gives the basic transformer equation:
where V = voltage, N = number of turns, I = current, P = primary, S = secondary. Note that current ratio is the inverse of the turns ratio — if voltage is stepped up, current is stepped down, and vice versa (power is conserved).
In an ideal transformer, input power = output power: \(V_P I_P = V_S I_S\). Rearranging: \(I_S / I_P = V_P / V_S = N_P / N_S\). A step-down transformer (V_P > V_S) therefore produces a larger secondary current than primary current — energy is conserved but delivered at lower voltage and higher current.
Step-Down Transformers
A step-down transformer has more primary turns than secondary turns (N_P > N_S), so V_S < V_P. The secondary current is proportionally higher. Used for: 11 kV → 400/230 V distribution substations, 230 V → 24 V control transformers, 230 V → 12 V bell transformers, 230 V → 55 V site safety transformers.
\[V_S = V_P \times \frac{N_S}{N_P}\]760 primary turns, 360 secondary turns, 230 V input, primary current 5 A:
\[V_S = 230 \times \frac{360}{760} = 109\text{ V}\] \[I_S = I_P \times \frac{N_P}{N_S} = 5 \times \frac{760}{360} = 10.56\text{ A}\]BS 7671 Regulation 411.8 requires 110 V centre-tapped transformers for portable tools on construction sites. Primary: 230 V. Secondary: 110 V centre-tapped to earth (55 V to earth on each side). Turns ratio needed:
\[\frac{N_P}{N_S} = \frac{V_P}{V_S} = \frac{230}{110} = 2.09:1\]If the secondary has 100 turns, the primary needs \(100 \times 2.09 = 209\) turns. The centre-tap means the maximum voltage to earth is only 55 V — significantly reducing shock risk if a tool is faulted.
Step-Up Transformers
A step-up transformer has more secondary turns than primary turns (N_S > N_P), raising voltage. Used for: power station generator output → grid transmission (25 kV → 400 kV), neon sign transformers, high-voltage test equipment.
Primary: 50 turns, 24 V input. Turns ratio 25:1 step-up (N_S/N_P = 25):
\[N_S = 50 \times 25 = 1250\text{ turns}\] \[V_S = 24 \times \frac{1250}{50} = 600\text{ V}\]If primary current is 10 A: \(I_S = 10 \times (50/1250) = 0.4\text{ A}\) — higher voltage, lower current (power ≈ the same).
Volts per Turn — A Useful Design Tool
Since both windings share the same flux, the volts-per-turn is identical in both windings. This provides a quick way to find secondary turns without using the full ratio equation:
\[\frac{V_P}{N_P} = \frac{V_S}{N_S} \quad \Rightarrow \quad N_S = N_P \times \frac{V_S}{V_P}\]A 230 V / 50 V transformer has 110 primary turns. Volts per turn: \(230/110 = 2.09\text{ V/turn}\). Secondary turns for 50 V output: \(50/2.09 = 23.9 ≈ 24\text{ turns}\).
The same transformer could have a second secondary winding for 12 V: \(12/2.09 = 5.74 ≈ 6\text{ turns}\). All secondaries share the same core and the same 2.09 V/turn — a key principle in multi-output power supplies.
Transformer kVA Rating
Transformers are rated in VA (volt-amperes), kVA, or MVA — not watts. This is because transformer losses depend on voltage (iron losses, constant) and current (copper losses ∝ I²), not on the power factor of the load. The kVA rating is the maximum apparent power the transformer can deliver continuously without overheating:
\[\text{kVA} = \frac{V \times I}{1000} \quad (\text{single-phase}) \qquad \text{kVA} = \frac{\sqrt{3} \times V_L \times I_L}{1000} \quad (\text{three-phase})\]A 230 V single-phase supply must feed a 100 A load. Required kVA:
\[\text{kVA} = \frac{230 \times 100}{1000} = 23\text{ kVA}\]Standard transformer sizes: 15, 30, 45, 75, 100, 150, 200, 315, 500 kVA. Select the next standard size above 23 kVA: 30 kVA. The spare capacity (30 − 23 = 7 kVA) allows for future load growth and derating for temperature.
A 400 V three-phase transformer supplies 144 A line current. kVA rating:
\[\text{kVA} = \frac{\sqrt{3} \times 400 \times 144}{1000} = \frac{1.732 \times 400 \times 144}{1000} = \frac{99.7 \times 10^3}{1000} = 99.7\text{ kVA}\]Select the next standard size: 100 kVA.
A step-down transformer with 1,000 primary turns and 250 secondary turns is supplied at 250 V ac. The secondary voltage:
\[V_S = 250 \times \frac{250}{1000} = 250 \times 0.25 = 62.5\text{ V}\]This confirms the self-assessment answer of 62.5 V ✓.
Isolating Transformers (1:1 Ratio)
An isolating transformer has equal numbers of turns on primary and secondary (turns ratio 1:1), so input and output voltages are equal. Its purpose is not to change voltage but to provide electrical isolation — the secondary winding has no conductive connection to the primary, to earth, or to any other circuit. A person touching the live terminal of the secondary does not complete an earth-return path and therefore receives no shock — the current cannot flow because the circuit is not complete.
Applications of isolating transformers:
- Shaver sockets (BS EN 61558-2-5): Mandatory in bathrooms (BS 7671 Section 701). The 230 V shaver socket output is isolated from mains earth, preventing shock if the user touches a live part while standing in water or touching a grounded surface. Rated typically 20–25 VA.
- Medical equipment (BS EN 61558-2-15): In operating theatres and cardiac care units, even a few microamps through the heart can cause fibrillation. Isolated supplies prevent fault currents through the patient.
- IT and data equipment: Isolation transformers prevent earth loop currents that cause hum in audio equipment and data corruption in sensitive electronics.
- Maintenance work on live equipment: Portable isolation transformers allow maintenance with reduced shock risk.
A standard 230 V mains socket in a bathroom: person touches live terminal + is grounded → 230/body_resistance A fault current flows → potentially lethal. With an isolating transformer: person touches secondary live terminal → no earth return path via the transformer (secondary is floating) → no current flows → no shock. The secondary is effectively a "floating" supply with no reference to earth potential.
Auto-Transformers
A conventional (double-wound) transformer has two completely separate windings sharing a common core. An auto-transformer has a single winding — part of which serves as the primary and part (or all, in a step-up) as the secondary. The two "windings" share a common section, called the common winding.
Advantages:
- The common winding carries only the difference current (I_P − I_S or I_S − I_P), so it can be wound in smaller wire — significant cost and weight saving for large transformers where the ratio is close to 1:1
- Better voltage regulation (lower leakage reactance)
- Lower losses for the same kVA
Disadvantages:
- No isolation — primary and secondary share a common conductor, so a fault on the primary can place full primary voltage on the secondary circuit
- Cannot be used where isolation is required for safety (e.g. not suitable as a bathroom shaver socket)
- If the common (neutral) terminal disconnects, the full input voltage appears across the output — dangerous if the step-down ratio is large
Applications: Motor star-delta soft starters and autotransformer starters (stepping voltage down for reduced-voltage starting), voltage regulators, variac (variable autotransformer) laboratory supplies, railway traction supplies.
A large induction motor requires reduced-voltage starting to limit starting current. An autotransformer starter with a 65% voltage tap reduces the motor terminal voltage to:
\[V_\text{motor} = 0.65 \times 400 = 260\text{ V}\]The starting current is proportional to V²: \(I_\text{start} \propto (0.65)^2 = 0.4225\) — so the motor starting current is reduced to 42.25% of its DOL starting current, and the supply current is reduced even further (by the transformer ratio squared): \(0.4225 \times 0.65 = 27.5\%\) of DOL. After the motor accelerates, the starter switches to full voltage.
Three-Phase Transformers
All transmission and most distribution transformers are three-phase. The windings can be connected in either star (Y) or delta (Δ) on each side, giving four possible configurations: Yy, Yd, Dy, Dd. The most common for 11 kV/LV distribution is Dyn11 — delta primary, star secondary with neutral, 30° phase shift between primary and secondary (the "11" refers to the clock-face hour position of the secondary voltage vector relative to the primary).
The basic transformer turns ratio still applies to each winding pair. However, the terminal voltage depends on the connection:
- Delta winding: line voltage = phase (winding) voltage
- Star winding: line voltage = √3 × phase (winding) voltage
A three-phase transformer has a turns ratio of 13:1. Primary is delta-connected at 22.52 kV. Secondary is star-connected.
Each primary winding (delta): V_P = 22.52 kV (line voltage = winding voltage in delta).
Each secondary winding voltage: \(V_{\text{S winding}} = 22520 / 13 = 1732\text{ V}\)
Star secondary line voltage: \(V_L = \sqrt{3} \times 1732 = 1.732 \times 1732 = 3000\text{ V} = 3\text{ kV}\)
Primary delta at 11 kV; secondary star, neutral earthed, rated 400/230 V. The transformation ratio is:
\[\frac{V_{\text{P winding}}}{V_{\text{S winding}}} = \frac{11000}{230} = 47.8:1\]Primary winding voltage (delta) = line voltage = 11,000 V. Secondary winding voltage = 11,000 / 47.8 = 230 V. Secondary line voltage = 230 × √3 = 398 ≈ 400 V ✓.
Transformer Efficiency and Losses
A real transformer is not 100% efficient because it has two categories of internal losses:
| Loss type | Cause | How it varies with load | How to reduce |
|---|---|---|---|
| Iron losses (core losses) | Eddy currents (circulating currents in the laminations) and hysteresis (energy to repeatedly magnetise/demagnetise the core) | Essentially constant — depends on applied voltage and frequency, not on load current | Thinner laminations; grain-oriented silicon steel; amorphous metal cores (very low loss) |
| Copper losses (I²R losses) | Resistive heating in the primary and secondary winding conductors | Vary with load squared — double the load current → four times the copper losses | Larger conductor cross-section; lower-resistance winding wire |
Total input = Output + Iron losses + Copper losses
\[\eta = \frac{P_\text{out}}{P_\text{in}} = \frac{P_\text{out}}{P_\text{out} + P_\text{iron} + P_\text{copper}} \times 100\%\]Because iron losses are constant and copper losses vary with load, efficiency is not constant — a transformer is most efficient at the load where iron losses = copper losses (this is the design point for distribution transformers, typically around 50–75% of rated load since transformers are rarely fully loaded).
A transformer takes 16 A at 180 V on the primary and supplies 36 A at 75.2 V on the secondary:
\[P_\text{in} = 16 \times 180 = 2880\text{ W} \qquad P_\text{out} = 36 \times 75.2 = 2707.2\text{ W}\] \[\eta = \frac{2707.2}{2880} = 0.94 = 94\%\]Losses = 2880 − 2707.2 = 172.8 W — this heat must be dissipated by the transformer cooling system.
A transformer has copper losses of 15 kW and iron losses of 10 kW at full load. Full-load output is 900 kW:
\[\text{Total losses} = 15 + 10 = 25\text{ kW}\] \[\eta = \frac{900}{900 + 25} \times 100 = \frac{900}{925} \times 100 = 97.3\%\]Full-load output = 750 kW, total losses = 20 kW:
\[\eta = \frac{750}{750 + 20} \times 100 = \frac{750}{770} \times 100 = 97.4\%\]This confirms the self-assessment answer of 97.4% ✓.
A 200 kVA distribution transformer has iron losses of 400 W and copper losses of 2 kW at full load. It operates at 60% load for 8,760 hours per year. Iron losses are constant; copper losses at 60% load = \(2000 \times 0.6^2 = 720\text{ W}\). Annual energy lost:
\[E_\text{losses} = (400 + 720) \times 8760 = 1120 \times 8760 = 9.81\text{ MWh/year}\]At 28 p/kWh: cost of losses = \(9810 \times 0.28 = £2747/\text{year}\). Over a 30-year transformer life, cumulative loss cost exceeds £82,000 — often exceeding the transformer's capital cost and justifying investment in a more efficient unit.
Instrument Transformers
Instrument transformers extend the range of measuring instruments to high voltages and currents that would be unsafe or impractical to measure directly. They also isolate the low-voltage measurement circuit from the high-voltage primary system.
Current Transformers (CTs)
A CT has a ring-shaped (toroidal) core with many turns of the secondary winding wound around it. The primary "winding" is a single conductor — a cable or busbar — passing through the hole in the ring. The primary current creates a flux in the toroidal core; this induces a proportional but scaled-down current in the secondary.
CTs are specified by their current ratio (e.g. 100:5 A — 100 A primary, 5 A secondary) and accuracy class (class 0.5 for billing meters, class 5P for protection relays). Secondary rated current is usually 1 A or 5 A — standardised to allow interchangeable metering equipment.
Critical safety rule: NEVER open-circuit a CT secondary while current flows in the primary. With no secondary load, the flux is not limited by the secondary ampere-turns reaction — it rises to saturation on every half-cycle, inducing very high voltage spikes on the secondary terminals (potentially thousands of volts) that can destroy the CT insulation and cause a lethal flashover. Always short-circuit CT secondary terminals before disconnecting instruments.
A 400:5 A CT (ratio 80:1) is installed on a 11 kV feeder carrying 200 A. Secondary current:
\[I_S = \frac{I_P}{\text{ratio}} = \frac{200}{80} = 2.5\text{ A}\]The 2.5 A drives the ammeter on the operator's panel at a safe 11 V (rather than 11 kV) — a direct reading that the operator can interpret without any danger.
Voltage Transformers (VTs) / Potential Transformers (PTs)
A VT is a conventional double-wound transformer with accurately known turns ratio, used to step down high voltages to a safe measurement level. Standard secondary voltages are 110 V or 63.5 V. Like a normal transformer, VT ratio = turns ratio.
Critical safety rule: NEVER short-circuit VT secondary terminals — a short circuit appears as a fault on the primary side, causing extremely high primary currents and potentially destroying the VT. A fuse on the VT primary provides protection.
A 33 kV/110 V VT (ratio 300:1) is used to measure the 33 kV busbar voltage. The voltmeter connected to the secondary reads 109.5 V. Actual busbar voltage:
\[V_P = V_S \times \text{ratio} = 109.5 \times 300 = 32\,850\text{ V} = 32.85\text{ kV}\]The 32.85 kV reading indicates the busbar is within the ±6% tolerance of the 33 kV nominal.
- CT secondary: never open-circuit — always short first before disconnecting
- VT secondary: never short-circuit — fuses must be fitted on the primary
- Both types provide isolation — never defeat this by connecting secondary ground to primary ground outside the designated earth point
🃏 Chapter 6 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Covers power generation, the National Grid, distribution, earthing, renewables, batteries and transformers.
📝 Chapter 6 — Multiple Choice Questions
Select your answer and click Submit. Two attempts per question. First-attempt score shown at the end.
1A CCGT (Combined Cycle Gas Turbine) power station has an overall efficiency of approximately:
2Why is electricity transmitted at very high voltage (400 kV) rather than at low voltage?
3In the UK voltage hierarchy, at what voltage does the 11 kV/LV distribution transformer deliver power to final consumers?
4The UK supply voltage tolerance is 230 V +10% / −6%. What is the minimum permitted supply voltage at the consumer's terminals?
5Which earthing system uses a combined neutral and protective earth (PEN) conductor in the supply cable, split at the consumer's earth block?
6A wind turbine doubles its wind speed. By what factor does its power output change?
7Six 2 V lead-acid cells are connected in series. What is the terminal voltage and what happens to the capacity (Ah)?
8A transformer has 1000 primary turns and 250 secondary turns supplied at 250 V. What is the secondary voltage?
9Why are transformers rated in kVA rather than kW?
10A transformer has copper losses of 15 kW and iron losses of 10 kW at full load. Full-load output is 900 kW. What is the efficiency?
11Why is an isolating transformer used in a bathroom shaver socket (BS 7671 Section 701)?
12What is the critical safety rule for a current transformer (CT) secondary circuit?
Sends your Chapter 6 MCQ score summary from your registered email to your tutor.
AC Circuits and Systems
- Explain the relationship between resistance, inductance, capacitance and impedance in ac circuits
- Calculate unknown values of R, XL, XC and Z; explain kW, kVA, kVAr and power factor
- Explain power factor correction and ways of improving power factor
- Explain load balancing and determine the neutral current in three-phase supply
- Calculate values of voltage and current in star and delta systems
AC circuits differ fundamentally from dc circuits because two additional components — inductors and capacitors — oppose current flow in ways that depend on frequency. Unlike resistance, this opposition does not dissipate energy as heat. This Part consolidates the key formulae before applying them to full circuit calculations.
Resistance (R)
Resistance is the same in ac as in dc: it opposes current flow and dissipates energy as heat (P = I²R). In an ac circuit, the voltage across a pure resistor is always exactly in phase with the current through it. Resistance does not change with frequency.
Inductive Reactance (XL)
An inductor (coil) stores energy in its magnetic field and opposes changes in current (Lenz's Law). In an ac circuit the current is continuously changing, so the inductor continuously opposes it. This opposition is the inductive reactance XL, measured in ohms:
\[X_L = 2\pi f L\]where f = frequency (Hz) and L = inductance (H). XL increases with frequency — a choke that offers 10 Ω at 50 Hz offers 20 Ω at 100 Hz. The voltage across a pure inductor leads the current by 90°. No energy is permanently dissipated — it is stored and returned each half-cycle.
Capacitive Reactance (XC)
A capacitor stores energy in its electric field. It opposes changes in voltage. In ac it repeatedly charges and discharges, blocking dc entirely but allowing ac to pass (the higher the frequency, the more easily it passes). The opposition is capacitive reactance XC, measured in ohms:
\[X_C = \frac{1}{2\pi f C}\]where C = capacitance (F). XC decreases with frequency (opposite to XL) — a capacitor becomes a better conductor at higher frequencies. The current through a pure capacitor leads the voltage by 90°. No energy is permanently dissipated.
Impedance (Z)
Impedance is the total opposition to ac current flow in a circuit combining resistance and reactance. Unlike resistance and reactance, R and X cannot simply be added — they are 90° apart in phase and must be added as vectors (using Pythagoras):
| Quantity | Symbol | Unit | Formula | Phase relationship | Dissipates energy? |
|---|---|---|---|---|---|
| Resistance | R | Ω | V/I (Ohm's Law) | V in phase with I | Yes (heat) |
| Inductive reactance | XL | Ω | 2πfL | V leads I by 90° | No (stores/returns) |
| Capacitive reactance | XC | Ω | 1/(2πfC) | I leads V by 90° | No (stores/returns) |
| Impedance | Z | Ω | √(R² + X²) | Depends on R and X values | Partially (R component only) |
A motor winding has inductance L = 0.1 H. At 50 Hz:
\[X_L = 2\pi \times 50 \times 0.1 = 31.4\text{ Ω}\]At 60 Hz (if the motor were used on a US supply): \(X_L = 2\pi \times 60 \times 0.1 = 37.7\text{ Ω}\) — 20% higher, which changes the motor's operating current and torque characteristics.
A power factor correction capacitor must have XC = 50 Ω at 50 Hz. Required capacitance:
\[C = \frac{10^6}{2\pi \times 50 \times 50} = \frac{10^6}{15708} = 63.7\text{ \mu F}\]A fluorescent lamp ballast has XL = 100 Ω at 50 Hz. Its inductance:
\[L = \frac{X_L}{2\pi f} = \frac{100}{2\pi \times 50} = \frac{100}{314} = 0.318\text{ H} = 318\text{ mH}\]Phasor Diagram Representation
A phasor diagram (vector diagram) represents ac voltages and currents as arrows (vectors). The length of the arrow represents the magnitude (rms value), and the angle of the arrow represents the phase angle relative to a chosen reference. Phasors are far more convenient than waveform diagrams because they allow the relative phase between quantities to be seen at a glance, and they allow out-of-phase quantities to be combined correctly using vector addition.
The key rule: ac quantities that are out of phase cannot be added or subtracted arithmetically. Adding 60 V across a resistor to 80 V across an inductor does not give 140 V — it gives 100 V (if the two voltages are 90° apart). This is one of the most important distinctions between dc and ac circuit analysis.
Choosing the Reference Phasor
- Series circuits: Current I is the reference phasor (drawn horizontally to the right at 0°). This is because current is the same throughout a series circuit — it is the common quantity. Voltage phasors for each component are then drawn at the appropriate angle to I.
- Parallel circuits: Voltage V is the reference phasor (drawn horizontally). Voltage is the same across all parallel branches — it is the common quantity. Branch current phasors are drawn at the appropriate angle to V.
Phase Relationships — Pure Components
| Component | Phase relationship | Phasor direction (V relative to I) | Memory aid |
|---|---|---|---|
| Pure resistor | V and I in phase (0°) | V drawn at 0° (same direction as I) | Resistor = no phase shift |
| Pure inductor | V leads I by 90°; or I lags V by 90° | VL drawn at +90° (vertically up) | CIVIL: In a Capacitor, I leads V; in an L (inductor), V leads I. |
| Pure capacitor | I leads V by 90°; or V lags I by 90° | VC drawn at −90° (vertically down) |
Physical Explanation of Phase Shifts
Why does current lag voltage in an inductor? When ac voltage is applied to an inductor, the inductor's self-induced emf (Lenz's Law) resists the build-up of current. The current cannot reach its peak instantaneously — it lags behind the applied voltage by 90° in a perfect (zero resistance) inductor. In practice, real inductors have some resistance, so the lag is between 0° and 90°.
Why does current lead voltage in a capacitor? Current flows into a capacitor to charge it before the voltage builds up. The current peaks (when the capacitor is most rapidly charging) before the voltage peaks (when the capacitor is fully charged). So current leads voltage by 90° in a perfect capacitor.
Phasor Addition — Voltage Triangle
When two phasors are at right angles (90°) to each other, Pythagoras applies directly. In an R-L series circuit with reference phasor I at 0°: VR is horizontal (in phase with I) and VL is vertical (90° ahead of I). The supply voltage VT is the hypotenuse of the right-angled voltage triangle:
\[V_T = \sqrt{V_R^2 + V_L^2}\]The phase angle θ (by which I lags VT) is found from:
\[\tan\theta = \frac{V_L}{V_R} \qquad \cos\theta = \frac{V_R}{V_T} \qquad \sin\theta = \frac{V_L}{V_T}\]VR = 40 V, VL = 60 V in a series R-L circuit:
\[V_T = \sqrt{40^2 + 60^2} = \sqrt{1600 + 3600} = \sqrt{5200} = 72.1\text{ V}\] \[\theta = \tan^{-1}\left(\frac{60}{40}\right) = \tan^{-1}(1.5) = 56.3°\]Current lags supply voltage by 56.3°. Power factor = cos 56.3° = 0.555 lagging.
VR = 90 V, VC = 120 V in a series R-C circuit. VC is drawn vertically downward (−90° from I reference). Supply voltage:
\[V_T = \sqrt{90^2 + 120^2} = \sqrt{8100 + 14400} = \sqrt{22500} = 150\text{ V}\] \[\theta = \tan^{-1}\left(\frac{120}{90}\right) = \tan^{-1}(1.33) = 53.1°\]Current leads supply voltage by 53.1° (capacitive circuit). Power factor = cos 53.1° = 0.6 leading.
A student measures 60 V across a resistor and 80 V across a series inductor. They add them: "the supply must be 140 V." But the supply voltage is:
\[V_T = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100\text{ V}\]The actual supply is 100 V — 40 V less than the student's arithmetic. This is why individual component voltages in an ac circuit can appear to exceed the supply voltage and why voltmeters must be used rather than arithmetic addition.
The Impedance Triangle
The impedance triangle is the key tool for all ac circuit calculations. Because R and X are 90° apart in phase, they form the two perpendicular sides of a right-angled triangle with Z as the hypotenuse. The angle θ at the base is the phase angle between supply voltage and current:
\[Z = \sqrt{R^2 + X^2} \qquad \cos\theta = \frac{R}{Z} \qquad \tan\theta = \frac{X}{R}\]Every ac circuit calculation follows the same systematic method: (1) find the reactances; (2) build the impedance triangle; (3) find Z; (4) use I = V/Z to find current; (5) find individual voltages from V = I × component.
R-L Series Circuit
In an R-L series circuit, resistance R and inductive reactance XL both oppose current — but R dissipates energy while XL only delays it. The phasor diagram shows VR horizontal (in phase with I) and VL vertical (leading I by 90°). The impedance is:
\[Z = \sqrt{R^2 + X_L^2}\]This confirms the self-assessment answer of 15.8 Ω ✓.
\[I = \frac{230}{15.81} = 14.55\text{ A} \qquad \theta = \tan^{-1}(15/5) = 71.6° \qquad \text{PF} = \cos 71.6° = 0.316\text{ lagging}\]A coil on a 230 V, 50 Hz supply draws 5 A. The resistance of the coil is measured at 20 Ω with a dc ohmmeter. Find the inductance.
\[Z = \frac{V}{I} = \frac{230}{5} = 46\text{ Ω}\] \[X_L = \sqrt{Z^2 - R^2} = \sqrt{46^2 - 20^2} = \sqrt{2116 - 400} = \sqrt{1716} = 41.4\text{ Ω}\] \[L = \frac{X_L}{2\pi f} = \frac{41.4}{314.2} = 0.132\text{ H} = 132\text{ mH}\]The impedance triangle also directly gives: \(Z = V_T/I\), \(X_L = V_L/I\), \(R = V_R/I\), and \(I = V_T/Z\).
R-C Series Circuit
In an R-C series circuit, VC lags I by 90° (drawn vertically downward on the phasor diagram). The impedance triangle has XC pointing downward but the Pythagorean magnitude is the same:
\[Z = \sqrt{R^2 + X_C^2}\]An R-C circuit on 230 V, 50 Hz has R = 30 Ω and draws 5 A. Find the capacitance.
\[Z = \frac{230}{5} = 46\text{ Ω} \qquad X_C = \sqrt{46^2 - 30^2} = \sqrt{2116 - 900} = \sqrt{1216} = 34.9\text{ Ω}\] \[C = \frac{10^6}{2\pi \times 50 \times 34.9} = \frac{10^6}{10965} = 91.2\text{ \mu F}\]R-L-C Series Circuit
With all three components in series, XL and XC act in opposition — XL pulls the phasor upward, XC pulls it downward. The net reactance is their difference. The larger one determines whether the circuit is overall inductive or capacitive:
- If XL > XC: net reactance = XL − XC (inductive, current lags V)
- If XC > XL: net reactance = XC − XL (capacitive, current leads V)
- If XL = XC: net reactance = 0, Z = R (purely resistive — this is resonance)
At resonance the impedance is minimum (= R only), current is maximum, and the power factor is unity. This is used in tuned circuits, radio receivers, and power factor correction.
Series circuit: R = 10 Ω, L = 50 mH, C = 100 µF on 230 V, 50 Hz supply.
\[X_L = 2\pi \times 50 \times 0.05 = 15.71\text{ Ω}\] \[X_C = \frac{10^6}{2\pi \times 50 \times 100} = 31.83\text{ Ω}\] \[X_\text{net} = X_C - X_L = 31.83 - 15.71 = 16.12\text{ Ω (capacitive)}\] \[Z = \sqrt{10^2 + 16.12^2} = \sqrt{100 + 259.9} = \sqrt{359.9} = 19.0\text{ Ω}\] \[I = \frac{230}{19.0} = 12.1\text{ A} \qquad \text{PF} = \frac{10}{19} = 0.526\text{ leading}\]At resonance, XL = XC: \(2\pi f L = \frac{1}{2\pi f C}\). Solving for f:
\[f_0 = \frac{1}{2\pi\sqrt{LC}}\]For L = 50 mH and C = 100 µF:
\[f_0 = \frac{1}{2\pi\sqrt{0.05 \times 100 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{5 \times 10^{-6}}} = \frac{1}{2\pi \times 2.236 \times 10^{-3}} = \frac{1}{0.01405} = 71.2\text{ Hz}\]Above 71.2 Hz the circuit is inductive; below it is capacitive. This resonant frequency concept underlies the design of filter circuits in motor drives and power electronics.
The Three Types of Power in AC Circuits
In a dc circuit, power = V × I (watts). In an ac circuit containing reactance, this simple product overestimates useful power because current and voltage are not in phase — at some instants they have opposite signs and power is momentarily returned to the supply rather than consumed. Three distinct power quantities are needed to fully describe ac power:
| Power type | Symbol | Unit | Definition | Physical meaning |
|---|---|---|---|---|
| True (active) power | P | W or kW | \(P = V \times I \times \cos\theta\) | Energy actually converted to heat, light, or mechanical work per second — what you pay for on the electricity bill |
| Reactive power | Q | VAr or kVAr | \(Q = V \times I \times \sin\theta\) | Energy stored and returned each cycle by inductors and capacitors — oscillates between source and load, doing no net useful work but flowing through the cables and heating them |
| Apparent power | S | VA or kVA | \(S = V \times I\) | The total volt-ampere product — what the supply cables and transformer must carry regardless of power factor. Used for rating generators, transformers, and cables. |
The three are related by the power triangle (a right-angled triangle identical in structure to the impedance and voltage triangles):
\[S = \sqrt{P^2 + Q^2} \qquad \text{(kVA)} = \sqrt{\text{kW}^2 + \text{kVAr}^2}\]Power Factor
Power factor (PF) is the ratio of true power to apparent power — the fraction of the apparent power that does useful work:
\[\text{PF} = \cos\theta = \frac{P}{S} = \frac{\text{kW}}{\text{kVA}} = \frac{R}{Z}\]Power factor ranges from 0 (purely reactive — no useful work done) to 1 (purely resistive — all power is useful). It is described as lagging (inductive loads: motors, transformers, fluorescent lamps) or leading (capacitive loads: capacitor banks, some variable speed drives under light loading).
A power factor of 0.8 lagging means only 80% of the apparent power is doing useful work — the remaining 20% is reactive power circulating in the cables.
Single-Phase Power Calculations
\[P = V \times I \times \cos\theta \quad (\text{W or kW})\] \[Q = V \times I \times \sin\theta \quad (\text{VAr or kVAr})\] \[S = V \times I \quad (\text{VA or kVA})\] \[I = \frac{P}{V \times \cos\theta} = \frac{S}{V}\]A motor has true power 3 kW and apparent power 5 kVA:
\[\text{PF} = \frac{3}{5} = 0.6\text{ lagging} \qquad \theta = \cos^{-1}(0.6) = 53.1°\] \[Q = \sqrt{5^2 - 3^2} = \sqrt{16} = 4\text{ kVAr}\]The 4 kVAr of reactive power flows uselessly through the supply cables — heating them without doing work. This is why poor power factor increases energy bills and requires larger conductors.
This confirms the self-assessment answer of 0.8 ✓.
This confirms the self-assessment answer of 10 kVA ✓. Note: PF = 8/10 = 0.8 lagging.
An industrial installation consumes 50 kW at a power factor of 0.65 lagging from a 400 V three-phase supply. Line current:
\[S = \frac{P}{\text{PF}} = \frac{50}{0.65} = 76.9\text{ kVA}\] \[I_L = \frac{S}{\sqrt{3} \times V_L} = \frac{76900}{1.732 \times 400} = \frac{76900}{692.8} = 111\text{ A}\]If the PF were corrected to 0.95: \(S = 50/0.95 = 52.6\text{ kVA}\) → \(I_L = 52600/692.8 = 75.9\text{ A}\). The cable carries 35 A less — a major saving in cable size and energy losses.
A factory runs 50 kW of load at PF = 0.65. The supply cable has resistance 0.5 Ω (line + neutral). Line current = 111 A (from above). Cable losses:
\[P_\text{loss} = 3 \times I_L^2 \times R = 3 \times 111^2 \times 0.5 = 3 \times 12321 \times 0.5 = 18.5\text{ kW}\]At PF = 0.95: I = 75.9 A, losses = \(3 \times 75.9^2 \times 0.5 = 8.65\text{ kW}\). Correcting PF reduces cable losses by nearly 10 kW — significant running cost savings.
Why Improve the Power Factor?
A low power factor is costly for both the supply authority and the consumer. Since cables, transformers, and switchgear must carry the total apparent current (kVA), not just the useful current (kW), a poor power factor means:
- Larger cables and switchgear than the true power alone would require — increasing installation capital cost
- Higher I²R losses in cables and transformers — wasting energy and increasing running costs
- Higher electricity bills — large industrial consumers are charged for kVA (maximum demand) not just kWh, so low PF directly increases the demand charge
- Reduced capacity of supply infrastructure — a transformer loaded to its kVA limit by a low-PF load could supply much more useful power if the PF were corrected
- Greater voltage drop — higher current at the same power means more volt drop in the distribution cables
UK electricity supply companies impose power factor penalties on large consumers whose PF falls below a threshold (typically 0.9 or 0.95). Many industrial energy contracts include a maximum demand kVA charge specifically to incentivise PF correction.
Methods of Power Factor Correction
PF correction works by introducing a leading reactive current to cancel the lagging reactive current drawn by inductive loads. The most common methods are:
| Method | How it works | Advantages | Disadvantages / limitations |
|---|---|---|---|
| Shunt capacitors (most common) | Capacitors connected in parallel with the inductive load draw leading current that cancels the lagging reactive current | Cheap, passive, no moving parts, can be fixed or switched in banks | Fixed capacitors can over-correct at light load, causing leading PF and voltage rise |
| Synchronous motors / condensers | Synchronous motors running overexcited draw leading current from the supply — acting as a rotating capacitor | Continuously variable, also provides useful mechanical output | Expensive, requires maintenance, needs excitation control |
| Static VAr compensators (SVC) | Electronically controlled thyristor-switched capacitors and inductors provide fast, variable reactive power compensation | Fast response, continuously variable, used on transmission grid | Expensive, complex power electronics |
| Individual motor capacitors | Small capacitors fitted directly at the motor terminals correct the motor's own lagging PF | Reduces reactive current in the distribution cable from motor to DB | Capacitor must be sized for the motor — risk of self-excitation if motor is switched with capacitor still connected |
Power Factor Correction Calculation — Capacitor Sizing
The required capacitor current IC is found from the phasor diagram. The original load current is split into active component (I cos θ₁, horizontal) and reactive component (I sin θ₁, vertical downward for lagging). The capacitor provides a leading current IC (vertical upward) that cancels some or all of the reactive component.
For correction to a target power factor, the required reactive current to be supplied by the capacitor is:
\[I_C = I_1 \sin\theta_1 - I_2 \sin\theta_2\]where θ₁ = original phase angle, θ₂ = corrected phase angle. The capacitor rating in kVAr:
\[Q_C = P(\tan\theta_1 - \tan\theta_2)\]230 V, 50 Hz motor, 50 A at PF = 0.8 lagging.
Original phase angle: \(\theta_1 = \cos^{-1}(0.8) = 36.87°\)
Active component: \(I\cos\theta_1 = 50 \times 0.8 = 40\text{ A}\)
Reactive component: \(I\sin\theta_1 = 50 \times 0.6 = 30\text{ A}\)
For unity PF (θ₂ = 0°), the capacitor must supply IC = 30 A:
\[X_C = \frac{V}{I_C} = \frac{230}{30} = 7.67\text{ Ω}\] \[C = \frac{10^6}{2\pi \times 50 \times 7.67} = \frac{10^6}{2409} = 415\text{ \mu F}\]After correction: supply current = 40 A (active component only) vs original 50 A — a 20% current reduction, saving both energy losses and cable capacity.
A factory draws 80 kW at PF = 0.7 lagging from a 400 V three-phase supply. Correct to 0.95 lagging.
\[\theta_1 = \cos^{-1}(0.7) = 45.57° \qquad \tan\theta_1 = 1.020\] \[\theta_2 = \cos^{-1}(0.95) = 18.19° \qquad \tan\theta_2 = 0.329\] \[Q_C = P(\tan\theta_1 - \tan\theta_2) = 80 \times (1.020 - 0.329) = 80 \times 0.691 = 55.3\text{ kVAr}\]A 55.3 kVAr three-phase capacitor bank is required. This reduces the apparent power from \(80/0.7 = 114.3\text{ kVA}\) to \(80/0.95 = 84.2\text{ kVA}\) — a 26% reduction in kVA demand charges.
It is generally not economic to correct all the way to unity PF. The final 5–10% of correction requires a disproportionately large capacitor bank for diminishing returns. Most tariffs and engineering standards target PF of 0.9–0.95 as the optimum economic correction level. Over-correction (leading PF) causes voltage rise on the network and can trigger protection systems.
Why Three-Phase?
Three-phase ac is the dominant form of electrical power generation, transmission, and industrial utilisation because it is more efficient than single-phase for the same power level. Understanding star and delta connections, and the relationships between line and phase quantities, is essential for sizing motors, calculating transformer ratings, and verifying supply configurations during inspection and testing.
Star (Y) Connection — Voltage and Current Relationships
In a star connection, each phase winding (or load) connects between one line conductor and the neutral (star) point. The star point may or may not be connected to neutral (4-wire vs 3-wire star).
The line voltage (VL, between any two line conductors) is √3 times the phase voltage (VP, from line to neutral) because the two phase voltages are 120° apart and combine as vectors:
Phase voltage VP = 230 V → Line voltage:
\[V_L = 230 \times 1.732 = 398.4 \approx 400\text{ V}\]Line voltage VL = 400 V → Phase voltage:
\[V_P = \frac{400}{1.732} = 231\text{ V}\]Three impedances of 30 Ω (PF = 0.9 lagging) connected in star on a 400 V, 50 Hz supply:
\[V_P = \frac{400}{1.732} = 231\text{ V} \qquad I_L = I_P = \frac{231}{30} = 7.7\text{ A}\] \[P = \sqrt{3} \times V_L \times I_L \times \cos\theta = 1.732 \times 400 \times 7.7 \times 0.9 = 4.79\text{ kW}\]Delta (Δ) Connection — Voltage and Current Relationships
In a delta connection, each phase winding (or load) connects directly between two line conductors — there is no neutral point. The phase voltage equals the line voltage. The line current is √3 times the phase current because two phase currents contribute to each line at 120° to each other:
Three impedances of 20 Ω (PF = 0.85 lagging) connected in delta on a 400 V supply:
\[V_P = V_L = 400\text{ V} \qquad I_P = \frac{400}{20} = 20\text{ A}\] \[I_L = \sqrt{3} \times 20 = 1.732 \times 20 = 34.64\text{ A}\] \[P = \sqrt{3} \times 400 \times 34.64 \times 0.85 = 20.4\text{ kW}\]Star vs Delta — Key Comparison
| Parameter | Star (Y) | Delta (Δ) |
|---|---|---|
| Phase voltage | \(V_P = V_L / \sqrt{3}\) | \(V_P = V_L\) |
| Phase current | \(I_P = I_L\) | \(I_P = I_L / \sqrt{3}\) |
| Line voltage | \(V_L = \sqrt{3} \times V_P\) | \(V_L = V_P\) |
| Line current | \(I_L = I_P\) | \(I_L = \sqrt{3} \times I_P\) |
| Neutral available? | Yes (4-wire) | No |
| Power (same impedances) | P_star | \(P_{\Delta} = 3 \times P_{\text{star}}\) |
Three-Phase Power — Formulae
For any balanced three-phase load (star or delta), the total true power, reactive power, and apparent power are:
\[P = \sqrt{3} \, V_L I_L \cos\theta = 3 \, V_P I_P \cos\theta \quad (\text{kW})\] \[Q = \sqrt{3} \, V_L I_L \sin\theta \quad (\text{kVAr})\] \[S = \sqrt{3} \, V_L I_L \quad (\text{kVA})\]Note: √3 × V_L × I_L equals 3 × V_P × I_P for both star and delta (they are mathematically identical — use whichever is more convenient given the available data).
Three 20 Ω impedances, PF = 0.85, 400 V supply:
Star: \(V_P = 231\text{ V},\; I_P = 11.55\text{ A},\; P = \sqrt{3} \times 400 \times 11.55 \times 0.85 = 6.8\text{ kW}\)
Delta: \(V_P = 400\text{ V},\; I_P = 20\text{ A},\; P = \sqrt{3} \times 400 \times 34.64 \times 0.85 = 20.4\text{ kW}\)
Delta power is exactly 3× star power because the phase voltage is √3 times larger in delta, and both voltage and current are √3 times larger, giving (√3)² = 3 times the power per phase. This is why motor starters use star-delta starting — the motor starts in star (1/3 of delta power and current), then switches to delta at speed.
A three-phase motor develops 15 kW at 85% efficiency. PF = 0.88 lagging, 400 V supply. Find the line current.
\[P_\text{input} = \frac{P_\text{output}}{\eta} = \frac{15000}{0.85} = 17647\text{ W}\] \[I_L = \frac{P}{\sqrt{3} \times V_L \times \cos\theta} = \frac{17647}{1.732 \times 400 \times 0.88} = \frac{17647}{609.7} = 28.9\text{ A}\]This is the information needed to select the correct overload relay setting on the motor starter and to verify the supply cable is adequately rated.
A 400 V three-phase load draws 40 A at PF = 0.75 lagging:
\[S = \sqrt{3} \times 400 \times 40 = 27.7\text{ kVA}\] \[P = 27.7 \times 0.75 = 20.8\text{ kW}\] \[Q = \sqrt{27.7^2 - 20.8^2} = \sqrt{767.3 - 432.6} = \sqrt{334.7} = 18.3\text{ kVAr}\]Balanced Three-Phase Loads
A balanced three-phase load has equal impedances on all three phases — each phase carries exactly the same current magnitude, and the three currents are equally spaced at 120° from each other. Three-phase motors, heaters, and other symmetrical three-phase equipment are inherently balanced.
When three equal current phasors at 120° are added vectorially, they form a closed triangle and their sum is zero. This means the neutral current is zero in a perfectly balanced system — the neutral conductor carries no current and could theoretically be omitted (as it is in three-phase three-wire delta systems). This is a major advantage of balanced three-phase: the neutral cable is unnecessary, saving cost in long distribution runs.
Three resistive loads each drawing 20 A on L1, L2, L3 respectively. Each current phasor is 20 A at 0°, 120°, 240°. Sum:
\[I_N = 20\angle 0° + 20\angle 120° + 20\angle 240° = 0\text{ A}\]This confirms the self-assessment answer — neutral current = 0 A when all three phases carry equal currents ✓.
Unbalanced Single-Phase Loads — Why They Occur
In practice, the loads on a three-phase LV supply are rarely perfectly balanced because individual consumers are connected to single phases. A street with 30 houses might have 10 houses connected to L1, 10 to L2, and 10 to L3 — balanced overall — but at any given moment the loads are different (some have their kettles on, some don't). Commercial and industrial premises with a mix of single-phase and three-phase equipment also create imbalance.
Distribution engineers aim to connect single-phase loads across the three phases as equally as possible when designing the LV network — this is the purpose of the phase rotation assignment in distribution feeder designs.
Neutral Current in an Unbalanced Load
When the three-phase currents are unequal, their vector sum is not zero — the residual phasor represents the neutral current IN that must flow in the neutral conductor to complete the circuit. The neutral current is found by phasor addition:
For a graphical solution: draw the three phase current phasors to scale, head-to-tail, each at 120° to the previous. The closing line (from the start of the first phasor to the end of the last) is IN.
IL1 = 150 A, IL2 = 200 A, IL3 = 100 A (all resistive, 230 V phase). Drawing to scale at 1 cm = 50 A and measuring the closing phasor gives IN ≈ 85 A.
Only two phases are loaded: L1 carries 40 A, L3 carries 30 A, L2 carries 0 A. All loads are resistive (in phase with their respective phase voltages). The phasors are at 0° and 240°:
\[I_N^2 = 40^2 + 30^2 + 2 \times 40 \times 30 \times \cos(240°-0°)\] \[= 1600 + 900 + 2400 \times \cos(240°) = 2500 + 2400 \times (-0.5) = 2500 - 1200 = 1300\] \[I_N = \sqrt{1300} = 36.1\text{ A}\]This illustrates that even with only two phases loaded, a significant neutral current flows. The neutral conductor must be rated to carry this current safely.
A commercial distribution board has 9 single-phase circuits. Planned loads: 3 × 6 kW on L1, 2 × 8 kW + 1 × 2 kW on L2, 3 × 6 kW on L3. Current per phase at 230 V unity PF:
\[I_{L1} = \frac{18000}{230} = 78.3\text{ A} \qquad I_{L2} = \frac{18000}{230} = 78.3\text{ A} \qquad I_{L3} = \frac{18000}{230} = 78.3\text{ A}\]The loads are perfectly balanced (18 kW per phase). IN = 0. If instead L2 had 4 × 8 kW = 32 kW and L3 had only 4 kW:
\[I_{L1} = 78.3\text{ A}, \quad I_{L2} = \frac{32000}{230} = 139.1\text{ A}, \quad I_{L3} = \frac{4000}{230} = 17.4\text{ A}\]L2 is heavily overloaded relative to L1 and L3. The cable to L2 circuits would need to be significantly larger, and a large neutral current would flow. Rebalancing the loads across phases before installation avoids these problems.
Consequences of Load Imbalance — Why Balance Matters
| Consequence | Technical explanation | Practical impact |
|---|---|---|
| Neutral current flows | Unequal phase currents don't cancel; residual flows in neutral | Neutral conductor must be sized to carry IN — potentially as large as the phase conductors in severe imbalance |
| Unequal phase voltages at load | Different volt drops on each phase due to different currents | Some equipment gets higher-than-rated voltage, some gets lower — equipment may malfunction or be damaged |
| Overloaded phase conductor | One phase carries more than its rated current | Cable overheating, insulation damage, premature failure |
| Increased power losses | \(P_{\text{loss}} = I_N^2 R_N\) in neutral; \(I^2 R\) losses higher on overloaded phase | Increased energy bills and wasted capacity |
| Transformer and generator stress | Unbalanced currents create uneven loading on transformer windings and alternator phases | Increased heating in transformer, reduced efficiency, shorter equipment life |
Rules for Balancing Single-Phase Loads
When designing a distribution board or LV network fed from a three-phase supply, single-phase circuits should be allocated to phases to achieve the most equal loading possible:
- Group circuits by estimated load and spread them as evenly as possible across L1, L2, and L3
- For very large single-phase loads (e.g. single-phase EV chargers, large AHUs), consider whether a three-phase version of the equipment is available to avoid imbalance
- In a distribution board serving a block of flats, alternate phase connections — flat 1 on L1, flat 2 on L2, flat 3 on L3, flat 4 on L1, etc.
- Provision for an oversized neutral conductor (up to the same size as the phase conductors) where significant imbalance is likely or where significant harmonic currents are expected (non-linear loads like computers, VFDs, and LED drivers produce triplen harmonics that add in the neutral rather than cancelling)
Three single-phase computer power supplies each draw 10 A fundamental (50 Hz) current plus 8 A of third harmonic (150 Hz) current. For the fundamental: \(I_N = 0\) (balanced). For the third harmonic: all three phases produce the same phase angle (they are "triplen" harmonics — 3rd, 6th, 9th — which are in phase rather than 120° apart). The neutral harmonic currents add:
\[I_{N,3\text{rd}} = 8 + 8 + 8 = 24\text{ A}\]The neutral carries 24 A of harmonic current even though the fundamental neutral current is zero — a neutral current that can exceed the phase current in installations with large numbers of computers, VFDs, and electronic loads. This is why BS 7671 requires consideration of harmonic content when selecting neutral conductor sizes.
🃏 Chapter 7 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Covers reactance, phasor diagrams, R-L-C circuits, power factor, PF correction, three-phase calculations and neutral current.
📝 Chapter 7 — Multiple Choice Questions
Select your answer and click Submit. You have two attempts per question. First-attempt score shown at the end.
1A coil has an inductance of 0.2 H. What is its inductive reactance XL at a frequency of 50 Hz?
2A 100 µF capacitor is connected to a 50 Hz supply. What is its capacitive reactance XC? (Use 2π = 6.283)
3A series R-L circuit has R = 5 Ω and XL = 12 Ω. What is the impedance Z?
4In a phasor diagram for a series R-L-C circuit, which quantity is used as the reference phasor (drawn horizontally at 0°)?
5In a purely inductive ac circuit, what is the phase relationship between voltage and current?
6In a series R-L-C circuit at resonance, which statement is correct?
7A single-phase motor draws 10 A from a 230 V supply and consumes 1840 W of true power. What is the power factor?
8An installation has a true power of 15 kW and a reactive power of 20 kVAr. What is the apparent power in kVA?
9What is the most common method of power factor correction for industrial installations?
10In a balanced three-phase star (Y) connected system, the line voltage is 400 V. What is the phase voltage?
11A balanced three-phase load draws a line current of 20 A at a line voltage of 400 V with a power factor of 0.85. What is the total true power consumed?
12A three-phase installation has large numbers of single-phase computer loads. Even when the fundamental phase currents are balanced, the neutral conductor may carry significant current because:
Sends your Chapter 7 MCQ score summary from your registered email to your tutor.
AC Motors and DC Machines
- Describe the operating principles of dc machines (motors and generators)
- State the basic types and applications of dc machines
- Explain the operating principles of single-phase and three-phase ac motors
- Describe the operation of variable frequency drives and inverters
- Explain the operation of synchronous motors
- Describe operating principles, limitations and applications of motor control (DOL, star-delta, soft-start)
The Motor Effect — Rotation from a Single Coil
A dc motor converts electrical energy into mechanical (rotational) energy using the electromagnetic force on a current-carrying conductor in a magnetic field (F = BIl, Chapter 5). The basic model is a single rectangular coil placed inside a fixed magnetic field between two poles. When current flows through the coil, two coil sides each experience a force (Fleming's left-hand rule): one side is pushed up, the other is pushed down. These equal and opposite forces create a turning moment (torque) that rotates the coil.
However, a single coil would only rotate to the position where its plane is perpendicular to the field (90° from start), because at that point the forces act radially (through the axis) rather than tangentially — they create no further torque. A mechanism to maintain torque beyond 90° is needed: the commutator.
Commutation — How Continuous Rotation is Achieved
The commutator is a rotating switch that reverses the current direction in each coil at precisely the moment the coil passes through the position of zero torque. This keeps the current in the correct direction relative to the field at all positions, maintaining continuous torque in the same rotational direction.
In practice, a dc motor has many armature coils (windings) distributed around the armature, each connected to a pair of commutator segments. Carbon brushes press against the commutator and remain stationary while the commutator rotates beneath them, making continuous electrical contact. More coils means smoother torque — practical motors have dozens to hundreds of coil segments.
In a motor: transfers supply current to armature coils and reverses current direction in each coil at the correct moment to maintain continuous torque in one direction.
In a generator: converts the alternating voltage induced in the rotating armature coils into a direct (unidirectional) voltage at the brushes — mechanical rectification.
A dc motor armature has 24 conductors, each 0.15 m long in a field of 0.9 T, carrying 8 A. Total electromagnetic force:
\[F = N \times B \times I \times l = 24 \times 0.9 \times 8 \times 0.15 = 25.9\text{ N}\]If the armature radius is 0.06 m, the torque developed:
\[T = F \times r = 25.9 \times 0.06 = 1.56\text{ Nm}\]Increasing either field strength or current directly increases torque — the basis of dc motor speed and torque control.
Back-EMF — Why DC Motors Don't Draw Infinite Current
As the armature rotates in the magnetic field, it acts simultaneously as a generator — its conductors cut through the flux and a voltage is induced (Faraday's Law). This induced voltage opposes the supply (Lenz's Law) and is called the back-emf (E_b). The net voltage driving current through the armature resistance is:
\[V_\text{supply} = E_b + I_a R_a \qquad \Rightarrow \qquad I_a = \frac{V - E_b}{R_a}\]At rest (start-up), E_b = 0, so all the supply voltage drives current through the armature resistance — producing a very high starting current. As the motor accelerates, E_b increases, reducing armature current. At full speed, E_b is close to the supply voltage and I_a is low (just enough to maintain speed against friction and load).
A 230 V dc motor has armature resistance R_a = 0.5 Ω and back-emf at full speed E_b = 215 V.
At start (E_b = 0): \(I_a = (230 - 0) / 0.5 = 460\text{ A}\) — dangerously high, would burn out the armature
At full speed (E_b = 215 V): \(I_a = (230 - 215) / 0.5 = 30\text{ A}\) — safe operating current
This is why dc motor starters insert series resistance to limit starting current, reducing it as speed (and E_b) builds up.
DC Generator — Construction and Operation
A dc generator has identical construction to a dc motor — both are electromagnetic energy converters. The key difference is the energy transfer direction: a motor receives electrical energy and produces mechanical rotation; a generator receives mechanical rotation and produces electrical energy.
When the armature coil rotates in the magnetic field, the induced emf varies sinusoidally with rotor angle (E = Blv sinθ). Without a commutator the output would be ac. The commutator acts as a mechanical full-wave rectifier: as each coil side reverses direction through the field, the commutator simultaneously reverses the brush connection to that coil, so the output at the brushes is always in the same direction — producing a pulsating dc voltage that smooths out as more coils are added.
The magnitude of the generated emf at any angle is:
| Coil angle (°) | Rate of flux cutting | Induced emf |
|---|---|---|
| 0° | Coil sides moving parallel to flux — zero cutting rate | Zero |
| 45° | Cutting at angle — moderate rate | 0.707 × E_max |
| 90° | Coil sides cutting at right angles — maximum rate | E_max |
| 135° | Same as 45° (by symmetry) | 0.707 × E_max |
| 180° | Parallel again | Zero (commutator reverses connection here) |
A generator armature has active conductor length 0.2 m in a field of 0.8 T, rotating at 25 rev/s. Peripheral velocity of armature (radius 0.05 m):
\[v = 2\pi r n = 2\pi \times 0.05 \times 25 = 7.85\text{ m/s}\]Maximum emf per conductor (at 90°): \(E = Blv = 0.8 \times 0.2 \times 7.85 = 1.26\text{ V}\). With 50 active conductors in series: \(E_\text{total} = 50 \times 1.26 = 63\text{ V}\).
Types of DC Motor
DC motors are classified by how their field windings (which create the main magnetic flux) are connected relative to the armature (rotor) windings. Each connection gives different speed-torque characteristics, making each type suited to specific applications. There are three main types: series, shunt, and compound wound.
| Type | Winding connection | Speed behaviour | Torque behaviour | Critical safety rule | Applications |
|---|---|---|---|---|---|
| Series motor | Field and armature windings connected in series — same current flows through both | Speed varies greatly with load: high speed at no-load, low speed at heavy load | Very high starting torque (torque ∝ I², and series field current = armature current) | Must NEVER run on no-load — with no load, current falls, field weakens, speed rises without limit (runaway to self-destruction). Belt drives must not be used — belt breakage = no-load. | Electric traction (trams, locomotives), cranes, hoists, lifts — where high starting torque under heavy load is essential |
| Shunt motor | Field winding connected in parallel with armature — field current is relatively constant regardless of armature current | Virtually constant speed — as load increases, armature current increases but field is constant; small speed drop only | Moderate starting torque; torque proportional to armature current only | Field circuit must never be open-circuited while running — field collapse causes speed runaway (same as series motor no-load) | Machine tools (lathes, milling), fans, conveyor belts, centrifugal pumps — where constant speed is required |
| Compound motor | Has both a series field winding and a shunt field winding — cumulative compounding (fields add) or differential (fields oppose) | Speed between series and shunt — some variation with load, but safe at no-load | Higher starting torque than shunt; moderate speed variation | Both field safety rules apply: no-load runaway is limited (shunt field maintains minimum flux) but field circuit must not be interrupted | Lifts, hoists, heavy intermittent loads — where high starting torque AND safe no-load behaviour are both needed |
Why the series motor must never run on no-load (explained)
In a series motor, field current = armature current. Under heavy load the current is high, producing strong field flux and high torque. At no-load, the current drops to near zero (only enough to overcome friction) — the field flux virtually disappears. With very weak flux, the back-emf = Blv becomes very small, so the supply voltage drives a large current, which accelerates the motor... which further reduces flux... which increases current further. This positive feedback runaway continues until the motor physically destroys itself (centrifugal forces burst the armature winding).
A 220 V dc shunt motor has its armature connected: A1 positive, A2 negative; field connected: F1 positive, F2 negative. To reverse: disconnect A1 and A2 and swap them (A2 becomes positive, A1 negative). The field connections F1/F2 are left unchanged. Reversing both armature AND field would leave the torque direction unchanged (both forces reverse, net direction same). Therefore: change either armature or field — never both.
DC Motor Speed Control
The speed of a dc motor is governed by the back-emf equation: \(n \propto (V - I_a R_a) / \Phi\). Speed can therefore be controlled by:
- Varying armature voltage (V): most efficient and common modern method — using a thyristor or MOSFET-based electronic speed controller (PWM drive). Reduces speed below base speed.
- Varying field flux (Φ) via field weakening: reducing field current weakens flux and increases speed above base speed. Speed above base speed but at reduced torque.
- Armature resistance control (older method): inserting a variable resistor in series with the armature reduces the effective voltage. Very inefficient — wastes energy in the resistor. Used only in older or simple applications.
A 220 V dc shunt motor runs at 1500 rpm at full voltage with back-emf 205 V and armature resistance 0.5 Ω. If the supply voltage is reduced to 180 V (armature current approximately constant at 30 A):
\[E_b^{\text{new}} = 180 - (30 \times 0.5) = 165\text{ V}\]Since speed is proportional to back-emf (flux constant): \(n = 1500 \times (165/205) = 1207\text{ rpm}\). Armature voltage control gives smooth, efficient speed reduction.
DC Motor Starting Methods
As shown in Part 1, a dc motor drawing 460 A at start (for a 30 A running motor) would immediately burn out the armature. Starting methods limit this inrush current:
- Face-plate starter (resistance starter): a bank of resistors inserted in series with the armature at start. The operator (or an automatic time-step controller) progressively shorts out sections of resistance as the motor accelerates and back-emf builds up. Simple and reliable but energy is wasted in the resistors.
- Electronic thyristor starter (modern): thyristors (SCRs) control the average voltage applied to the armature by phase-angle firing. Starting current can be precisely limited. Also provides speed control and regenerative braking. Now the standard approach for new installations.
Types of DC Generator
DC generators are classified by the source of their field excitation:
| Type | Field excitation source | Output characteristics | Applications |
|---|---|---|---|
| Separately excited | External dc supply (battery + variable resistor) feeds the field winding independently of the armature | Good voltage regulation; output controllable over wide range by adjusting field current; slight voltage drop under load due to armature resistance | Laboratory power supplies; Ward-Leonard speed control systems; excitation of large alternators |
| Shunt (self-excited) | Field winding connected in parallel with armature — generator feeds its own field | Requires residual magnetism in poles to "bootstrap" initial voltage; fairly constant voltage; poorer regulation than separately excited | Battery chargers; small dc power supplies; older welding generators |
| Series (self-excited) | Field winding in series with armature and load — field current = load current | Voltage rises with load (opposite to shunt); poor regulation; rarely used | Series arc lighting (obsolete) |
| Compound (self-excited) | Both series and shunt field windings — series field counteracts armature reaction | Flat compounding gives near-constant terminal voltage from no-load to full load | Where good voltage regulation is needed: emergency lighting systems, hospital standby generators |
Three-Phase vs Single-Phase AC Motors
Three-phase motors dominate industrial applications because they offer significant advantages over single-phase motors of the same power rating:
| Feature | Three-phase motor | Single-phase motor |
|---|---|---|
| Cost | Lower for same power output | Higher — more complex windings needed |
| Size / weight | Smaller and lighter | Larger for same rating |
| Efficiency | Higher (typically 85–95%) | Lower |
| Self-starting | Yes — rotating field starts the rotor automatically | No — needs starting winding, capacitor, or other means |
| Supply current | ¼ to ⅓ of equivalent single-phase motor current | High — limited by single-phase supply capacity |
| Torque | Smooth, continuous (three phases always producing torque) | Pulsating (twice per cycle on a single-phase supply) |
| Power factor | Typically 0.85–0.92 at full load | Lower, especially at part load |
Note on units: 1 hp (horsepower) = 746 W ≈ 0.75 kW. Older motor nameplates often quote output in hp; modern practice uses kW. Convert when comparing: a "10 hp" motor delivers approximately 7.5 kW of mechanical output.
Construction of the Cage-Rotor Induction Motor
The three-phase cage-rotor induction motor is the workhorse of industry — simple, robust, and requiring minimal maintenance. It consists of two main parts:
- Stator (fixed part): A laminated silicon-steel core with evenly spaced slots around the inner circumference. Three sets of windings (one per phase) are distributed in these slots. When three-phase current flows, the windings produce the rotating magnetic field. The stator core is laminated to reduce eddy current losses.
- Rotor (moving part — cage type): A laminated steel core with longitudinal slots around its outer circumference. Solid copper or aluminium rotor bars are pressed or cast into these slots. Each end of the cage is short-circuited by a heavy copper or aluminium end ring. The complete assembly resembles a squirrel cage (hence "squirrel cage motor" in older terminology). There are no brushes, slip rings, or external connections to the rotor — the rotor circuit is entirely self-contained.
How the Rotating Magnetic Field is Produced
This is the key to understanding all three-phase motors. When three-phase current flows in the three stator windings (120° apart in time), it creates a magnetic field that rotates around the stator bore at synchronous speed. This is not a mechanical rotation — it is an electromagnetic phenomenon: the resultant magnetic field vector sweeps continuously around the 360° of the stator bore, completing one full revolution per cycle of the supply (50 times per second on a 2-pole motor).
The rotating field spins at synchronous speed determined by supply frequency and number of pole pairs. It is this rotating field that induces emf in the rotor bars and drives the rotor.
How the Rotor is Driven — Induction Principle
The rotating stator field sweeps past the stationary rotor bars, inducing an emf in each bar (Faraday's Law — moving field, stationary conductor). Since the rotor bars are short-circuited by the end rings, the induced emf drives large currents through the rotor cage. These rotor currents create their own magnetic field, which interacts with the stator's rotating field and exerts a force (F = BIl) on the rotor bars, accelerating the rotor in the direction of the rotating field.
This is the induction principle: the motor is called an induction motor because the rotor current is induced by the rotating field — there is no direct electrical connection to the rotor. The motor is essentially a rotating transformer: stator = primary, rotor cage = short-circuited secondary.
Slip — Why the Rotor Never Reaches Synchronous Speed
If the rotor were to spin at exactly synchronous speed, there would be no relative motion between the rotor bars and the rotating field. No relative motion → no induced emf → no rotor current → no force → no torque. The rotor would decelerate. A stable equilibrium is only possible when the rotor runs slightly below synchronous speed — enough below that the relative motion induces sufficient current and torque to maintain that speed against the load. This speed difference is the slip.
At full load, typical cage motors run at 2–5% slip. At no-load (only friction to overcome) slip is very small (≈0.5–1%).
Motor Terminal Box — Star and Delta Connections
Three-phase motors with six terminal box connections (U1, V1, W1, U2, V2, W2 — the two ends of each of the three phase windings) can be connected in either star or delta:
- Star connection: U2, V2, W2 linked together (two copper links in an upside-down triangle pattern). Supply connected to U1, V1, W1. Each winding sees phase voltage (230 V on a 400 V supply). Used for star-delta starting or on 230 V three-phase supplies.
- Delta connection: Three links connect U2-V1, V2-W1, W2-U1 (each link connects the end of one winding to the start of the next). Each winding sees line voltage (400 V). Used for normal running on 400 V supplies.
A motor nameplate marked "230/400 V Δ/Y" means: connect in delta for a 230 V supply; connect in star for a 400 V supply. Both give the same winding voltage (230 V) and the same output power.
A motor nameplate reads: 400 V Δ, 5.5 kW, 50 Hz, 1450 rpm, PF 0.87, 12.5 A. On a 400 V three-phase supply connected in delta:
Line current = 12.5 A (as stated). Input power: \(P_{\text{in}} = \sqrt{3} \times 400 \times 12.5 \times 0.87 = 7.54\text{ kW}\). Output = 5.5 kW. Efficiency: \(\eta = 5500/7540 = 72.9\%\). Check against nameplate — this seems low; most motors of this type are 85–90% efficient. The nameplate current of 12.5 A corresponds to: \(\eta = 5500/(\sqrt{3} \times 400 \times 12.5 \times 0.87) = 72.9\%\). (In practice, the stated current would give a higher efficiency — this example illustrates the calculation method).
Reversal of Rotation
Reversing any two of the three incoming phase connections reverses the direction of the rotating magnetic field — and therefore the rotor direction. This is done at the terminal box (or by a reversing contactor in the starter circuit). It is the simplest and most reliable method of motor reversal, and requires no change to the motor windings or internal connections.
Advantages and Limitations of Cage-Rotor Induction Motors
| Advantages | Limitations |
|---|---|
| Simple, rugged construction — no brushes, slip rings, or commutator to maintain | High starting current (6–10 × full-load current) — causes supply voltage dips |
| Self-starting on three-phase supply | Fixed speed (approximately) — cannot easily vary speed without VFD |
| Long service life (30+ years with basic maintenance) | Poor power factor at light loads and during starting |
| Wide range of standard frame sizes and powers (BS EN 60034) | Limited starting torque in standard single-cage design |
| Low cost per kW | Motor overheating if stalled or run at very low speed without forced cooling |
Double-Cage Rotor Motor — Improved Starting
The single-cage motor faces a design conflict: low rotor resistance is needed for efficient running (minimal I²R losses in rotor) but high rotor resistance is needed for high starting torque (R-L effect in rotor circuit at start). The double-cage rotor resolves this with two concentric sets of rotor bars:
- Outer cage: high-resistance bars (bronze or brass), close to the rotor surface. At starting (high slip, high rotor frequency), the skin effect and high leakage reactance of the inner cage force current into the outer cage — where the high resistance produces high starting torque.
- Inner cage: low-resistance bars (copper), deep in the rotor. At running speed (low slip, low rotor frequency), the inner cage carries most of the current — low resistance means efficient operation.
The result: high starting torque with low starting current, transitioning to efficient low-loss running. This is particularly important for conveyor drives, compressors, and any load that starts fully loaded.
| Parameter | Single cage | Double cage |
|---|---|---|
| Starting torque | 100–150% of full-load torque | 200–250% of full-load torque |
| Starting current | 6–10 × full-load current | 4–6 × full-load current |
| Running efficiency | High | Slightly lower than single cage |
| Cost | Lower | Higher (more complex rotor) |
Wound-Rotor (Slip-Ring) Induction Motor
The stator is identical to a cage-rotor motor. The rotor has three proper windings (distributed coils, not just bars) connected in star, with the free ends brought out through slip rings to external resistors. This allows the rotor circuit resistance to be controlled externally:
Starting with maximum external resistance: high rotor resistance → high starting torque (as in double cage outer cage), low starting current. As the motor accelerates, resistance is gradually reduced in steps (manually or automatically). At full speed all external resistance is shorted out — the motor runs as a standard induction motor.
Speed control: leaving some resistance in the rotor circuit increases slip and reduces speed — useful for cranes, hoists, and large fans where speed control is needed. This is inefficient (energy wasted in resistors) but simple and inexpensive for intermittent duty.
A 75 kW slip-ring motor starts a large centrifugal pump against full head. Starting with maximum external resistance (10 Ω per phase), rotor circuit resistance limits starting current to 150% of full-load current while delivering 150% of full-load starting torque — enough to overcome the pump's inertia and static head. A standard cage-rotor motor would either stall or draw excessive current at start.
Three-Phase Synchronous Motor
A synchronous motor has the same stator as an induction motor (producing a rotating magnetic field) but uses a dc-excited rotor winding (not a cage). The rotor is wound with field coils fed through slip rings from a dc supply (the exciter). The rotor's magnetic poles lock onto and follow the stator's rotating field, spinning at exactly synchronous speed — regardless of load (as long as it is not overloaded).
Starting limitation: A synchronous motor cannot start from rest on its own — the rotor has inertia and cannot accelerate from zero to synchronous speed in one half-cycle. Starting methods include: damper windings (short-circuited rotor conductors that act as a cage during starting, then the dc excitation is applied as the motor approaches synchronous speed) or using a separate starting motor.
Power factor control: The major advantage of the synchronous motor over induction motors is that its power factor is controllable by adjusting the dc excitation current:
- Under-excited: motor draws lagging reactive current (like an induction motor)
- Normal excitation: motor operates at unity power factor
- Over-excited: motor draws leading reactive current — acts as a large capacitor, supplying reactive power to the network and correcting the lagging PF of nearby induction motors
A factory's induction motor bank draws 500 kVA at PF = 0.75 lagging. A 200 kVA synchronous motor (over-excited) running at unity PF supplies 150 kVAr of leading reactive power. This reduces the induction motor reactive demand from 330 kVAr to 180 kVAr, improving the overall installation PF to approximately 0.90 lagging — saving on tariff demand charges.
Synchronous Speed — How Poles and Frequency Determine Speed
The speed at which the stator's rotating magnetic field sweeps around the motor bore is called the synchronous speed (Ns). It depends entirely on two quantities: the supply frequency and the number of pole pairs wound into the stator. More poles → lower synchronous speed; higher frequency → higher speed.
\[N_s = \frac{f}{p} \times 60 \text{ rev/min} \qquad \text{or} \qquad n_s = \frac{f}{p} \text{ rev/s}\]where f = supply frequency (Hz) and p = number of pairs of poles (total poles ÷ 2).
Synchronous Speeds at 50 Hz
| Total poles | Pole pairs (p) | Synchronous speed (rev/min) | Typical full-load rotor speed (rev/min) | Typical slip (%) |
|---|---|---|---|---|
| 2 | 1 | 3000 | 2900 | 3.3% |
| 4 | 2 | 1500 | 1440 | 4.0% |
| 6 | 3 | 1000 | 960 | 4.0% |
| 8 | 4 | 750 | 720 | 4.0% |
| 10 | 5 | 600 | 580 | 3.3% |
| 12 | 6 | 500 | 480 | 4.0% |
| 16 | 8 | 375 | 360 | 4.0% |
Notice that full-load slip is typically 3–5% for standard cage-rotor motors regardless of pole count. This is because the slip percentage is determined by the rotor resistance design, not by the number of poles.
Slip — Why It Exists and How to Calculate It
The slip (S) is the difference between the synchronous speed of the rotating field and the actual rotor speed, expressed as a fraction (per-unit) or percentage of synchronous speed:
\[S = \frac{N_s - N_r}{N_s} \qquad S(\%) = \frac{N_s - N_r}{N_s} \times 100\]Why slip must exist: If the rotor ran at synchronous speed there would be zero relative motion between the rotor bars and the rotating field — no flux cutting, no induced emf, no rotor current, no electromagnetic force, and therefore no torque. The motor would decelerate. Slip is not a design flaw; it is a physical necessity for torque production. Higher load → more torque needed → more rotor current needed → more slip required to increase relative motion and induced emf.
The rotor speed can also be expressed directly from slip:
\[N_r = N_s (1 - S) \qquad \text{or} \qquad n_r = n_s(1-S)\]This confirms the self-assessment answer of 1000 rev/min ✓.
A 16-pole motor is used where very slow shaft speeds are needed (e.g. conveyor drives, direct-coupled fans) to avoid a gearbox.
A 12-pole, 50 Hz motor runs at 475 rev/min at full load:
\[N_s = \frac{50}{6} \times 60 = 500\text{ rev/min}\] \[S = \frac{500 - 475}{500} \times 100 = 5\%\]5% slip is slightly high for a standard motor (normal is 3–4%) — this motor may be heavily loaded or have higher-than-standard rotor resistance.
A 6-pole motor runs at 5% slip on a 50 Hz supply:
\[n_s = \frac{50}{3} = 16.67\text{ rev/s}\] \[n_r = n_s(1-S) = 16.67 \times 0.95 = 15.84\text{ rev/s} = 15.84 \times 60 = 950\text{ rev/min}\]A motor connected to a 50 Hz supply runs at 960 rev/min at full load. Identify the number of poles.
960 rpm is close to the 1000 rpm synchronous speed (6-pole). Check: slip = (1000−960)/1000 = 4% — typical full-load slip. Therefore the motor is a 6-pole machine.
This calculation is used when testing motors with unknown nameplate data — measure the speed, identify which synchronous speed it runs below, and determine pole count.
A 4-pole motor (p=2) is supplied at variable frequency by a VFD. At 50 Hz its synchronous speed is 1500 rpm (rotor ≈ 1440 rpm). At 30 Hz:
\[N_s = \frac{30}{2} \times 60 = 900\text{ rev/min} \qquad N_r \approx 900 \times 0.96 = 864\text{ rev/min}\]At 70 Hz: \(N_s = (70/2) \times 60 = 2100\text{ rev/min}\), rotor ≈ 2016 rev/min — above the 50 Hz synchronous speed. This shows how a VFD achieves speed control by changing frequency rather than by inserting resistance or voltage reduction.
Why Single-Phase Motors Are Not Self-Starting
A three-phase motor produces a rotating magnetic field automatically because the three current waveforms are 120° apart in time — as one phase rises, the next falls, creating a continuously rotating resultant field. A single-phase supply produces only a single sinusoidally alternating field — it pulses back and forth along one axis. This pulsating field can be resolved into two equal counter-rotating fields, both spinning at synchronous speed in opposite directions. At standstill these two fields produce equal and opposite torques that cancel exactly — the rotor receives zero net starting torque.
Once the rotor is spinning (by hand or by other means) it couples preferentially with the field rotating in the same direction, and continues to accelerate to near synchronous speed. Therefore, all single-phase induction motors require an additional starting mechanism to break the symmetry and produce an initial net torque.
Split-Phase Motor
The split-phase motor solves the starting problem by using two stator windings whose currents are out of phase with each other, producing a crude rotating field:
- Run winding: heavy gauge wire (low resistance), wound with many turns (high inductance). Current lags voltage by a large angle (inductive).
- Start winding: fine gauge wire (high resistance), fewer turns (low inductance). Current lags voltage by a much smaller angle (resistive dominant).
The two winding currents are out of phase by approximately 30° — enough to create a weak rotating component in the stator field, producing starting torque. A centrifugal switch (speed-sensitive switch mounted on the rotor shaft) disconnects the start winding automatically when the motor reaches approximately 75% of synchronous speed. This is essential because the start winding is designed for short-term use only — it would overheat and burn out if left connected at full speed.
A 230 V, 50 Hz split-phase motor connected to a pump: at start, both windings are energised; the 30° phase difference produces a rotating field that starts the motor. At approximately 1125 rpm (75% of 1500 rpm synchronous speed for a 4-pole motor), the centrifugal switch opens, disconnecting the start winding. The motor continues to accelerate on the run winding alone to approximately 1440–1460 rpm full-load speed. If the centrifugal switch fails to open, the start winding will overheat and eventually open-circuit due to thermal overload — a common fault mode.
| Parameter | Split-phase | Rationale |
|---|---|---|
| Phase displacement | ~30° | Produced by resistance/inductance ratio difference between windings |
| Starting torque | 175–200% of full-load torque | Moderate — adequate for light loads |
| Starting current | 600–900% of full-load current | High — both windings in parallel at start |
| Centrifugal switch disconnects at | ~75% of synchronous speed | Protects start winding from overheating |
| Typical applications | Washing machine agitators, tumble dryers, small fans | Light starting load — adequate torque, low cost |
| Reversal method | Reverse start OR run winding — not both | Reversing both leaves rotation unchanged |
Capacitor-Start, Induction-Run Motor
A capacitor is connected in series with the start winding. The capacitor's leading current effect shifts the start winding current by approximately 90° relative to the run winding current — much closer to the ideal 90° for maximum starting torque (compare with the ~30° of a split-phase motor). The closer the phase split to 90°, the stronger the rotating field component and therefore the higher the starting torque.
As with the split-phase motor, a centrifugal switch disconnects the start winding (and its capacitor) at approximately 75% speed. The motor then runs on the single run winding as a conventional single-phase induction motor.
The two rotating field components (from the two winding currents) combine. Maximum resultant rotating field (and therefore maximum torque) occurs when the two mmf vectors are 90° apart in both space and time. The run winding and start winding are wound 90° apart in the stator slots (spatial). The capacitor provides the ~90° time displacement. This "two-phase" starting principle gives starting torques of 250–350% of full-load torque — sufficient for compressors and refrigerators starting against back-pressure.
| Parameter | Capacitor-start | Advantage over split-phase |
|---|---|---|
| Phase displacement | ~90° | Approaches ideal — maximum starting torque |
| Starting torque | 250–350% of full-load torque | ~2× the starting torque of split-phase |
| Starting current | 500–700% of full-load current | Slightly lower than split-phase |
| Applications | Compressors, refrigerators, air-conditioning, hard-starting loads | Required where high starting torque is essential |
| Reversal method | Reverse start OR run winding connections — not both | Same rule as split-phase |
Capacitor-Start, Capacitor-Run Motor
A variant where a second (smaller) capacitor remains in circuit after the centrifugal switch opens. The run capacitor improves the running power factor and efficiency (giving more nearly a two-phase operation at full speed). Used in higher-efficiency applications such as commercial refrigeration, air-source heat pumps, and high-efficiency fan motors.
A refrigerator compressor motor hums but fails to start. Likely causes: (1) failed starting capacitor (open-circuit) — motor has run winding energised but no starting torque; (2) seized centrifugal switch in open position — start winding not connected; (3) low supply voltage — insufficient torque to start against compressor back-pressure. Diagnosis: measure start capacitor with capacitance meter; check centrifugal switch continuity with rotor stationary; measure supply voltage. A failed start capacitor is the most common cause.
Universal (AC/DC) Motor
The universal motor is constructed almost identically to a dc series-wound motor — the armature and field windings are connected in series, and a commutator and brushes transfer current to the rotating armature. On ac supply, both the armature current and the field current reverse simultaneously on each half-cycle. Since torque ∝ Φ × I_a, and both Φ and I_a reverse together, the torque always acts in the same direction — the motor runs continuously in one direction on ac.
The laminated stator core (essential for ac operation) reduces eddy current losses that would be severe in a solid iron core at mains frequency.
| Parameter | Universal motor | Rationale |
|---|---|---|
| Speed range | 8,000–30,000 rpm (no load) | No fixed synchronous speed — speed rises with decreasing load like dc series motor |
| Starting torque | Very high (series characteristic) | High current at start, same series field behaviour as dc series motor |
| Speed on no-load | Very high — potential runaway | Series characteristic: I falls → Φ falls → speed rises without limit. In practice the fan/friction load limits the maximum speed |
| Supply | ac or dc (hence "universal") | Useful for battery-operated tools that also work on mains |
| Maintenance | Brushes and commutator wear — periodic replacement needed | Brush life typically 500–1000 hours; more frequent on ac due to sparking |
| Applications | Vacuum cleaners, food mixers, hand drills, jigsaws, circular saws | High speed-to-weight ratio, compact, cheap, high starting torque |
Power tool speed control: a triac (semiconductor switch) is used to chop the ac waveform — by varying the firing angle, the average voltage across the motor is varied, changing speed. At full trigger angle (full voltage) the drill runs at maximum speed. At half voltage the speed is approximately halved. This is why variable-speed drills have a trigger that varies the triac firing angle — the same principle as a light dimmer, but rated for motor loads.
Why Motor Starters Are Necessary
A three-phase cage-rotor induction motor connected directly to the supply at standstill draws a starting current of 6–10 times its full-load current. This high inrush occurs because at standstill the rotor is stationary, slip = 1 (100%), and the motor behaves like a short-circuited transformer — very low impedance. The problems caused by high starting current include:
- Voltage dips: large starting current causes a momentary volt drop on the supply — enough to dim lights, trip sensitive equipment, and cause other running motors to lose torque. Distribution Network Operators (DNOs) specify maximum voltage dip limits.
- Mechanical stress: the high starting torque (jerk) can damage couplings, gearboxes, and driven equipment, especially if the connected load has high inertia.
- Thermal stress on motor windings: high current produces I²R heating — repeated starting cycles can overheat the windings if the motor is frequently started.
All motor starters must provide the following functions:
| Function | Component | Regulation / Rationale |
|---|---|---|
| Safe connection and disconnection of motor | Contactor (main contacts) | Contactor rated for motor duty (AC3 utilisation category) |
| Overload protection (sustained overcurrent) | Thermal overload relay or electronic overload | BS 7671 Regulation 552 — all motors >0.37 kW must have overload protection |
| Starting current limitation (larger motors) | Star-delta, soft start, autotransformer, or rotor resistance | Required where starting current would cause unacceptable voltage dip |
| No-volt (undervoltage) protection | Hold-on (self-latching) contact in contactor circuit | Prevents unexpected automatic restart — critical safety requirement |
| Short-circuit protection | Fuses or MCB upstream of starter | Starter contactors are not rated for fault current interruption |
Direct-On-Line (DOL) Starter
The DOL starter connects the motor directly to the full supply voltage. It is the simplest, cheapest, and most common starter for small motors and for large motors where the supply can absorb the starting current without causing unacceptable voltage dips. DNOs typically permit DOL starting for motors up to 7.5 kW without prior approval on standard domestic/commercial supplies.
Typical starting characteristics:
- Starting current: 6–10 × full-load current
- Starting torque: ~150% of full-load torque
- Starting time: typically 2–10 seconds for a lightly loaded motor
Control circuit sequence:
- Press Start (normally-open pushbutton): energises the contactor coil via the overload contact and stop button
- Main contacts close: motor receives full supply voltage and accelerates
- Hold-on (auxiliary) contact closes in parallel with start button: maintains coil energised when start button is released
- Press Stop (normally-closed pushbutton): breaks the control circuit, de-energises the coil, main contacts open, motor stops
- Supply failure: contactor de-energises, hold-on contact opens. Motor will NOT restart when power returns — operator must press Start again (no-volt protection)
- Overload trip: overload relay opens its contact in the control circuit, de-energising the contactor. Must be reset manually after the motor cools
A 7.5 kW, 400 V three-phase motor (full-load current 15 A, PF 0.85) starts DOL. Starting current at 7× full-load:
\[I_\text{start} = 7 \times 15 = 105\text{ A}\]This 105 A demand for 3–5 seconds causes a voltage dip on the local supply. If the supply transformer impedance is 4% and rated for 100 kVA, the voltage dip caused by the 105 A inrush can be estimated — on small rural transformers this may be unacceptably large, requiring a reduced-voltage starter.
Three emergency stop stations and two start stations are needed for a conveyor. Wiring rule: all stop buttons (normally-closed) in series — any one stop breaks the circuit. All start buttons (normally-open) in parallel — any one start closes the circuit. Only three cores are needed to the remote panel: common, start parallel chain, stop series chain.
Star-Delta (Y-Δ) Starting
The star-delta starter is the most widely used reduced-voltage starting method for motors between 7.5 kW and approximately 30 kW. The motor windings (which must be delta-connected for normal running on 400 V) are temporarily connected in star during starting, then switched to delta at near full speed.
Why star connection reduces voltage and current: In star, each winding receives the phase voltage (400/√3 = 230 V) — only 58% of the 400 V it would receive in delta. Since current ∝ voltage in a linear impedance, and since current is proportional to voltage at start (the motor is almost a pure impedance at standstill), the winding current is 58% of DOL. However, the LINE current is the same as winding current in star, whereas in delta I_L = √3 × I_winding. The combined effect:
| Parameter | DOL (delta direct) | Star-delta (star start) | Ratio |
|---|---|---|---|
| Winding voltage | 400 V | 230 V (= 400/√3) | 58% of DOL |
| Winding current | I_dol | I_dol/3 | 33% of DOL |
| Line (supply) current | √3 × I_dol | I_dol/3 | 33% of DOL |
| Starting torque | T_dol | T_dol/3 | 33% of DOL |
DOL winding voltage = 400 V, winding current = I. In star: winding voltage = 230 V, winding current = I × (230/400) = 0.575 I. Since torque ∝ V²: torque in star = (230/400)² × T_DOL = (1/√3)² × T_DOL = 1/3 × T_DOL. This confirms the factor of 1/3 for both current and torque in star connection.
An 18.5 kW conveyor motor uses an automatic star-delta starter with a 10-second timer. Start sequence: (1) Start pressed → main contactor + star contactor energise simultaneously → motor starts in star at 1/3 DOL current. (2) After 10 s, timer output de-energises star contactor and energises delta contactor → motor now at full 400 V delta. (3) Hold-on contact maintains main contactor. The 10-second star period must be sufficient for the lightly loaded conveyor to reach near-synchronous speed — switching to delta while still at low speed would produce a current surge almost as large as DOL starting.
Limitation: Star-delta starting gives only 1/3 of DOL starting torque. If the motor must start against significant load (e.g. a loaded conveyor or a pump against full head), the reduced torque may be insufficient to accelerate the load — the motor stalls in star and overheats. Star-delta is only suitable for light-load starting.
Rotor Resistance Starting (Wound-Rotor Slip-Ring Motor)
This method is used exclusively with wound-rotor (slip-ring) induction motors. External resistors are connected via the slip rings to the rotor circuit. By Thevenin's theorem, adding resistance to the rotor circuit shifts the maximum torque point to occur at higher slip (lower speed) — the motor can produce maximum torque at starting (slip = 1) rather than at near-synchronous speed.
Starting sequence: All rotor resistance in circuit at start → high torque, limited current. As motor accelerates, resistance is progressively cut out in steps (manually or automatically by a time-stepping controller). At full speed, all resistance is shorted out — the motor runs with its natural low-slip characteristic.
A 45 kW slip-ring motor drives a hoist. Starting with full rotor resistance gives starting current ≈ 150% of full-load current (compared to 600–800% for DOL on a cage motor) while delivering 200% of full-load torque — enough to start the fully loaded hoist. As the hoist accelerates, resistance is cut in five steps over 15 seconds. At full speed, slip rings are short-circuited and the motor runs as a standard induction motor at 4% slip.
Solid-State Soft-Start Controllers
A soft starter uses three pairs of back-to-back thyristors (SCRs), one pair per phase, connected in series with the supply lines. By controlling the firing angle of the thyristors (phase-angle control), the effective voltage applied to the motor terminals is varied from a preset pedestal voltage (typically 30–40% of supply) up to full voltage over a programmable ramp time.
What the soft starter does:
- Ramps up voltage smoothly from the pedestal to full voltage → smooth mechanical acceleration with limited torque surge
- Limits starting current to a programmable level (typically 2–4 × full-load current)
- Can also perform soft stop — ramping down voltage for smooth deceleration (important for pump systems to avoid water hammer)
- Provides overload protection, phase-loss detection, and thermistor input for motor thermal protection
Advantages over star-delta:
- No transition current surge when switching from star to delta
- Programmable acceleration time and current limit
- Works with all motor types (not limited to motors wound for star-delta)
- Compact — no multiple contactors required
- Soft stop capability
A 22 kW pump motor (full-load current 42 A) is started via a soft starter programmed for: pedestal voltage 40%, ramp time 8 seconds, current limit 300% (126 A). Starting sequence: thyristors fire at high phase angle (low voltage) → pedestal voltage applied → ramp begins → voltage increases linearly over 8 s → full voltage reached → motor running at full speed. The current never exceeds 126 A (3 × full-load) — compared with up to 420 A DOL. The pump accelerates smoothly, eliminating water-hammer pressure surges in the pipework.
| Starter type | Starting current | Starting torque | Best application | Typical cost |
|---|---|---|---|---|
| DOL | 6–10 × FLC | ~150% FLT | Small motors ≤7.5 kW, light loads | Lowest |
| Star-delta | 2–4 × FLC (1/3 DOL) | ~50% FLT (1/3 DOL) | Medium motors, light-load starting | Low-medium |
| Rotor resistance | ~1.5 × FLC | Up to 200% FLT | Heavy loads, slip-ring motors only | Medium |
| Soft start | 2–4 × FLC (programmable) | Programmable | Pumps, conveyors, fans — smooth start/stop needed | Medium-high |
| VFD | <1.5 × FLC | Up to 150% FLT | Where speed control is also needed | Highest |
Why Speed Control Matters
Controlling the speed of a motor-driven machine is often more important than the motor itself. Fans, pumps, and compressors have power consumption that varies with the cube of speed (P ∝ n³ for centrifugal machines) — running a pump at 80% of full speed reduces power consumption to approximately 0.8³ = 51% of full-load power. Over a year of operation, this energy saving typically pays for the VFD within 1–3 years. The VFD is now the dominant method of motor speed control in HVAC, water treatment, process industry, and manufacturing.
Factors Controlling Induction Motor Speed
From the synchronous speed formula \(N_s = (f/p) \times 60\), motor speed depends on:
| Factor | Method of control | Practical use |
|---|---|---|
| Number of poles (p) | Pole-changing (two-speed motor with separate windings) | Two fixed speeds only — fans, machine tools. No intermediate speeds. |
| Supply frequency (f) | Variable Frequency Drive (VFD/inverter) | Smooth, continuous speed control over wide range — by far the most common modern method |
| Supply voltage | Soft starter (limited range) | Useful for starting only — significant speed control by voltage alone wastes energy in the motor and gives poor torque |
| Load (slip) | Rotor resistance (wound-rotor motors only) | Limited range, inefficient — energy wasted in resistors. Only used where VFD not viable. |
How a Variable Frequency Drive Works
A VFD (also called an inverter drive, AC drive, or frequency converter) converts the fixed-frequency mains supply (50 Hz, 400 V) into a variable-frequency, variable-voltage output that feeds the motor. It operates in three stages:
- Rectifier (AC → DC): A diode bridge rectifier converts the 400 V, 50 Hz three-phase supply to a dc voltage (approximately 565 V dc for a 400 V ac input). This dc bus is smoothed by large electrolytic capacitors.
- DC bus: The smoothed dc bus acts as an energy reservoir. It also allows the VFD to accept regenerated energy from the motor during braking.
- Inverter (DC → variable-frequency AC): Six IGBT (Insulated Gate Bipolar Transistor) switches are switched on and off rapidly (typically 2–16 kHz carrier frequency) using Pulse Width Modulation (PWM). By varying the PWM pattern, the inverter synthesises a quasi-sinusoidal output voltage at any frequency from near-zero to 120 Hz or above.
V/f (Voltage-to-Frequency) Control
To maintain constant motor torque across the speed range, the VFD must maintain a constant ratio of voltage to frequency (V/f = constant). This is because the motor's flux is proportional to V/f — reducing frequency without reducing voltage would cause the core to saturate. Below base speed (50 Hz), the VFD reduces voltage proportionally with frequency. Above base speed (field weakening region), voltage is held constant at maximum while frequency continues to increase — the motor flux weakens and maximum torque falls.
A 400 V, 50 Hz motor. V/f ratio at base speed = 400/50 = 8 V/Hz. At 25 Hz (half speed), VFD output voltage = 25 × 8 = 200 V. At 75 Hz (field weakening), output voltage = 400 V (maximum), flux reduced to 2/3 of rated.
A 30 kW fan runs at full speed 50% of the time and at 70% speed the other 50%. Power at 70% speed: \(P = 30 \times 0.7^3 = 30 \times 0.343 = 10.3\text{ kW}\). Annual energy without VFD: \(30 \times 8760 = 262800\text{ kWh}\). With VFD: \((30 \times 4380) + (10.3 \times 4380) = 131400 + 45114 = 176514\text{ kWh}\). Energy saved: \(262800 - 176514 = 86286\text{ kWh/year}\). At 28 p/kWh = £24,160/year saving.
Advantages of VFD Speed Control
| Advantage | Explanation |
|---|---|
| Smooth, continuous speed control | Any speed from near-zero to above synchronous — not limited to fixed steps |
| Accurate speed control | Speed set to within ±0.5% with open-loop V/f control; ±0.01% with closed-loop encoder feedback |
| Very low starting current | Typically <1.5 × full-load current — eliminates supply voltage dip problems |
| Controlled acceleration and deceleration | Programmable ramp times prevent mechanical shock and reduce maintenance |
| Speeds above 50 Hz (field weakening) | Motor can run to 100 Hz or above — useful for high-speed machining |
| Three-phase motor from single-phase supply | Single-phase input rectifier + three-phase inverter output — useful in locations without three-phase supply |
| Regenerative braking | During deceleration, motor acts as generator; energy returned to dc bus and (with brake resistor or active front-end) to mains — reduces energy costs on cranes, lifts, downhill conveyors |
| Reduced mechanical wear | Soft starting eliminates belt slippage, coupling shock, and water hammer in pumping systems |
| Process optimisation | Speed can be matched exactly to process requirements rather than using throttling valves or dampers |
Limitations and Installation Considerations
| Limitation | Explanation and mitigation |
|---|---|
| Harmonics | The diode rectifier draws non-sinusoidal current from the supply — generating 5th, 7th, 11th, and 13th harmonic currents. These cause transformer heating, neutral current overload, and interference. Mitigation: AC or DC line chokes, 12-pulse rectifiers, active front-end drives. BS EN 61000-3-2/3-12 compliance required. |
| Motor insulation stress | PWM switching causes high-frequency voltage spikes at the motor terminals (dV/dt effect). Standard motor insulation may degrade over time. Mitigation: use inverter-duty rated motors or fit du/dt filters on the output. |
| Bearing current damage | Common-mode voltage from PWM can drive currents through motor bearings, causing pitting (fluting). Mitigation: shaft grounding rings, insulated bearings on larger motors. |
| Cogging at low speeds | At very low frequencies the PWM output approximates a stepped waveform rather than a smooth sinusoid — motor jerks. Mitigation: vector control drive (more sophisticated algorithm); encoder feedback. |
| Motor overheating at low speed | Standard motor fans are shaft-driven — at low speed, cooling airflow is reduced. Motor can overheat if run slowly for extended periods. Mitigation: independently-powered forced ventilation fan, or a motor rated for inverter duty with separately powered cooling. |
| Cable length limits | Long cables between VFD and motor increase voltage reflection and PWM spike amplitude. Recommended maximum without output filter: 50–100 m depending on drive make. Mitigation: output choke or sine-wave filter. |
| EMC and screening | VFDs generate significant electromagnetic interference (EMI). Screened cable must be used between VFD and motor. VFD must be correctly earthed and fitted with appropriate EMC filters. BS EN 61800-3 compliance required. |
A 22 kW, 400 V, 4-pole pump motor is to be speed-controlled by a VFD:
- Select VFD rated ≥ 22 kW (allow 10–20% margin if motor will run above 50 Hz)
- Fit AC line choke on input (reduces harmonic current and protects VFD from supply transients)
- Use screened (SY) cable between VFD output and motor — earthed at VFD end only (to avoid circulating currents). Maximum 50 m without output filter.
- Fit an EMC filter on VFD input if required by BS EN 61800-3 environment class
- Program ramp-up time (e.g. 10 s) and ramp-down time (e.g. 15 s) to match pump inertia and avoid water hammer
- Set minimum frequency ≥ 20 Hz to prevent motor overheating (or fit forced cooling)
- Enable motor thermistor input in VFD to protect motor at low speeds
Applications of VFDs
- HVAC fans and AHUs: Largest single application — variable airflow saves enormous energy in large buildings. BS EN 15232 building energy efficiency standards drive adoption.
- Pumps: Variable flow without throttling valves — major energy saving in water treatment, district heating, chiller circuits
- Compressors: Variable-speed compressors maintain constant pressure without load/unload cycling — more efficient, less mechanical wear
- Conveyors and production lines: Speed can be matched to product flow rate
- Machine tools and CNC: Precise spindle speed control without gearboxes
- Lifts and hoists: Active-front-end VFDs return regenerated energy to the supply during descent
- Electric vehicles: VFD technology (though at higher voltages) drives traction motors in EVs and hybrid vehicles
🃏 Chapter 8 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Covers DC machines, three-phase induction motors, single-phase motors, starters and VFDs.
📝 Chapter 8 — Multiple Choice Questions
Select your answer and click Submit. You have two attempts per question. First-attempt score shown at the end.
1A 230 V dc motor has an armature resistance of 0.5 Ω and a back-EMF of 218 V at full speed. What is the armature current at full speed?
2What is the function of the commutator in a dc motor?
3A dc series motor must never be run on no-load because:
4Which type of dc motor maintains an almost constant speed regardless of load and is best suited to machine tools and centrifugal pumps?
5In a three-phase cage-rotor induction motor, how is the rotor driven to rotate?
6A 4-pole induction motor is connected to a 50 Hz supply and runs at 1440 rev/min at full load. What is the percentage slip?
7A motor connected to a 50 Hz supply runs at approximately 960 rev/min at full load. How many poles does it have?
8A capacitor-start induction-run motor is used instead of a basic split-phase motor for a compressor because:
9A double-cage rotor motor resolves a fundamental design conflict in single-cage motors by:
10During the star phase of a star-delta starter, the voltage across each motor winding and the starting current are each reduced to what fraction of their direct-on-line values?
11A VFD maintains a constant V/f ratio when reducing motor speed. On a 400 V, 50 Hz motor running at half speed (25 Hz), the VFD output voltage should be:
12A centrifugal fan motor rated at 40 kW is run at 80% of full speed by a VFD. Approximately what power does the fan consume at this reduced speed?
Sends your Chapter 8 MCQ score summary from your registered email to your tutor.
Electrical Components
- Specify the main types of stated electrical components
- Explain the operating principles of relays, contactors and solenoids
- Describe overcurrent protection devices: fuses and circuit breakers
- Explain residual current devices (RCDs) and RCBOs — operation and applications
- State the advantages and disadvantages/limitations of each component type
Relays
A relay is an electrically operated switch — a small electrical signal energises a coil, creating a magnetic field that mechanically moves a set of contacts, thereby switching one or more separate circuits. The fundamental advantage is electrical isolation: the control circuit (which operates the coil) and the switched circuit (which the contacts control) share no electrical connection. This makes relays essential wherever a low-power signal must control a high-power load, or where circuits of different voltage levels or types (ac/dc) must interact.
Operating Principle
The relay coil (typically wound on an iron core to concentrate flux) is energised by the control signal. The resulting electromagnetic force attracts the armature (a pivoted iron lever) toward the pole-piece, overcoming a return spring. As the armature moves, it carries the moving contact(s) with it, either closing normally-open (N/O) contacts, opening normally-closed (N/C) contacts, or both simultaneously (changeover). When the coil is de-energised, the spring returns the armature and contacts to their rest position.
Operate voltage and drop-out voltage: The coil must be energised above the operate voltage (typically 80–85% of rated coil voltage) to attract the armature. The coil current can fall to a much lower level — the drop-out voltage (typically 10–20% of rated) — before the spring overcomes the residual magnetic force and releases the armature. This hysteresis gives relays stability against minor supply fluctuations.
Why Use a Relay? — Four Key Reasons
| Reason | Explanation | Practical example |
|---|---|---|
| Current amplification | A few milliamps in the coil switches amps (or hundreds of amps via contactors) in the load circuit | A PLC output (20 mA) controls a relay that switches a 230 V, 10 A lighting circuit |
| Voltage level conversion | Control circuit and load circuit can be at entirely different voltages | 24 V dc control system switching a 400 V ac motor circuit via a relay |
| Circuit isolation (safety) | No electrical connection between control and load circuits — coil failure cannot energise the load accidentally | ELV control wiring in a damp location switching a 230 V socket circuit |
| Logic functions | Multiple relay contacts can be combined to implement AND, OR, NOT logic without electronics | Conveyor interlock: motor only runs if both guard switches AND the start button are operated (AND logic using N/O contacts in series) |
Relay Contact Configurations
| Code | Full name | Description | Use |
|---|---|---|---|
| SPST N/O | Single-pole single-throw, normally open | One contact that closes when coil is energised | Simple on/off switching; hold-on contact in DOL starter |
| SPST N/C | Single-pole single-throw, normally closed | One contact that opens when coil is energised | Overload trip contact in motor starter control circuit |
| SPDT | Single-pole double-throw (changeover) | Common contact moves between N/O and N/C | Directing signal to one of two circuits; changeover between duty and standby pump |
| DPST | Double-pole single-throw | Two independent contacts both N/O or both N/C, operated simultaneously | Switching both L and N simultaneously for double-pole isolation |
| DPDT | Double-pole double-throw | Two changeover contacts operated simultaneously | Motor reversing relay — swapping two phases while maintaining third |
A 24 V dc relay has a coil resistance of 240 Ω. Coil current when energised:
\[I_{\text{coil}} = \frac{24}{240} = 0.1\text{ A} = 100\text{ mA}\]This 100 mA switches a 230 V, 6 A lighting circuit through the relay's N/O contacts — a current ratio of 60:1. The PLC output driving the relay coil is rated at 500 mA maximum, easily handling the 100 mA coil current.
A machine has two guard switches (A and B), both N/O. The machine motor contactor can only be energised if both guards are closed AND the start button is pressed. Logic: contactor coil circuit = Guard A (N/O) in series with Guard B (N/O) in series with Start button (N/O). This is relay AND logic — all three must be closed for the circuit to complete. If either guard opens, the contactor de-energises immediately and the motor stops.
Flyback Diode Protection
Relay coils are inductive loads. When the coil is de-energised, the collapsing magnetic field induces a large voltage spike (back-emf, Chapter 5) that can damage the transistor or PLC output driving the coil. A flyback diode (freewheeling diode) is connected in reverse-bias across the coil: during normal operation it is reverse-biased and carries no current; when the coil is switched off, the spike drives current through the diode in the forward direction, clamping the voltage to ~0.7 V and safely dissipating the stored energy. This is mandatory in all electronic relay drive circuits.
Solid-State Relay (SSR)
An SSR performs the same switching function as an electromechanical relay but uses semiconductor switching elements (triacs for ac loads, MOSFETs or IGBTs for dc loads) instead of mechanical contacts. Optocoupler isolation provides the same electrical isolation as a relay coil/contact arrangement.
| Feature | Electromechanical relay | Solid-state relay (SSR) |
|---|---|---|
| Moving parts | Yes — armature, contacts, spring | No — purely electronic |
| Switching speed | 5–15 ms | <1 ms (zero-crossing SSRs switch at the ac zero-crossing) |
| Noise / EMI | Contact bounce causes noise spikes | Silent switching; zero-crossing reduces EMI |
| Contact wear | Limited mechanical life (~10⁶ operations) | No wear — effectively unlimited life |
| Heat dissipation | Low (resistive contacts) | Higher — triac drops ~1 V in conduction; heat sink required |
| Short-circuit protection | Fuse external | Often none — semiconductor can be destroyed by overcurrent |
| Cost | Lower for simple applications | Higher, but no maintenance cost |
| Applications | General switching; motor contactors; infrequent operation | High-cycle-rate (temperature controllers, PID heating); where noise-free switching is essential |
Contactors
A contactor is a heavy-duty relay designed specifically for switching high-current power circuits repeatedly over millions of operations. While a relay might switch 10 A at 230 V, a contactor may switch hundreds of amperes at 400 V — the same electromagnetic principle but with much more robust construction.
Key constructional differences from a relay:
- Main contacts are silver-cadmium oxide or silver-tin oxide for long contact life and resistance to welding under high make/break currents
- Arc extinction chambers (arc chutes or arc splitters) fitted to each main contact pair — essential when breaking hundreds of amperes at 400 V ac
- The ac electromagnet uses a shading ring (a short-circuited copper ring on the pole face) to prevent contact chatter at 50 Hz — without it the 50 Hz alternating flux would attract and release the armature 100 times per second
- Three main contacts plus auxiliary contacts (N/O and N/C) for control circuit use
Contactor Utilisation Categories (BS EN 60947-4)
Contactors are rated for different types of load — the most demanding duty is switching inductive loads (motors) where breaking current can be much higher than the nominal load current:
| Category | Load type | Typical application |
|---|---|---|
| AC1 | Non-inductive or slightly inductive loads; resistance furnaces | Heating elements, incandescent lighting |
| AC2 | Slip-ring motors: starting and switching off | Starting/stopping slip-ring (wound rotor) induction motors |
| AC3 | Squirrel-cage motors: starting and switching off during running | Normal three-phase cage-rotor motor control — the most common category |
| AC4 | Squirrel-cage motors: starting, plugging, inching (jogging) | Cranes, hoists, pressing machines where motor is reversed or inched frequently |
A 15 kW motor (full-load current 30 A) is started DOL. Starting current = 7 × 30 = 210 A. An AC1-rated contactor may only be rated to interrupt 30 A (1× rated current). Under AC3 duty (making and breaking at up to 6× rated current = 180 A), the contactor must be specifically rated for this duty — an AC1 contactor on motor duty would suffer contact burning and welding within a few hundred operations.
Motor: 22 kW, 400 V, 42 A full-load current. Application: DOL starting (AC3 duty). Select a contactor rated at minimum AC3 42 A, coil voltage to match control circuit (e.g. 230 V ac or 24 V dc), with at least one auxiliary N/O contact (for hold-on) and one auxiliary N/C contact (for overload relay interlocking).
Solenoids and Electromagnets
A solenoid is a coil of wire wound in a helix (corkscrew shape). When current flows through it, a magnetic field is produced along the coil's axis. Winding the coil around an iron core greatly increases the flux produced (the core has much higher permeability than air) — this is an electromagnet.
The key property of an electromagnet is that it can be switched on and off by controlling the current — making it far more useful than a permanent magnet for industrial applications. The force produced is proportional to the square of the coil's MMF (NI)² and to the core material's permeability.
Applications of Solenoids and Electromagnets
| Device | How solenoid is used | Example |
|---|---|---|
| Contactor / relay | Solenoid coil attracts iron armature to close contacts | Motor starter contactor, control relay |
| Solenoid valve | Current through coil attracts plunger, opening or closing a valve port in a pipe | Gas shut-off valve in a boiler; hydraulic valve in a press; water control in a dishwasher |
| Magnetic overload trip | Excess current through the solenoid coil creates sufficient force to trip the mechanism; three separate solenoids (one per phase) in a three-phase DOL starter | Instantaneous overcurrent trip element in an MCCB |
| Electric lock / door strike | Energising the solenoid retracts the latch, allowing the door to open | Access control systems, fire-rated door holders |
| Lifting electromagnet | Large dc electromagnet lifts ferrous scrap or steel plates in a steelworks | Scrap metal handling cranes |
| Electrical bell / buzzer | Solenoid attracts iron armature which strikes the bell; this breaks the circuit, releasing the armature, which re-makes the circuit — oscillating at audible frequency | Trembler bell, door chime |
A 24 V dc solenoid valve coil has resistance 40 Ω and must produce a plunger force of 15 N. Coil current:
\[I = \frac{24}{40} = 0.6\text{ A}\]MMF = N × I. If the coil has 500 turns: MMF = 500 × 0.6 = 300 At. The electromagnetic force depends on the air gap and core dimensions, but the 300 At is sufficient to actuate most standard industrial solenoid valves. If the supply voltage drops to 18 V (75%), current falls to 0.45 A, MMF = 225 At — the solenoid may fail to open fully, which is why solenoid valves should be specified to operate down to 85% of nominal supply voltage.
BS 7671 Regulation Group 431 requires that overcurrent protective devices be installed to protect conductors from damage by excessive heating and electromechanical stresses. There are two distinct types of overcurrent that any protective device must deal with:
- Overload currents: Currents exceeding the circuit's rated current but flowing through conductors with no fault — e.g. too many appliances on a circuit, a motor running overloaded. These build up gradually and must be cleared before the conductor reaches its maximum permitted temperature. Overloads are typically 1.5–5× rated current.
- Short-circuit (fault) currents: Very large currents — potentially thousands of amperes — resulting from a fault between live conductors or between a live conductor and earth. These must be cleared within milliseconds to prevent conductor and insulation damage.
A good protective device must respond appropriately to both types: slow enough not to trip on harmless transients (motor starting surge), fast enough to clear faults before damage occurs.
Key Terminology
| Term | Definition | Significance |
|---|---|---|
| Current rating (In) | Maximum continuous current the device can carry without deteriorating | Must be ≥ design current of circuit (Ib) and ≤ cable current-carrying capacity (It) |
| Fusing current (If) | Minimum current that will cause the fuse element to melt and interrupt the circuit | Must be reached within the permitted disconnection time |
| Fusing factor (FF) | Ratio of fusing current to current rating: FF = If / In | Lower fusing factor = closer protection = device operates at current closer to its rating |
| Breaking capacity | Maximum fault current the device can safely interrupt without self-destruction | Must exceed the prospective short-circuit current (PSCC) at the point of installation |
| Prospective short-circuit current (PSCC) | The maximum fault current that could flow if a zero-impedance fault occurred at that point | Determined by supply impedance; must be measured or calculated at design stage |
Why a Lower Fusing Factor Gives Better Protection
A fusing factor of 2 means the fuse only operates when current reaches 2× its rated current — an overloaded cable could overheat significantly before the fuse acts. A fusing factor of 1.25 means the fuse trips at only 25% above rated current, catching overloads much earlier and protecting the cable more closely.
| Device type | Standard | Typical fusing factor | Protection quality |
|---|---|---|---|
| Semi-enclosed rewireable fuse | BS 3036 | 1.5–2.0 | Poorest — trips at up to 2× rated current |
| Cartridge fuse (consumer unit) | BS 88-3 (BS 1361) | 1.25–1.75 | Better than rewireable |
| 13 A plug-top fuse | BS 1362 | ~1.5 | Moderate — specific to appliance protection |
| HBC industrial fuse | BS EN 60269 / BS 88 | <1.5 (often <1.3) | Best — closest protection; up to 80 kA breaking capacity |
| Miniature circuit breaker (MCB) | BS EN 60898 | Effectively 1.0–1.45 (current-sensing) | Very good — tripping characteristic is calibrated |
15 A BS 3036 (fusing factor 2.0): \(I_f = 2.0 \times 15 = 30\text{ A}\) — the cable carries 30 A before the fuse acts.
15 A BS 88-3 (fusing factor 1.5): \(I_f = 1.5 \times 15 = 22.5\text{ A}\) — much closer protection; trips at only 7.5 A over the rated current.
Consequence: a 15 A BS 3036 fuse protecting a cable rated at 15 A allows the cable to carry 30 A — double its rating — before the fuse blows. The cable would be dangerously hot. A BS 88-3 fuse would act at 22.5 A, giving far better protection.
BS 7671 Table 53.1 requires that when semi-enclosed fuses (BS 3036) are used, the cable's current-carrying capacity must be at least 1.45/0.725 = 2× the fuse rating, because of the high fusing factor. Alternatively, BS 7671 Regulation 433.1.201 states that for BS 3036 fuses, the cable current-carrying capacity It ≥ 1.45 × If / fusing factor ≥ 1.45 × rated current × fusing factor = effectively specifying a larger cable. This is why modern installations use MCBs or HBC fuses instead of rewireable fuses — they allow smaller, less expensive cables.
Breaking Capacity
The breaking capacity is the maximum fault current a device can safely interrupt. During a short circuit, fault current can reach thousands or tens of thousands of amperes for the first few milliseconds. A device with insufficient breaking capacity will be destroyed attempting to interrupt this current — potentially causing fire, explosion, or injury from expelled hot arc products.
The breaking capacity of any overcurrent device must exceed the prospective short-circuit current (PSCC) at the point of installation. PSCC is measured using a loop impedance tester or calculated from the supply transformer impedance. Typical PSCC values:
| Location | Typical PSCC | Required device breaking capacity |
|---|---|---|
| Consumer unit (domestic, 230 V) | 1–6 kA | MCB rated 6 kA (most consumer CBs) |
| Sub-distribution board (commercial) | 3–10 kA | MCB rated 10 kA or MCCB |
| Main LV distribution board (industrial) | 10–50 kA | HBC fuses or MCCBs rated ≥50 kA |
| 11 kV/LV transformer secondary (400 V) | Up to 25 kA | ACB or HRC fuses rated accordingly |
A fuse is the simplest overcurrent protective device: a conductor with a deliberately small cross-sectional area installed in series with the circuit. When excessive current flows, \(P = I^2R\) heating in the fuse element causes it to melt — opening the circuit. Once blown, the fuse must be replaced. The fuse's characteristics (time-current curve) are determined by the element material, shape, and size, and the filler material surrounding the element.
Types of Fuse — Detailed Comparison
| Type | Standard | Construction | Advantages | Limitations |
|---|---|---|---|---|
| Semi-enclosed (rewireable) | BS 3036 | Tinned copper or zinc alloy wire element stretched between two screw terminals in a porcelain or Bakelite holder. Element is visible and replaceable. | Very cheap; easily repaired on-site with replacement wire; easy to see when blown; no stock of spares needed (wire available) | High fusing factor (up to 2.0) — poor overload protection; easily abused by fitting wrong wire gauge; element deteriorates by oxidation over years; unpredictable operation at high fault currents (element may scatter molten metal); must be derated by factor per BS 7671 requiring larger cables |
| Cartridge (consumer unit) | BS 88-3 / BS 1361 (consumer unit) BS 1362 (plug top) | Silver or tin alloy element in a ceramic tube filled with silica sand (quartz). The sand quenches the arc and absorbs the energy when the element blows. | Lower fusing factor (1.25–1.75); better short-circuit performance than rewireable; element enclosed — no scatter risk; difficult to abuse (incorrect rating immediately obvious by physical size difference); cheap replacement | Cannot be repaired — must replace whole cartridge; not immediately obvious if blown; spare stock must be maintained; not suitable for very high fault currents |
| High Breaking Capacity (HBC / HRC) | BS EN 60269 (BS 88) parts 1, 2, 4, 6 | Multiple shaped silver strip elements inside a high-strength ceramic body packed with silica sand. Precise manufacturing gives accurate, consistent time-current characteristics. | Very low fusing factor (<1.5, often <1.3); can safely interrupt up to 80–100 kA fault current; consistent, accurate, repeatable operation; silver element gives 'avalanche' operation; suitable for discrimination; colour-coded by rating | Expensive to purchase; must maintain stock of all required sizes; must replace with IDENTICAL type and rating (different manufacturers' fuses may have different time-current characteristics even at the same rating); cannot be repaired |
HBC Fuse Operation — Avalanche Effect
HBC fuses use multiple shaped silver strip elements rather than a single wire. Each strip has a narrow waist section with reduced cross-sectional area. Under overload, the element melts progressively at the waist. Under short-circuit conditions, the extremely rapid I²R heating causes the elements to vaporise almost simultaneously — current limiting action cuts off the fault current before it reaches its prospective peak. This is the fuse's most important protective function: it limits the actual peak current that flows, protecting downstream equipment (contactors, cables, motors) from the mechanical and thermal stresses of the full fault current.
A fault occurs on a 400 V circuit with PSCC of 20 kA. An HBC fuse (BS 88) rated 100 A operates in 5 ms (half-cycle at 50 Hz). The peak prospective fault current (at the first current peak) would be \(20000 \times \sqrt{2} = 28.3\text{ kA}\). The HBC fuse operates before this peak is reached, limiting the actual let-through current to perhaps 8–10 kA. Downstream equipment (cables, motor windings, busbars) is protected from the full 28 kA peak — a major safety advantage over rewireable or standard cartridge fuses.
HBC Fuse Utilisation Categories
- gG (formerly gL): General purpose — for protection of cables and conductors in all circuits except motor circuits. The 'g' means the fuse provides protection in the overload range AND the short-circuit range (full-range fuse).
- gM: Motor circuit — has a higher time-current characteristic to tolerate motor starting current surges (6–10× FLC for several seconds) without blowing during normal starting, while still protecting the cable and motor against sustained overload and fault.
- aM: Partial-range motor — protects only against short circuits (not overload); must be used with a separate overload relay (thermal overload in a motor starter).
A 15 kW motor (FLC = 30 A) starts DOL with a starting current of 7 × 30 = 210 A for 5–8 seconds. A gG fuse rated at 32 A would blow under this starting current (210 A >> 32 A × 1.3 fusing factor = 41.6 A). A gM fuse has a higher short-time withstand current — it will tolerate 210 A for 8 seconds without blowing, then protect the cable from sustained overload. Alternatively, an aM fuse (partial range) of larger rating can be used paired with the starter's thermal overload relay which handles the overload protection role.
Fuse Selection Rules (BS 7671 Regulations 432.1, 533.1)
- Rated current In ≥ design current Ib (fuse must carry the full load continuously)
- Rated current In ≤ current-carrying capacity of cable It (fuse must blow before the cable overheats)
- Breaking capacity ≥ PSCC at the point of installation
- For overload protection: \(I_f \leq 1.45 \times I_t\) (the effective tripping current must not exceed 1.45 × cable rating)
Design current Ib = 20 A. Select a 20 A BS 88-2 gG fuse (In = 20 A). Check: \(I_n = 20\text{ A} \geq I_b = 20\text{ A}\) ✓. \(I_n = 20\text{ A} \leq I_t = 27\text{ A}\) ✓. Fusing current \(= 1.45 \times 20 = 29\text{ A} \leq 1.45 \times 27 = 39.15\text{ A}\) ✓ — all conditions met. If a 25 A fuse were chosen instead: \(I_n = 25\text{ A} \leq I_t = 27\text{ A}\) ✓ but effective tripping current = 1.45 × 25 = 36.25 A — still ≤ 39.15 A, so this also complies.
Circuit breakers are electromechanical switches that open automatically when overcurrent is detected, and can be reset and reused — unlike fuses which must be replaced after operation. Two main types are used in building electrical installations:
- Miniature Circuit Breakers (MCBs) — BS EN 60898: Used in consumer units and small distribution boards for final circuits up to typically 125 A. Current ratings: 6 A, 10 A, 16 A, 20 A, 25 A, 32 A, 40 A, 50 A, 63 A.
- Moulded Case Circuit Breakers (MCCBs) — BS EN 60947-2: Used in industrial distribution boards for larger currents (up to several thousand amps). Have adjustable trip settings and higher breaking capacities. May also be used as main switches.
Operating Principle — Thermal and Magnetic Trips
Most MCBs use two separate trip mechanisms working together:
| Trip mechanism | How it works | What it protects against | Typical response |
|---|---|---|---|
| Thermal (bimetal strip) | Current heats a bimetal strip (two metals with different expansion coefficients bonded together). As it heats, it bends and eventually trips the mechanism. The higher the overload current, the faster the bending. | Sustained overloads (1.13–1.45× rated current and above) | Minutes at 1.45× rated current; seconds at 2× rated current |
| Magnetic (solenoid) | A solenoid coil carries the circuit current. At high fault currents, the magnetic force becomes large enough to trip the mechanism without waiting for thermal effects. | Short-circuit currents — operates almost instantaneously | <10 ms at rated magnetic trip current |
MCB Trip Characteristics — Types B, C, D
Different applications require different levels of tolerance to short-duration current surges (motor starting, transformer magnetising inrush, capacitor charging). BS EN 60898 defines MCB types by their instantaneous magnetic trip threshold:
| Type | Instantaneous trip range | Typical applications | Rationale |
|---|---|---|---|
| Type B | 3–5 × In | Domestic final circuits, resistive loads, socket outlets (low inrush loads) | Trips quickly on even moderate overloads — best protection for cables but may nuisance-trip on motor starting |
| Type C | 5–10 × In | Commercial and light industrial circuits; fluorescent lighting (higher inrush); small motors | Tolerates moderate inrush currents without nuisance tripping |
| Type D | 10–20 × In | Heavy inductive loads — X-ray equipment, transformers, large motors, welding equipment | Tolerate very high inrush; cable must still be protected by thermal element; used where Type C would nuisance-trip on start |
A 3-phase 5.5 kW motor (FLC = 12 A) starts DOL with a starting current of 7 × 12 = 84 A. A Type B 16 A MCB would have an instantaneous trip at 3–5 × 16 = 48–80 A — the 84 A starting current would trip it immediately. A Type C 16 A MCB trips at 5–10 × 16 = 80–160 A — the 84 A starting current is near the lower end, risking nuisance tripping. A Type D 16 A MCB trips at 10–20 × 16 = 160–320 A — comfortably above the 84 A starting current, preventing nuisance tripping while still protecting against genuine short circuits above 160 A.
A 32 A Type B MCB must disconnect within 0.4 s (socket outlet circuit, 230 V TN supply). Maximum permitted ZS from BS 7671 Table 41.3 = 230 / (5 × 32) = 1.44 Ω. The measured ZS must be ≤ 1.44 Ω to ensure the magnetic trip (at 5× rated = 160 A) will operate within 0.4 s.
Advantages of MCBs over Fuses
| Advantage | Explanation |
|---|---|
| Resettable — no replacement stock | After tripping and fault clearance, simply reset. No need for spare fuses in stock. Significant advantage in remote locations. |
| Clear fault indication | Tripped MCB is immediately visible (handle in intermediate or down position); no need to check each fuse individually |
| Cannot be abused | Rating is fixed — unlike rewireable fuses where the element can be replaced with incorrect wire; the MCB cannot be uprated in the field |
| Dual protection in one device | Thermal and magnetic trips in a single unit provide both overload and short-circuit protection |
| Can act as isolator | MCBs can be manually switched off for isolation — fuses cannot (they must be removed from the circuit to isolate) |
| Type selection for inrush | B/C/D trip characteristics allow selection to suit the load's inrush current without changing cable size |
Limitations of MCBs
- More expensive than fuses for equivalent rating
- Lower breaking capacity than HBC fuses — standard MCBs are rated at 6 kA, 10 kA, or 25 kA. At high PSCC locations, HBC fuses are still preferred.
- Mechanical wear — contacts wear with each operation. An MCB that has interrupted many fault currents may need replacement even if it appears to operate correctly.
- May not discriminate as well as HBC fuses — two MCBs in series need careful selection to ensure the downstream device trips without the upstream device also tripping.
Arc Quenching in Circuit Breakers
When the MCB contacts separate under load or fault current, an electric arc forms between them. The arc can sustain the current even with a significant air gap. If not extinguished quickly, the arc generates intense heat, damages contacts, and may cause fire. Two arc-quenching methods are used:
- Magnetic arc blowout: The arc current creates a magnetic field that interacts with a permanent magnet or field coil to push the arc into a set of arc splitter plates (metal fins). Each splitter plate cuts the arc into shorter sections, each requiring a higher voltage to sustain — the supply voltage can no longer maintain the arc and it extinguishes. Used in MCBs and MCCBs.
- Contact separation distance: For lower-current MCBs (up to ~63 A at 230 V), simply opening the contacts wide enough extinguishes the arc by cooling and stretching it beyond its self-sustaining length.
Discrimination (Selectivity)
Discrimination (or selectivity) means that when a fault occurs, only the protective device closest (upstream) to the fault operates — all other devices in the system remain closed and their circuits remain energised. Without discrimination, a fault on a single socket outlet could trip the main incomer and black out the entire building.
Discrimination is achieved through careful coordination of:
- Current rating: upstream devices must have higher ratings than downstream devices (e.g. 100 A main fuse, 32 A final circuit MCB)
- Time-current characteristics: upstream devices must operate more slowly than downstream devices at any given fault current level
- Type: using HBC fuses upstream and MCBs downstream gives good discrimination because fuses are current-limiting (faster at high fault currents) while MCBs are not
A fault on a table lamp (plugged into a ring circuit socket) draws 30 A through a 13 A BS 1362 plug fuse and a 32 A Type B MCB (which protects the ring circuit). Both devices see 30 A. The plug fuse trips at ≈ 1.5 × 13 = 19.5 A — so at 30 A it will blow quickly. The 32 A MCB's thermal element takes longer to trip at 30 A (close to but below its rating). The plug fuse clears the fault first, leaving the ring circuit MCB intact and all other sockets energised. This is correct discrimination.
Device Selection Rules — Summary
When selecting an overcurrent protective device (fuse or MCB), BS 7671 requires:
- \(I_n \geq I_b\) — device rating ≥ circuit design current
- \(I_n \leq I_t\) — device rating ≤ cable current-carrying capacity (adjusted for installation conditions)
- \(I_2 \leq 1.45 \times I_t\) — effective operating current ≤ 1.45 × cable rating
- Breaking capacity ≥ PSCC at point of installation
- Device type appropriate for load (B/C/D for MCBs; gG/gM for fuses)
- Devices coordinated for discrimination at every level
Why RCDs are Needed — the Limitation of Overcurrent Devices
Fuses and MCBs protect cables against overcurrent — they operate when current is large enough to cause damaging heating. But a person receiving an electric shock may experience only 30–50 mA of current through their body — far too small to blow a fuse or trip an MCB (a 32 A MCB will not trip on 30 mA). A 50 mA earth-fault current is life-threatening but invisible to an overcurrent device.
An RCD detects current imbalance between line and neutral — which occurs when any current flows to earth (whether through a person, a fault in equipment, or damaged cable insulation). It can respond to earth leakage currents as low as 5 mA and disconnect the supply within 25–40 milliseconds — fast enough to prevent ventricular fibrillation from a 30 mA shock current.
Operating Principle of the RCD
All current that flows out of the supply on the line conductor must return on the neutral conductor (Kirchhoff's Current Law — in a healthy circuit there is no other path). The RCD exploits this by measuring both currents simultaneously using a toroidal iron core:
- The line conductor passes through the toroid, wound as the "line coil" (or simply threaded through).
- The neutral conductor passes through the same toroid in the opposite direction, wound as the "neutral coil".
- In a healthy circuit: \(I_L = I_N\) — equal and opposite currents produce equal and opposite fluxes that cancel exactly. Net flux in the toroid = zero. No voltage is induced in the search coil.
- During an earth fault: \(I_L = I_N + I_f\) — line current exceeds neutral current by the fault current I_f. The fluxes no longer cancel — a residual flux exists in the toroid. This induces a voltage in the search coil (sensing coil) wound on the toroid.
- The search coil voltage drives current through the trip coil (or electronic amplifier circuit). When the residual current reaches the rated sensitivity (e.g. 30 mA), the trip coil releases a powerful spring, forcibly opening the main contacts and disconnecting the circuit.
A person standing on damp ground touches the live terminal of a 230 V socket. Body resistance ≈ 1000 Ω. Current through person: \(I = 230/1000 = 230\text{ mA}\). This current flows: live terminal → person → ground → earth electrode → supply transformer neutral (via earth) — it does NOT return via the neutral conductor. The RCD detects the imbalance (I_L = I_N + 230 mA). A 30 mA RCD trips in <40 ms. At 230 mA the RCD trips almost instantaneously (<10 ms). The person receives a very brief shock before disconnection — much less than the 100 ms threshold for cardiac effects at this current level.
RCD Sensitivity Ratings (IΔn)
| Sensitivity (IΔn) | Response current | Required disconnection time | Applications |
|---|---|---|---|
| 6 mA | 6 mA residual | <300 ms | Medical locations (cardiac-sensitive patients); protection against cardiac effects at very low currents |
| 10 mA | 10 mA residual | <300 ms | High-risk locations; individual circuit protection where increased sensitivity is required |
| 30 mA | 30 mA residual | <300 ms (must not exceed 40 ms in practice per BS 7671) | Standard personal protection — mandatory for socket outlets ≤20 A (BS 7671 Reg 411.3.3); portable equipment outdoors; bathrooms; construction sites |
| 100 mA | 100 mA residual | <500 ms | Fire protection (not personal protection); detecting significant earth leakage before it causes overheating in cables or equipment |
| 300 mA / 500 mA | 300–500 mA residual | <1 s | Fire protection at distribution board / sub-main level; upstream in a discriminating RCD scheme |
When RCDs are Mandatory (BS 7671)
- All socket outlets ≤ 20 A in domestic and similar installations — Regulation 411.3.3 (mandatory since 17th Edition Amendment 3, 2015)
- Outdoor socket outlets used by portable equipment
- Circuits in bathrooms and shower rooms (Part 7, Section 701)
- Circuits in swimming pools and hot tubs (Section 702)
- Circuits in agricultural and horticultural premises (Section 705)
- Construction sites (Section 704) — all socket outlets
- TT earthing systems — where earth electrode resistance prevents sufficient fault current to trip overcurrent devices; RCD is the primary protection method
- Cables buried less than 50 mm deep without protection (mechanical damage risk)
RCD Types — AC, A, F, B
RCDs are classified by the type of residual current waveform they can detect:
| Type | Detects | Applications |
|---|---|---|
| Type AC | Sinusoidal ac residual current only | Simple resistive and inductive ac loads; not suitable where any dc component may be present |
| Type A | Sinusoidal ac AND pulsating dc residual current | Circuits supplying equipment with single-phase rectifiers (washing machines, dishwashers, EV chargers single-phase). The most common type in modern UK domestic installations. |
| Type F | Type A plus high-frequency residual currents | Circuits with frequency converters (VFDs) on single-phase supplies |
| Type B | All of Type F plus smooth dc residual current | Three-phase VFD circuits, EV charging points with three-phase connection, solar PV inverters; where dc fault currents can occur and would blind a Type A or AC device |
A standard Type AC RCD operates on sinusoidal ac earth fault currents. An EV charger contains a single-phase rectifier — if a fault develops in the rectifier, the earth fault current has a pulsating dc component. A Type AC RCD may not detect this correctly and fail to trip (the smooth portion of the dc fault current saturates the toroid core, preventing correct flux sensing). Type A RCDs are specified for Mode 2 and Mode 3 EV charging circuits; Type B for Mode 4 (three-phase rapid charging). This requirement is specified in BS 7671 Section 722 (Electric Vehicle Charging Installations).
Test Button and Regular Testing
The test button creates a controlled imbalance by connecting a resistor from the line conductor (upstream of the line coil) to the neutral conductor (downstream of the neutral coil). Current through this resistor bypasses the line coil, creating an artificial imbalance equal to approximately the rated sensitivity current. The trip operates — confirming the mechanical and electronic parts of the RCD are functioning.
BS 7671 Regulation 514.12.2 requires that a notice be fixed near every RCD, instructing users to test it quarterly. The test button check does NOT verify that the RCD will trip at exactly the rated sensitivity — only a calibrated RCD tester (loop impedance tester with RCD function) can do this during periodic inspection.
Nuisance Tripping — Causes and Solutions
RCDs can trip unexpectedly due to natural earth leakage currents in healthy equipment:
- All electrical equipment has some capacitance between live conductors and earth — this "leakage current" flows continuously. A 30 mA RCD protecting 10 socket circuits, each with equipment drawing 3 mA leakage, will see 30 mA total — right at the trip threshold and prone to nuisance tripping.
- Long cable runs have significant capacitance to earth — especially in industrial installations with metal conduit or screened cables.
- Inrush currents in motors and transformers create transient leakage.
Solutions: Use time-delayed RCDs upstream (S-type, which have 60–200 ms delay, coordinating with 30 mA instant RCDs downstream for discrimination); split circuits across multiple RCDs so total leakage on each device is limited; use higher-sensitivity (100 mA) devices at distribution level with 30 mA devices on individual circuits.
RCBO — Residual Current Circuit Breaker with Overcurrent Protection
An RCBO combines the functions of a 30 mA RCD and an MCB in a single DIN-rail-mounted device, occupying one module-width in a consumer unit. It provides:
- Earth fault protection (RCD function — 30 mA): trips on earth leakage currents ≥ 30 mA
- Overload protection (thermal element): trips after a time inversely proportional to the overload magnitude
- Short-circuit protection (magnetic element): trips instantaneously on fault currents above the magnetic threshold (Type B: 3–5×, Type C: 5–10×, Type D: 10–20×)
Main advantage over a shared RCD feeding multiple circuits: if a fault occurs on one circuit, only that RCBO trips — all other circuits remain energised. A single RCD protecting 8 circuits would disconnect all 8 on a fault on any one of them. In a household, this might extinguish all lights and power simultaneously.
Consumer unit wiring: The RCBO has a line connection from the busbar (same as an MCB), plus two flying leads: a blue (neutral) flying lead connected to the neutral terminal bar, and a cream/green-yellow (functional earth) flying lead connected to the earthing terminal bar inside the consumer unit. This double connection to the neutral bar is what allows the RCBO to compare line and neutral currents.
A 16-circuit consumer unit protects 16 final circuits. Option A: one 30 mA 80 A RCD protecting all 16 circuits. A fault on the kitchen socket circuit trips the RCD — all 16 circuits lose power including freezer, smoke alarms, security system, stair lights. Option B: 16 individual RCBOs (e.g. 16 A Type B 30 mA each). A fault on the kitchen circuit trips only that RCBO. All other 15 circuits remain energised. Option B costs more initially but provides much better selectivity and avoids complete loss of supply for a single fault. Modern best practice (18th Edition consumer units) favours RCBOs for final circuit protection.
- 30 mA RCDs are mandatory for all socket outlets ≤20 A in UK domestic installations (BS 7671 Reg 411.3.3)
- RCDs detect earth fault current — they do NOT protect against overloads or short circuits between L and N (the overcurrent device provides this)
- Test monthly (quarterly minimum per BS 7671 notice requirement)
- Type A RCDs required where equipment with single-phase rectifiers is used (washing machines, dishwashers, EV chargers)
- Type B required for three-phase VFDs and EV rapid chargers (can have smooth dc fault current)
- RCBO = RCD + MCB in one device; provides individual circuit selectivity
🃏 Chapter 9 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Covers relays, contactors, solenoids, overcurrent protection, fuses, circuit breakers and RCDs.
📝 Chapter 9 — Multiple Choice Questions
Select your answer and click Submit. You have two attempts per question. First-attempt score shown at the end.
1What is the fundamental advantage of using a relay to switch a load rather than connecting the control signal directly to the load circuit?
2A machine guarding circuit requires the motor to run only when both guard switch A AND guard switch B are closed. Using relay N/O contacts, this AND logic is achieved by connecting the two contacts:
3A contactor is used to start and stop a standard three-phase cage-rotor induction motor (DOL). Which BS EN 60947-4 utilisation category must it be rated for?
4Which fuse type can safely interrupt fault currents up to 80–100 kA by limiting peak let-through current before it reaches its prospective maximum — protecting downstream equipment from full fault energy?
5A motor circuit fuse must tolerate a starting current surge of 7× full-load current for several seconds without blowing, while still protecting against sustained overload and faults. Which HBC fuse category is correct?
6A Type D MCB has an instantaneous magnetic trip threshold of 10–20 × In. In which situation would it be specified in preference to Type B or Type C?
7Which is a genuine advantage of an MCB over a semi-enclosed (rewireable) fuse?
8Why will a standard 32 A MCB NOT protect a person receiving a 50 mA electric shock through a line-to-earth fault?
9A socket outlet circuit supplies a washing machine. Which RCD type does BS 7671 require, and why?
10What is the key advantage of fitting individual RCBOs to each final circuit rather than one shared RCD protecting all circuits?
11A fault on a table lamp (plugged into a ring circuit socket) draws 30 A — seen by both a 13 A plug fuse AND the 32 A Type B ring circuit MCB. What demonstrates correct discrimination?
12A flyback (freewheeling) diode is connected in reverse-bias across a dc relay coil driven by a transistor. What is its purpose?
Sends your Chapter 9 MCQ score summary from your registered email to your tutor.
Lighting Systems
- Explain the basic principles of illumination: lux, lumens, candelas, efficacy
- Apply the inverse square law and cosine law to illuminance calculations
- Apply the lumen method for interior lighting design
- Explain the operating principles and types of incandescent lamps
- Explain discharge lighting: fluorescent, HPMV, metal halide, sodium lamps
- Describe energy-saving lamps: CFL and LED principles and applications
Visible Light and the Electromagnetic Spectrum
Visible light is a narrow band of the electromagnetic spectrum — electromagnetic radiation that human eyes can detect. It sits between infrared radiation (longer wavelength, felt as heat) and ultraviolet radiation (shorter wavelength, causes tanning and germicidal effects). The visible range spans approximately 380 nm (violet) to 780 nm (red), with the full colour sequence: violet → blue → green → yellow → orange → red.
Different lamp types emit light across different portions of this spectrum, which affects how colours of objects and surfaces appear under that lighting — this is colour rendering. A lamp that only emits yellow light (like a low-pressure sodium street lamp) makes red cars appear brown and skin look grey. A lamp with a broad, balanced spectrum (like a high-quality LED) shows colours naturally. Colour rendering is specified by the Colour Rendering Index (CRI or Ra): Ra = 100 means perfect colour rendering (like daylight); Ra < 60 is poor.
Colour temperature (K — Kelvin) describes the "warmth" or "coolness" of white light:
- < 3000 K — warm white (yellow-red tones) — domestic, hospitality, restaurants
- 3000–4000 K — neutral white — offices, retail
- 4000–6500 K — cool white / daylight — industrial, healthcare, task areas
- > 6500 K — very cool "blue-white" — specialist applications
Lighting Quantities and Units
Five quantities describe different aspects of light — each measures something different, and confusing them is a very common examination error:
| Quantity | Symbol | Unit | What it measures | Analogy |
|---|---|---|---|---|
| Luminous Intensity | I | candela (cd) | The power of a light source in a specific direction — how concentrated the beam is | The strength of a fire hose jet — it measures pressure at the source |
| Luminous Flux | Φ | lumen (lm) | Total light output emitted by the source in all directions — the complete "quantity" of light | Total water flow from the fire hose in all directions |
| Illuminance | E | lux (lx) = lm/m² | The amount of light arriving at and falling on a surface — what a light meter measures | How wet the ground gets — water per unit area |
| Luminance | L | cd/m² | Light reflected or emitted from a surface in a particular direction — what the eye actually perceives as "brightness" | How much water bounces off a surface — depends on surface texture and colour |
| Luminous Efficacy | η (or K) | lm/W | Light output per watt of electrical input — the "efficiency" of the lamp as a light source | Miles per gallon for a car — useful output per unit energy |
A 100 W lamp emits 1,400 lm total flux. It is mounted 2 m above a desk. At the desk surface, the illuminance is measured as 350 lux. The lamp's intensity in the downward direction is 350 cd. The white desk surface reflects 80% of incident light, giving a luminance of approximately 280 cd/m² — which is what your eyes perceive as the desk's brightness. Efficacy = 1400/100 = 14 lm/W.
Recommended Illuminance Levels (SLL/CIBSE)
The Society of Light and Lighting (SLL) Code for Lighting specifies target illuminance values (in lux) for different tasks and environments:
| Environment / task | Target illuminance (lux) |
|---|---|
| Corridor, circulation areas | 100 |
| Staircase, storage | 150 |
| General office, classroom | 300–500 |
| Drawing office, detailed work | 500–750 |
| Precision engineering, surgery | 1000–2000 |
| Bright sunlight (outdoor) | ~100,000 |
| Overcast day (outdoor) | ~10,000 |
Luminous Efficacy — The Lamp Efficiency Measure
\[\text{Efficacy} = \frac{\text{Luminous flux output (lm)}}{\text{Electrical power input (W)}}\]| Lamp type | Efficacy (lm/W) | Relative energy use for same light |
|---|---|---|
| GLS tungsten filament | 10–18 | Baseline (100%) |
| Tungsten-halogen | 12–22 | ~80% of GLS |
| HP Mercury (HPMV/MBF) | 32–58 | ~40% of GLS |
| Fluorescent (MCF) | 60–78 | ~20% of GLS |
| HP Sodium (SON) | 55–120 | ~15% of GLS |
| LP Sodium (SOX) | 70–160 | ~10% of GLS |
| CFL (compact fluorescent) | 55–70 | ~20% of GLS |
| LED (modern, 2024) | 80–200+ | ~8–10% of GLS |
A 150 W tungsten-filament lamp produces 1,960 lm:
\[\text{Efficacy} = \frac{1960}{150} = 13.1\text{ lm/W}\]A factory needs 100,000 lm of light output (1,000 lx × 100 m²) for 4,000 hours/year. Energy consumed:
GLS (13 lm/W): power = 100,000/13 = 7,692 W → 7.692 kW × 4,000 h = 30,769 kWh/year
LED (150 lm/W): power = 100,000/150 = 667 W → 0.667 kW × 4,000 h = 2,667 kWh/year
LED saving vs GLS: 28,102 kWh/year. At 28p/kWh = £7,869/year. Modern LED installations pay back within 1–2 years in commercial applications.
Three Sources of Lighting from Electricity
All electric lamps convert electrical energy to light by one of three physical mechanisms, each with fundamentally different efficiency and characteristics:
| Source | Mechanism | Examples | Key characteristics |
|---|---|---|---|
| Incandescent | Electrical resistive heating of a tungsten filament to ~2500°C until it glows white-hot — thermal radiation | GLS, tungsten-halogen | Simple, cheap, good colour rendering; very inefficient (90%+ of energy becomes heat); being phased out by EU/UK regulations |
| Discharge (arc/gas) | Electrical discharge through a gas or vapour; electrons collide with gas atoms, exciting them to emit photons at characteristic wavelengths | Fluorescent, HPMV, metal halide, SOX, SON | Much more efficient than incandescent; requires control gear (ballast/choke); most types need warm-up time; some cannot restrike when hot |
| LED (solid-state) | Semiconductor p-n junction — electrons recombine with holes, releasing energy directly as photons (electroluminescence) | LED lamps, LED drivers, LED modules | Highest efficacy available; instant start; dimmable; very long life; no UV or IR in beam; requires electronic driver; rapidly falling in cost |
Inverse Square Law
When light radiates from a point source, it spreads outward in all directions — the same total flux illuminates a larger and larger area as distance increases. The area illuminated by a cone of light increases with the square of the distance (area of a sphere = 4πr²), so the illuminance (flux per unit area) must decrease by the inverse of the square of the distance:
\[E = \frac{I}{d^2}\]where E = illuminance (lux), I = luminous intensity in that direction (candela), d = distance from source to the illuminated surface (m). This formula applies when the light strikes the surface at right angles (perpendicular incidence).
Practical implication: doubling the mounting height of a luminaire reduces the illuminance at the working plane to one quarter. This is why high bay industrial luminaires must have much higher lumen output than low bay luminaires to achieve the same floor illuminance.
Doubling distance → illuminance falls by factor of 4. This is the inverse square law in action.
This confirms the self-assessment answer of 200 lux ✓.
A downlight must provide 400 lux at a working plane 2.5 m below it. Required intensity directly below:
\[I = E \times d^2 = 400 \times 2.5^2 = 400 \times 6.25 = 2500\text{ cd}\]This intensity figure is used to select an appropriate luminaire from a manufacturer's photometric data.
Cosine Law (for Oblique Light Incidence)
When light strikes a surface at an angle θ to the perpendicular (normal), the same flux is spread over a larger area (the beam footprint is elongated). The effective illuminance is reduced by the factor cos θ:
\[E = \frac{I}{d^2} \times \cos\theta\]where θ is the angle between the light direction and the normal (perpendicular) to the surface. Since \(\cos\theta = h/d\) (vertical height ÷ slant distance), the formula can also be written:
\[E = \frac{I \times h}{d^3} \qquad (\text{useful alternative form})\]Compare with the direct-below illuminance at the same height: \(E = 500/2^2 = 125\) lux. The off-axis illuminance (51.4 lux) is only 41% of the direct-below value — demonstrating the combined effect of greater distance and oblique incidence.
A 1000 cd lamp is mounted 3 m above the working plane. Find the illuminance at a point 4 m horizontally from directly below the lamp.
\[d = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5\text{ m}\] \[\cos\theta = \frac{3}{5} = 0.6\] \[E = \frac{1000}{5^2} \times 0.6 = \frac{1000}{25} \times 0.6 = 40 \times 0.6 = 24\text{ lux}\]Note: 3-4-5 is the classic Pythagorean triangle (Chapter 1) — recognising this avoids the need to calculate the square root.
Maintenance Factor (MF)
Over time, lighting installations deteriorate from their initial design values due to two simultaneous effects:
- Lamp lumen depreciation: All lamps lose light output as they age — typically 20–30% over their rated life due to filament evaporation (incandescent) or phosphor degradation (fluorescent, LED).
- Luminaire dirt depreciation: Dust, grease, and insects accumulate on luminaire optics and diffusers, reducing light transmission. The rate depends on the environment — clean offices vs. dirty foundries.
The maintenance factor (MF) accounts for both effects combined. It is always less than 1.0:
| Environment | Typical MF | Examples |
|---|---|---|
| Very clean (sealed luminaires, clean room) | 0.90–0.95 | Hospitals, clean manufacturing |
| Clean | 0.80–0.90 | Offices, shops, schools |
| Normal | 0.70–0.80 | General industrial, warehouses |
| Dirty | 0.60–0.70 | Metalworking, cement, chemical plants |
| Very dirty | 0.50–0.65 | Foundries, welding shops, mining |
The lighting design must provide sufficient initial illuminance that, after depreciation, the illuminance at end-of-maintenance-cycle still meets the target. This is why the lumen method divides by MF — it requires more lamps initially to compensate for future depreciation.
Utilisation Factor (UF)
Not all light leaving a lamp reaches the working plane usefully — some is absorbed by walls, ceiling, and luminaire components; some goes sideways or upward. The utilisation factor (UF) is the fraction of the total lamp flux that actually reaches the working plane:
\[UF = \frac{\text{Flux received on working plane}}{\text{Total flux emitted by lamps}}\]UF depends on several factors:
- Room geometry: expressed as the Room Index (RI) or Room Cavity Ratio: \(RI = \frac{L \times W}{H_m(L+W)}\) where L = room length, W = room width, H_m = height of luminaire above working plane. A larger RI (wider, lower room) gives a higher UF because more light reaches the floor directly.
- Reflectances: Light-coloured walls and ceilings reflect more light to the working plane — higher UF. Dark rooms need more lamps for the same illuminance.
- Luminaire type: Direct (all light downward) gives higher UF than indirect (light to ceiling then reflected down).
Manufacturers publish UF tables for their luminaires, indexed by room index and reflectance values. Practical UF values range from 0.1 (small, dark room, indirect luminaire) to 0.95 (large, white room, highly efficient direct luminaire).
Lumen Method — Number of Lamps Formula
The lumen method is the standard technique for interior lighting design. It calculates the number of lamps needed to achieve a target illuminance on the working plane, accounting for lamp output, room geometry, and depreciation:
\[N = \frac{E \times A}{F \times UF \times MF}\]where: N = number of lamps, E = required illuminance (lux), A = area of working plane (m²), F = initial luminous flux per lamp (lm), UF = utilisation factor, MF = maintenance factor.
The formula can be rearranged to check the achieved illuminance from a given number of lamps:
\[E = \frac{N \times F \times UF \times MF}{A}\]Factory: 72 m × 20 m, required illuminance E = 300 lux, UF = 0.4, MF = 0.75, each high-bay luminaire delivers F = 18,000 lm.
\[N = \frac{300 \times (72 \times 20)}{18000 \times 0.4 \times 0.75} = \frac{300 \times 1440}{5400} = \frac{432000}{5400} = 80\text{ lamps}\]Design the lighting for an open-plan office: 15 m × 10 m × 2.8 m high. Suspended luminaires at 2.5 m above working plane (0.3 m below ceiling, 0.3 m working plane height → H_m = 2.8 − 0.3 − 0.3 = 2.2 m). Required illuminance: 500 lux. Reflectances: ceiling 70%, walls 50%, floor 20%.
Step 1 — Room Index: \(RI = \frac{15 \times 10}{2.2 \times (15+10)} = \frac{150}{2.2 \times 25} = \frac{150}{55} = 2.73\)
Step 2 — UF from manufacturer's table (for RI ≈ 2.7, 70/50/20 reflectances): UF = 0.62 (typical value from a good direct fluorescent luminaire).
Step 3 — MF: clean office, lamps replaced every 3 years → MF = 0.80.
Step 4 — Lamp flux: each luminaire has two 36 W fluorescent tubes, each 3,250 lm → F = 2 × 3,250 = 6,500 lm per luminaire.
Step 5 — Number of luminaires:
\[N = \frac{500 \times 150}{6500 \times 0.62 \times 0.80} = \frac{75000}{3224} = 23.3 \approx 24\text{ luminaires}\]Step 6 — Arrange in grid: 24 luminaires in a 6 × 4 grid, spacing 15/6 = 2.5 m × 10/4 = 2.5 m. Check space-to-height ratio: SHR = 2.5/2.2 = 1.14 — typical recommended SHR for this luminaire type is ≤ 1.5, so ✓.
A corridor is to have 10 LED luminaires (each 1,800 lm), UF = 0.55, MF = 0.85 in a 12 m × 2 m space. Achieved illuminance:
\[E = \frac{10 \times 1800 \times 0.55 \times 0.85}{12 \times 2} = \frac{8415}{24} = 350.6\text{ lux}\]Target for a corridor is 100 lux (SLL). The design exceeds this — good. If the target were 500 lux (staircase with detailed tasks), the design would be insufficient and more luminaires would be needed.
Space-to-Height Ratio (SHR)
The lumen method calculates how many lamps are needed for average illuminance but does not guarantee uniform distribution. If lamps are spaced too far apart relative to their mounting height, the lighting will be uneven — bright patches under each luminaire with dark zones between them. The space-to-height ratio controls uniformity:
\[SHR = \frac{\text{Centre-to-centre spacing of luminaires (m)}}{\text{Height of luminaires above working plane (m)}}\]Each luminaire manufacturer specifies a maximum SHR (typically 1.0–1.5) above which uniformity becomes unacceptable. If the lumen method gives a number of lamps whose required spacing exceeds the maximum SHR × height, additional lamps must be added (increasing both number and cost) to achieve adequate uniformity.
An industrial workshop has luminaires 4 m above the working plane. The manufacturer specifies maximum SHR = 1.2. Maximum permitted spacing = 1.2 × 4 = 4.8 m. If the lumen method calculates 20 luminaires in a 24 m × 12 m room, a 5 × 4 grid gives spacing of 24/5 = 4.8 m × 12/4 = 3.0 m. The 4.8 m spacing in the long direction exactly meets the SHR limit ✓.
Lamp Designations
Lamp type codes appear on datasheets, luminaire labels, and maintenance schedules. Understanding them ensures correct lamp selection:
| Code | Meaning | Example |
|---|---|---|
| GLS | General Lighting Service — standard pear-shaped tungsten filament | GLS 60W BC |
| TH / QT | Tungsten-Halogen / Quartz-Tungsten | QT-LP 100W R7s (linear halogen) |
| PAR | Parabolic Aluminised Reflector — diameter in eighths of an inch | PAR38 = 38/8 = 4.75″ = 120 mm |
| MR | Multifaceted Reflector — precision faceted aluminium reflector | MR16 = 16/8 = 2″ = 51 mm, 12 V |
GLS Tungsten-Filament Lamp — Operating Principle
The GLS lamp operates on the incandescent principle: I²R heating raises a coiled tungsten filament to ~2500°C, at which point it radiates visible white light. Colour temperature ≈ 2700 K (warm white). CRI = 100 (perfect colour rendering). Efficacy: 10–18 lm/W — meaning over 90% of energy input is wasted as heat, not light.
Why tungsten? Tungsten has the highest melting point of any metal (3422°C), making it uniquely suitable for incandescence. Even so, it evaporates slowly at operating temperature, progressively thinning the filament until it breaks (typical life ~1000 hours).
- Vacuum-filled (<25 W): vacuum slows evaporation but limits temperature (~2000°C); lower efficacy (10–13 lm/W)
- Gas-filled (≥25 W): argon + nitrogen atmosphere allows higher temperature (~2500°C), better efficacy (13–18 lm/W), pearl coating diffuses light
A 60 W GLS lamp produces approximately 700 lm. Efficacy = 700/60 = 11.7 lm/W. Only ~5% of the electrical energy becomes visible light; 95% is radiated as infrared heat. This extreme inefficiency drove EU/UK regulations progressively banning high-wattage GLS lamps (from 2009). Non-directional mains-voltage incandescent lamps are now effectively banned for general sale in the UK under retained EU Regulation 244/2009 — meaning replacements for failed lamps in existing fittings must be LED or CFL.
Tungsten-Halogen Lamps — The Halogen Cycle
Tungsten-halogen lamps overcome the tungsten deposition problem of GLS lamps using a self-regenerative chemical cycle. A small amount of a halogen compound (typically a bromine compound) is added to the filling gas inside the compact quartz envelope:
- Tungsten evaporates from the hot filament (~3000°C)
- Near the hot quartz wall (>250°C), tungsten reacts with halogen to form tungsten halide
- Tungsten halide migrates back toward the filament
- Near the filament, high temperature decomposes the halide, redepositing tungsten back on the filament and releasing the halogen to repeat the cycle
Critical requirement: The quartz wall must be maintained above 250°C for the cycle to work — hence the small, closely-fitting quartz envelope. This makes tungsten-halogen lamps much hotter to touch than GLS lamps.
Skin oils deposited on the quartz envelope by fingerprints react chemically with the quartz at high temperature, causing localised crystallisation (devitrification) that weakens the envelope — potentially causing explosion. Always handle halogen capsule lamps with gloves or clean tissue. If touched, clean with isopropyl alcohol before switching on.
| Feature | GLS lamp | Tungsten-halogen |
|---|---|---|
| Envelope | Soft glass | Fused quartz |
| Filament temperature | ~2500°C | ~3000°C |
| Efficacy | 10–18 lm/W | 12–22 lm/W |
| Rated life | ~1000 h | 2000–4000 h |
| Colour temperature | ~2700 K | ~3000–3200 K |
| CRI | 100 | 100 |
| Applications | General domestic (now obsolete) | Display, accent, security, floodlighting |
Extra Low Voltage (ELV) Tungsten-Halogen Lamps
At 12 V, the same power requires ~19× more current than 230 V (P = VI: 35 W at 12 V = 2.9 A). The filament can therefore be shorter and thicker, giving a compact, intense arc-like source ideal for precision optical systems (spotlights, display lighting). A step-down transformer (electronic or wound, 230 V/12 V) is required. BS 7671 SELV requirements apply to the secondary wiring.
Six MR16 35 W lamps from one transformer. Total load = 6 × 35 = 210 W. Specify a 250 W transformer. Secondary current = 210/12 = 17.5 A — requiring 2.5 mm² cable even for short runs. This explains why 12 V halogen systems have been largely replaced by 230 V GU10 LED lamps — simpler wiring and no large secondary currents.
Lamp Caps and Bases
The cap determines the physical and electrical interface between lamp and luminaire. Correct cap identification is essential for specification and replacement:
| Code | Full name | Voltage | Typical lamp |
|---|---|---|---|
| BC / B22d | Bayonet cap, 22 mm, double-contact | 230 V | GLS, CFL, LED retrofit (UK standard) |
| SBC / B15d | Small bayonet cap, 15 mm | 230 V | Small candle/golf-ball lamps |
| ES / E27 | Edison screw, 27 mm | 230 V | GLS, LED (European/worldwide fittings) |
| SES / E14 | Small Edison screw, 14 mm | 230 V | Candle lamps in European fittings |
| GES / E40 | Goliath Edison screw, 40 mm | 230 V | SON, MBI high-wattage discharge lamps |
| GU10 | Bipin twist-and-lock, 10 mm spacing | 230 V | Halogen and LED downlights (mains voltage) |
| GU5.3 / MR16 | Bipin, 5.3 mm spacing | 12 V | 12 V halogen and LED MR16 lamps |
| R7s | Recessed contact, 7 mm spacing | 230 V | Linear double-ended halogen capsule |
The Discharge Principle
All discharge lamps work by passing an electric current through a gas or vapour. Electrons collide with gas atoms, raising them to excited energy states. When these atoms return to their ground state, they emit photons at specific wavelengths determined by the atomic structure of the gas — this is why sodium lamps emit characteristic yellow light and mercury vapour emits blue-green light. Because a gas offers negative resistance (current increases as voltage drops as the arc heats), all discharge lamps need a current-limiting device — the ballast.
The Inductor / Choke / Ballast
An inductor (choke) in series with the lamp provides two essential functions:
- Striking the arc: when the starter switch opens, the collapsing magnetic field induces a high-voltage surge (back-emf = L × ΔI/Δt) that ionises the gas, initiating the discharge
- Current limiting: the inductor's inductive reactance (X_L = 2πfL) limits the steady-state current to the lamp's rated operating value
Magnetic ballasts introduce a lagging power factor (typically 0.4–0.6). Electronic ballasts operate at high frequency (28–50 kHz) and achieve PF >0.95, smaller size, and eliminate stroboscopic effects.
Power Factor Correction
A capacitor connected in parallel with the luminaire supplies reactive current locally, preventing it from flowing through the distribution cables. A 400 W SON circuit at PF 0.45 draws:
\[I = \frac{P}{V \times \text{PF}} = \frac{400}{230 \times 0.45} = 3.86\text{ A}\]With PF correction to 0.9: \(I = 400/(230 \times 0.9) = 1.93\text{ A}\) — halving the cable current and reducing distribution losses by 75%.
Switches for Discharge Lighting (BS 7671 Reg 559.6.1.1)
Discharge circuits are inductive — contact arcing causes rapid erosion of unrated switches. If the switch is not specifically inductive-rated, its current rating must be at least twice the circuit current:
\[I_{\text{switch rated}} \geq 2 \times I_{\text{circuit}}\]Circuit current = 5 A (inductive, PF 0.5). If the switch is not inductive-rated: \(I_{\text{switch}} \geq 2 \times 5 = 10\text{ A}\). A 10 A switch must be fitted even though circuit current is only 5 A.
Low Pressure Mercury Vapour — Fluorescent Lamp (MCF)
A glass tube coated internally with phosphor powder, filled with mercury vapour at low pressure plus a small amount of argon. The arc discharge in mercury vapour produces predominantly UV at 253.7 nm — the phosphor coating absorbs this and re-emits it as visible light (fluorescence). By varying the phosphor mix, different colour temperatures and CRI values are produced.
Switch-start circuit sequence:
- Switch-on: current flows through ballast, both cathode filaments, and the glow-starter switch. Starter gas heats, closing bimetal contacts.
- Contacts close: current pre-heats both cathodes to electron-emission temperature.
- Bimetal cools, contacts open: inductive surge strikes the main arc.
- Ballast limits steady-state lamp current. Starter voltage is now too low to re-close it.
Electronic starters provide single reliable start, extending lamp life. High-frequency electronic ballasts (28–50 kHz) improve efficacy 10–15%, enable dimming (DALI, 1–10 V), and eliminate stroboscopic effects.
T8 36 W tube + 9 W magnetic ballast = 45 W circuit. T5 28 W tube + 3 W electronic ballast = 31 W circuit. Saving = 14 W per luminaire. In a school with 200 luminaires, 10 h/day, 250 days/year: 200 × 14 × 10 × 250 / 1000 = 7000 kWh/year = £1960/year at 28p/kWh.
High Pressure Mercury Vapour Lamp (HPMV / MBF)
A quartz arc tube (mercury vapour at 1–10 atm) inside an outer glass bulb with phosphor coating. An auxiliary electrode allows self-starting without an external ignitor — but 3–5 minutes warm-up to full brightness. Critical limitation: the hot high-pressure vapour requires much higher voltage to re-strike after switch-off — the lamp must cool 10–20 minutes before it can restart. A brief supply interruption darkens all HPMV lamps for up to 20 minutes — a major operational disadvantage now driving replacement by SON and LED.
| Feature | HPMV (MBF) |
|---|---|
| Efficacy | 32–58 lm/W |
| CRI | 40–60 (blue-green biased) |
| Warm-up / re-strike | 3–5 min / 10–20 min |
| Applications | Legacy street/industrial (being replaced by SON/LED) |
High Pressure Metal Halide Lamp (MBI / CDM)
Metal halide lamps add iodides/bromides of metals (dysprosium, holmium, thulium, indium, thallium) to the mercury arc tube. At arc temperature (~6000 K) these halides decompose, adding emission lines that produce a much more complete, white spectrum with higher CRI and efficacy than HPMV. An external ignitor (2–5 kV) is required.
| Feature | Metal halide (MBI/CDM) | Advantage over HPMV |
|---|---|---|
| Efficacy | 80–108 lm/W | ~2× better |
| CRI | 65–95 | Much better colour |
| Colour temp. | 3000–6000 K | Wide range of options |
| Rated life | 10,000–20,000 h | Similar |
| Re-strike | 5–15 min (hot) | Similar limitation |
| Applications | Retail, exhibitions, sports, TV studios | Where colour quality matters |
TV broadcasting requires CRI ≥ 90 and colour temperature 5600 K to match natural daylight. HPMV (CRI 40) would produce unacceptable colour distortion on camera. Metal halide CDM lamps at CRI 90+, 5600 K give broadcast-quality colour rendering — this is why all major TV-lit sports venues use metal halide or LED (which can also achieve CRI >90).
Stroboscopic Effect — Cause, Danger, and Solutions
At 50 Hz, the discharge arc extinguishes momentarily at each current zero-crossing — 100 times per second. Equipment rotating at speeds that are multiples of 3000 rpm (50 rev/s = 3000 rpm) can appear stationary. A lathe running at 1500 rpm with 50 Hz lighting appears to rotate at only apparent 0 rpm — catastrophically dangerous. This is a legal health and safety risk under the Health and Safety at Work Act 1974 and requires mitigation in machine shops and workshops.
| Method | Mechanism | Effectiveness |
|---|---|---|
| Different phases for adjacent luminaires | L1/L2/L3 on consecutive rows — peaks are offset by 120°, smoothing total illumination | Good — reduces but does not eliminate |
| Lead/lag paired luminaires | Capacitor in series with one tube (leading), inductor with other (lagging) — peaks offset by ~90° | Very good — near-elimination |
| Tungsten lamp supplement on rotating parts | Filament's thermal inertia smooths the flicker visible on the rotating surface | Effective but energy-inefficient |
| High-frequency electronic ballast | Arc operates at 28–50 kHz — zero-crossings 56,000–100,000 times/second; completely imperceptible | Eliminates stroboscopic effect completely — preferred solution |
Low Pressure Sodium Lamp (SOX)
The highest-efficacy light source ever developed (70–160 lm/W) — but emits nearly all light in a narrow monochromatic yellow band (sodium D-line at 589 nm). CRI ≈ 0 — colour rendering is non-existent. A neon/argon starter gas provides initial discharge while sodium heats up (5–10 min warm-up). Unlike HPMV, can re-strike when hot (~1 minute). Burning position: ±20° of horizontal. Now being replaced by LED street lighting, which offers similar or better efficacy with useful colour rendering for improved visibility and CCTV performance.
High Pressure Sodium Lamp (SON)
Uses a translucent sintered aluminium oxide (PCA — polycrystalline alumina) arc tube that withstands ~1200°C — far hotter than quartz is capable of. High-pressure sodium vapour at this temperature produces a broader emission spectrum than LP sodium, giving a warm golden-white light with acceptable CRI (Ra = 20–65). A high-voltage ignitor (2–4.5 kV) initiates the discharge. Outer glass envelope is evacuated to maintain arc tube temperature.
| Feature | SOX (LP sodium) | SON (HP sodium) |
|---|---|---|
| Efficacy | 70–160 lm/W | 55–120 lm/W |
| CRI | ~0 (monochromatic) | 20–65 |
| Colour temp. | N/A (monochromatic) | ~2000–2200 K (golden-warm) |
| Re-strike when hot | ~1 minute | 1–3 minutes |
| Burning position | ±20° of horizontal | Any position |
| Applications | Legacy street lighting — now replaced by LED | Car parks, sports, road — being replaced by LED |
- Fluorescent (MCF): best for interior — good CRI, dimming possible, high efficacy
- HPMV (MBF): mostly obsolete — poor CRI, 20-min re-strike delay
- Metal halide (MBI/CDM): best colour rendering of discharge types — sports, retail, TV
- SOX: highest efficacy ever — zero colour rendering; replaced by LED
- SON: good efficacy + acceptable CRI — now being replaced by LED street lighting
- All discharge lamps: require ballast, most have warm-up time and hot re-strike limitation
Compact Fluorescent Lamp (CFL)
CFLs operate on the same low-pressure mercury vapour / phosphor fluorescence principle as linear fluorescent tubes — but with the tube folded or spiralled into a compact form fitting standard BC or ES holders. They use approximately 20% of the power of the GLS lamp they replace:
| CFL wattage | GLS equivalent | Typical lumen output | Power saving |
|---|---|---|---|
| 7 W | 40 W GLS | ~400 lm | 83% |
| 11 W | 60 W GLS | ~600 lm | 82% |
| 15 W | 75 W GLS | ~900 lm | 80% |
| 20 W | 100 W GLS | ~1200 lm | 80% |
Average rated life: 10,000–15,000 hours (10× GLS). Efficacy: 60–72 lm/W.
Ballast types: Magnetic (may flicker on start; non-dimmable) vs Electronic (instant, flicker-free; dimmable types available; higher efficiency).
Construction types: Integrated (lamp + ballast as one unit — replace whole unit if ballast fails) vs Non-integrated (replaceable tube only; ballast stays in luminaire — better for commercial use).
A hotel replaces 500 × 60 W GLS with 11 W CFLs, running 4,000 h/year:
\[\text{Energy saved} = (60-11) \times 500 \times 4000 / 1000 = 98\,000\text{ kWh/year}\]At 28p/kWh = £27,440/year saving. Lamps cost ~£3 each (£1,500 total) — payback under 3 weeks.
CFL limitations: Some types slow to reach full output (1–3 min) — unsuitable for motion-detector circuits. Contain mercury (~3–5 mg) — must be recycled via WEEE scheme. Most non-dimmable. Now being phased out in favour of LED (most EU CFL types banned September 2023). Building Regulations Part L: ≥75% of fixed fittings in new builds must be high-efficacy type (≥45 lm/W) — verified at completion.
LED Lamps — Operating Principle
A Light Emitting Diode (LED) is a semiconductor p-n junction. When forward-biased, electrons recombine with holes at the junction, releasing energy directly as photons — electroluminescence. Unlike thermal radiation (incandescent) or arc ionisation (discharge), this is a cold, highly efficient process. The photon wavelength (colour) depends on the semiconductor bandgap energy:
- Red/orange: AlGaAs, GaAsP
- Green/yellow: GaP, AlGaP
- Blue: InGaN (indium gallium nitride) — Nobel Prize 2014; key to practical white LEDs
White LED production:
- Blue LED + yellow phosphor (most common): Blue InGaN chip coated with YAG:Ce phosphor emits yellow. Blue + yellow = white. CRI 80–98 achievable. Colour temperature set by phosphor formulation.
- RGB mixing: separate R, G, B chips controlled independently. Colour-changing and tunable white capability. Used in entertainment and architectural lighting.
LED Performance Characteristics
| Feature | LED | vs CFL / discharge |
|---|---|---|
| Efficacy | 80–200+ lm/W (2024) | 1.5–3× better than CFL; 10–14× better than GLS |
| Rated life (L70) | 25,000–50,000 h | 2–5× longer than CFL; 25–50× longer than GLS |
| CRI | 70–98 Ra | Specify Ra ≥ 80 general; Ra ≥ 90 for retail/healthcare/art |
| Colour temperature | 1800–6500 K | Widest range of any lamp type |
| Dimming | 0–100% with compatible driver/dimmer | Better than most discharge lamps |
| Warm-up | Instant — full output immediately | Better than CFL and all discharge lamps |
| UV/IR in beam | Negligible | Safe for UV-sensitive materials, artwork, food |
| Vibration resistance | Excellent — solid state, no filament | Suitable for vehicles, outdoor, harsh environments |
| Type | Power | Annual energy | Annual cost @28p | Lamp life |
|---|---|---|---|---|
| GLS 60 W | 60 W | 24,000 kWh | £6,720 | 1,000 h |
| CFL 11 W | 11 W | 4,400 kWh | £1,232 | 10,000 h |
| LED 8 W | 8 W | 3,200 kWh | £896 | 25,000 h |
LED saves £5,824/year vs GLS; £336/year vs CFL; with far less frequent replacement.
A red traffic signal: 196 LEDs consuming 10 W vs 150 W incandescent — 93% power saving. At a junction with 12 signal heads, 24 h/day: annual energy: incandescent = 12 × 150 × 8760 / 1000 = 15,768 kWh; LED = 12 × 10 × 8760 / 1000 = 1,051 kWh. Saving = 14,717 kWh = £4,121/year per junction. Plus: individual LED failure does not extinguish the whole signal head (safety); no warm-up time (safety); no lamp replacement needed for 10–15 years.
LED Drivers
LEDs are current-controlled devices requiring regulated dc current. An LED driver converts mains ac to the required dc conditions:
- Constant current (CC): fixed output current (e.g. 350 mA, 700 mA), variable voltage. LEDs in series. Used for high-power modules, streetlights, floodlights.
- Constant voltage (CV): fixed output voltage (12 V or 24 V dc), variable current. LEDs in parallel (as tapes/strips). Used for LED tape, under-cabinet, signage.
- Dimmable drivers: DALI (IEC 62386 — digital addressable, up to 64 devices/loop), 1–10 V analogue, PWM, or trailing-edge phase-cut (for domestic dimmer retrofits).
Eight 7 W GU10 LED downlights in a kitchen. Total load = 56 W, current = 56/230 = 0.24 A. If dimmer required: specify trailing-edge LED-compatible dimmer, minimum load rating ≤ 56 W, maximum load ≥ 56 W (e.g. 10–150 W range). Incompatible leading-edge (resistive/inductive) dimmers cause flicker, buzzing, or LED driver failure.
LED Heat Management
LEDs still convert ~70–80% of input to heat (only 20–30% becomes light). Unlike incandescent lamps, LED heat must be conducted away from the chip through substrate into a heat sink — not radiated outward. Excessive junction temperature shortens life and causes lumen depreciation. Key installation considerations:
- Do not cover LED luminaires with thermal insulation unless they are rated for IC (Insulation Contact) installation
- Check lamps are rated for enclosed-fitting use before fitting in sealed bulkheads
- High-power modules require external heat sinks
- Life (L70 rating) depends critically on operating temperature — a 10°C rise at the junction can halve LED life
Applications of LED Technology
| Application | Why LED is preferred |
|---|---|
| Traffic and railway signals | 93% power saving; individual LED failure doesn't extinguish signal; no warm-up; 10–15 year maintenance-free life |
| Emergency lighting | Instant full output on battery; low current extends backup duration; long life reduces maintenance |
| Street / road lighting | 100–160 lm/W efficacy; 50,000 h life; precise optics reduce light pollution; improves CCTV visibility (better CRI than SON/SOX) |
| Retail and display | CRI ≥ 90 for true colour rendering; no UV bleaching of merchandise; cool beam allows proximity to products; dimmable |
| Domestic and commercial | Retrofit into existing BC/ES/GU10 fittings; 25,000 h life = no lamp changes for 25+ years at 3 h/day domestic use |
| Automotive | Compact, low power, fast switching (indicators), vibration-resistant, long life without replacement |
| Horticulture grow lights | Specific red (660 nm) + blue (450 nm) wavelengths optimise photosynthesis without wasting energy on green spectrum |
🃏 Chapter 10 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Covers illumination principles, inverse square law, lumen method, incandescent, discharge, CFL and LED lamps.
📝 Chapter 10 — Multiple Choice Questions
Select your answer and click Submit. You have two attempts per question. First-attempt score shown at the end.
1A photometer measures 450 lux at a point on a work surface. What does this value represent?
2A lamp of 600 cd is suspended 3 m directly above a workbench. What is the illuminance at the point directly below it?
3A 1000 cd luminaire is mounted 3 m above a working plane. Find the illuminance at a point 4 m horizontally from directly below the lamp (use the cosine law).
4Using the lumen method, how many lamps are required to illuminate a 20 m × 15 m factory to 400 lux, if each lamp delivers 5,000 lm, UF = 0.5 and MF = 0.8?
5A lighting maintenance factor (MF) of 0.65 is appropriate for which type of environment?
6Which lamp type produces light by heating a tungsten filament to approximately 2500°C until it glows, and has a Colour Rendering Index (CRI) of 100?
7A stroboscopic effect occurs in fluorescent-lit machine shops because the discharge arc extinguishes 100 times per second (at 50 Hz). Which solution completely eliminates this hazard?
8A high-pressure mercury vapour (HPMV/MBF) lamp fails to restart immediately after a brief supply interruption. What is the technical reason?
9Low-pressure sodium (SOX) lamps have the highest efficacy of any lamp type (up to 160 lm/W). Why are they being replaced by LED for street lighting despite this efficiency advantage?
10White LED lamps are most commonly produced using which method?
11An inductive fluorescent circuit draws 5 A at a poor power factor. The switch controlling it is not inductive-rated. According to BS 7671 Regulation 559.6.1.1, what minimum current rating must the switch have?
12An LED luminaire is installed in a ceiling with thermal insulation directly above it. What installation risk does this create, and what is required to avoid it?
Sends your Chapter 10 MCQ score summary from your registered email to your tutor.
Electrical Heating
- Explain the basic principles of electrical space and water heating
- Identify the different types of space and water heating appliances and components
- Describe operating principles of space and water heating appliances and components
- Explain the applications and limitations of space and water heating appliances and components
Principles of Heat Transfer
All electric heating converts electrical energy (I²R heating) into thermal energy with 100% efficiency at the point of use — unlike fossil fuel heating, which always loses energy through combustion products and flue gases. The heat produced is then transferred to the space or object being heated by one or more of three fundamental mechanisms:
| Mechanism | How it works | Key characteristics | Heater types that use it |
|---|---|---|---|
| Radiation (infrared) | Electromagnetic waves travel from the hot source through air (or vacuum) and are absorbed by surfaces — which then warm the air by conduction and convection | Follows inverse square law (like light); heats people/objects directly without warming the air first; no heat loss to ventilation | Quartz infrared, ceramic infrared, electric fires, radiant ceiling panels |
| Convection | Hot air rises, cool air takes its place — creating a circulating current that distributes heat throughout a space | Warms the air (not people directly); creates dry atmosphere; effective for enclosed spaces; slow to heat large, draughty areas | Convector heaters, storage heaters, fan heaters, oil-filled radiators |
| Conduction | Heat flows through solid materials from hot to cold by direct contact between molecules | Most important for heating through floors (underfloor heating) or warming objects placed on a surface; rate depends on thermal conductivity of the material | Underfloor heating cables, heated mats, trace heating on pipes |
Most practical heaters use a combination of mechanisms — a storage heater releases heat by both radiation (from the warm surfaces) and convection (warm air rising from the casing).
Radiant Heaters
Radiant heaters emit infrared electromagnetic radiation from a hot element. The radiation travels through the air without warming it and is absorbed by surfaces (walls, floors, people, objects), which convert it to heat. This makes radiant heating highly effective for spot heating of occupied areas in otherwise cold, draughty, or high-ceiling spaces — the energy goes directly to the people rather than being wasted warming large volumes of cold air.
Types of radiant element:
- Tungsten quartz element: a coiled tungsten wire enclosed in a heat-resistant quartz tube. Glows orange-red at operating temperature — emits visible light as well as infrared. Fast response (seconds). Highly efficient (≥95% of input becomes infrared radiation).
- Ceramic infrared element: an electrical element heats a ceramic tile to ~300–500°C. The ceramic emits long-wave infrared radiation (invisible — no glare). Slower warm-up than quartz (minutes) but more comfortable for occupants who find visible glare unpleasant.
- Halogen infrared: similar to quartz but uses a halogen lamp element. Faster response, more compact.
A vehicle workshop is 20 m × 10 m × 5 m high with poor insulation. Heating the whole air volume would require enormous energy and most heat would escape when the roller doors open. Instead, two 3 kW quartz infrared heaters are mounted above the working bays, directed downward. They provide instant, localised warmth to the mechanics without wasting energy on unoccupied areas. PIR sensors switch them off when no one is present for more than 5 minutes.
Energy saving vs convective whole-space heating: the infrared system heats people directly — an equivalent convective system would need to heat the entire 1000 m³ air volume to perhaps 18°C from 5°C, requiring approximately 1000 × 1.2 × 1.0 × 13 / 3600 ≈ 4.3 kWh per heating cycle. The spot infrared approach uses only 2 × 3 = 6 kW while occupied, which is far less than a whole-space heater system on a continuous basis.
Applications: Warehouses, workshops, garages, churches (high-ceiling, intermittently occupied), outdoor areas (patio heaters, construction site welfare), sports changing rooms, food preparation areas.
Energy management: PIR (passive infrared) motion sensors and timer controls are essential — a radiant heater left on in an unoccupied workshop wastes energy just as quickly as a convective heater.
Convector Heaters
Convector heaters warm the air in a room by natural or forced convection. A heating element heats air that flows across it; the warmed air rises and is replaced by cooler air — establishing a circulating current that gradually raises the entire room temperature.
Natural convection sequence:
- Cool room air enters the base of the heater through inlet vents
- Air passes across the heating element (resistance wire or nichrome element)
- Heated air becomes less dense and rises through the outlet vent at the top
- Warm air spreads across the ceiling, cools, and descends at the opposite wall
- The cooled air returns to the base of the heater — completing the convection loop
- A thermostat switches the element off when the room reaches the set temperature
Fan-assisted convector heaters (fan heaters): A fan forces air through the heating element at a controlled rate, significantly speeding up heat distribution. Fan heaters produce room-temperature heat much faster than natural convection heaters — making them useful for rapid temperature recovery after opening external doors. However, they are noisier and circulate dust and allergens.
A bathroom requires supplementary heating. A 1.5 kW oil-filled radiator with an IP44 splash-proof rating is installed on the wall outside Zone 1 (more than 60 cm from the bath edge, more than 2.25 m from the floor if within 60 cm horizontally of the bath). The integral thermostat is set to 21°C. Supply is from an RCBO in the consumer unit. The radiator is permanently wired (not via a socket outlet — socket outlets are not permitted in bathrooms except for shaver sockets). Double-pole isolation is provided at the consumer unit or a fused connection unit outside the bathroom zones.
The oil fill (mineral oil) inside the radiator acts as a heat buffer — the element heats the oil, which then radiates and convects heat to the room. The oil's thermal mass means the radiator continues warming the room for some time after the element switches off — reducing short-cycling of the thermostat.
Storage Heaters
Storage heaters are designed to take advantage of Economy 7 (or similar off-peak) electricity tariffs, which offer significantly cheaper electricity during overnight periods (typically 11 pm–7 am) in exchange for agreeing not to draw heavy loads at peak times. The heater stores thermal energy in dense refractory bricks (high thermal mass) during the cheap-rate period and releases this stored heat during the day when electricity is more expensive.
Key components:
- Heating elements: embedded in the refractory brick core; energised only during the off-peak period via a time switch or radio teleswitch (RTS) controlled by the supply authority
- Refractory bricks (magnetite or similar): very dense, high specific heat capacity materials that store large amounts of thermal energy at relatively moderate temperatures (~700°C internal). The high density (2500–3000 kg/m³ vs wood ~500 kg/m³) means a compact unit stores a lot of heat.
- Thermal insulation (rockwool / ceramic fibre): surrounds the brick core to slow heat loss and allow the stored heat to be released in a controlled way throughout the day
- Input thermostat: measures the brick core temperature and limits maximum charge level to prevent overheating
- Output damper / damper control: a manually or electronically controlled air damper regulates how quickly stored heat is released; closing the damper retains heat for the evening
A living room (30 m³, poorly insulated) needs 2 kW average heating over 12 hours/day. Using Economy 7 (off-peak rate ~13p/kWh vs peak rate 28p/kWh). The storage heater must store enough heat during 7 off-peak hours to supply 2 kW for 12 hours: stored energy needed = 2 × 12 = 24 kWh. Charging at off-peak: 24 kWh × 13p = £3.12/day. Equivalent convective heating at peak rate: 2 × 12 = 24 kWh × 28p = £6.72/day. Storage heater daily saving = £3.60 (53% cheaper). Over a 150-day heating season: £540 saving.
Types of Storage Heater — Comparison
| Type | Key features | Energy efficiency | When to specify |
|---|---|---|---|
| Basic (manual) storage heater | Manual input and output controls; operator sets charge level based on expected weather; simple construction | Lowest — over-charging wastes energy; under-charging leaves rooms cold in the evening | Budget installations where operator will be present to manage controls; not suitable for new builds under Part L |
| Combination storage heater | Storage core + integral convector element; convector provides top-up heat at peak rates; ambient thermostat optimises charge | Better — top-up convector prevents rooms being cold if storage runs out | Homes where occupants are present in the evening and may need additional heat beyond stored capacity |
| Fan-assisted storage heater | More insulation than basic type; quiet thermostatically-controlled fan releases heat on demand; damper retains heat until needed | Good — delivers stored heat more efficiently and uniformly; less heat lost through radiation from case | Bedrooms and living rooms where even heat distribution is needed; better comfort than basic types |
| Automatic charge control (ACC) storage heater | Sensors measure room temperature and outside temperature during charging; automatically calculates required charge level for the coming day; may connect to smart meter/home energy management system | Best — prevents over-charging; adapts to weather; reduces waste. Required by Building Regulations Part L for new builds | All new-build and major renovation projects; most cost-effective over the lifetime of the installation |
Underfloor Heating
Underfloor heating (UFH) uses electrical resistance cables or mats embedded in or beneath the floor to heat the room by warming the floor surface, which then warms the air above it by natural convection. The floor itself becomes a large, low-temperature radiant surface — a very even, comfortable heat distribution with no visible equipment.
System types:
| System type | Construction | Floor finish | Power density |
|---|---|---|---|
| Loose cable (twin conductor) | Heating cable stapled in loops to the sub-floor, then embedded in screed; spacing determines power density | Stone, ceramic tile (after screed cures) | 100–200 W/m² (typically 150 W/m²) |
| Pre-formed cable mat | Twin-conductor cable pre-attached to fibreglass mesh at fixed spacing; unrolled directly onto sub-floor then tiled over | Stone or ceramic tile — mat laid directly in tile adhesive bed | 100–200 W/m² (mat wattage fixed by mat area) |
| Aluminium foil heating mat | Carbon heating element on aluminium foil backing; laid between sub-floor and floating floor | Laminate, engineered wood, vinyl — floating floor laid directly over mat | 80–160 W/m² |
| Carbon film mat | Carbon film elements on polyethylene film; very thin (0.5 mm) | Carpet (with suitable thermal resistance rating) | 80–140 W/m² |
Cable construction: All UFH cables have: twin or coaxial copper conductors (one acts as the return), high-temperature double insulation, a metallic earth screen (earthed to the consumer's earthing system for protection against cable damage), and an overall PVC sheath. Diameter 3–6 mm; thickness 2–3 mm — floor height raised negligibly.
A bathroom floor area is 6 m² (total), but a 1.2 m² area is occupied by the bath, WC, and vanity unit — heating cables must not be installed there (risk of overheating under fixed equipment). Effective heated area = 6 − 1.2 = 4.8 m².
Selecting 150 W/m² cable mat: total power = 4.8 × 150 = 720 W. Circuit current = 720/230 = 3.13 A — a 6 A MCB and 1.5 mm² cable are adequate. The mat is supplied from a dedicated circuit with a 30 mA RCD (bathrooms require RCD protection per BS 7671 Section 701).
Checking screed temperature: at 150 W/m² under 10 mm ceramic tile + 3 mm adhesive, the tile surface reaches approximately 26–28°C — pleasantly warm underfoot but well below the maximum permitted floor temperature (~35°C for most systems to avoid thermal discomfort and damage to floor coverings).
Before the floor screed is poured (or tiles are laid), the UFH cable is tested:
- Continuity: measure cable resistance using a low-resistance ohmmeter. Compare with manufacturer's rated resistance (e.g. 720 W cable on 230 V: \(R = V^2/P = 230^2/720 = 73.5\text{ Ω}\)). A measured value significantly different from rated indicates a cable fault.
- Insulation resistance: apply 500 V dc between the conductor and the earth screen. IR > 1 MΩ is normal. Low IR indicates insulation damage.
- Repeat both tests immediately after the screed is poured (while it is still wet) to check the pouring process did not damage the cable.
- Repeat before commissioning (after the screed has fully cured and before floor covering is laid).
A record of all test results must be retained and included in the installation's electrical documentation.
Control: A digital programmable thermostat with a floor probe (thermistor) controls the UFH. The floor probe limits maximum floor temperature (typically 27°C under carpet, 35°C under tiles) to protect the floor covering. Some systems also have an air temperature sensor for zone-based control.
- Never install cables under fixed equipment (baths, WC pans, basin pedestals, kitchen units) — risk of overheating
- Always mark the cable positions on a plan (measured from fixed reference points) for future reference during building modifications
- Test before, immediately after, and before commissioning (three separate tests)
- Allow screed to cure fully (minimum 28 days for concrete screed) before switching UFH on — avoid thermal cracking
- Use RCD protection on all UFH circuits (earth fault to metallic screen is detected as a current imbalance)
Electric water heaters fall into two main categories: storage (heat water in advance and keep it hot) and instantaneous (heat water only when needed). Each has distinct advantages and limitations that determine the appropriate application.
Storage Types — Immersion Heater
An immersion heater is a resistance heating element (similar to a kettle element but larger) inserted directly into a hot water cylinder. It is the most common form of standalone electric domestic water heating, typically used as a backup or supplement to central heating boilers. The cylinder is heavily insulated (modern foam-insulated tanks lose <1°C per hour) to maintain water temperature with minimal heat loss.
Element construction: Incoloy (nickel-chrome alloy) sheathed element, typically 2.5–3 kW for domestic use, fitted with a threaded flange that seals into the cylinder boss. The element must be selected for the type of water supply — hard water areas require special anti-scale alloys or titanium elements because limescale builds up rapidly on standard incoloy elements.
Temperature setting: The rod thermostat (see below) is set to 60–65°C. 60°C is the minimum recommended setting to kill Legionella bacteria — lower settings risk growth of this dangerous pathogen in the stored water. For frost protection only, 45°C is acceptable. Never set above 65°C as this creates a scalding risk at taps.
A 3 kW immersion heater heats a 150-litre cylinder from 15°C (cold mains) to 60°C. Energy required:
\[Q = m \times c_p \times \Delta T = 150 \times 4.18 \times (60-15) = 150 \times 4.18 \times 45 = 28\,215\text{ kJ} = 7.84\text{ kWh}\]Time to heat: \(t = 7.84/3 = 2.6\text{ hours}\). Running cost: \(7.84 \times 28\text{p} = £2.19\). On Economy 7 off-peak (13p/kWh): \(7.84 \times 13\text{p} = £1.02\) — saving £1.17 per heating cycle. Over a year (once daily): \(£1.17 \times 365 = £427\).
Dual-element cylinder (two immersion heaters):
- Bottom element (long): connected to the off-peak supply via a time switch or radio teleswitch. Heats the whole cylinder during off-peak hours (typically 11 pm–7 am). This provides a full tank of hot water for the following day.
- Top element (short): connected to the normal (peak rate) supply. Heats only the top portion of the cylinder (approximately 50 litres) for a quick top-up if the off-peak supply has not restored sufficient hot water. The top element should only be used sparingly — it operates at peak tariff rates.
A 3 kW, 230 V immersion heater. Current = 3000/230 = 13 A. BS 7671 requires a dedicated radial circuit with:
- 16 A MCB (Type B) or 16 A fuse to protect the circuit
- 2.5 mm² copper cable (current rating 27 A — adequate with 13 A load)
- Double-pole isolator switch (to allow safe isolation for maintenance) adjacent to the heater
- The circuit must not be connected via a socket outlet
- If in a bathroom, RCD protection (30 mA) is required per BS 7671 Section 701
Instantaneous Water Heaters — Electric Showers
An instantaneous electric shower heats cold mains water as it flows through a compact heating chamber containing a powerful heating element (typically 7–10.8 kW). Cold water enters continuously; the element heats it as it passes through; hot water exits immediately. There is no stored water.
Why such high power? The flow rate and temperature rise are related by:
\[P = \dot{m} \times c_p \times \Delta T\]where \(\dot{m}\) = mass flow rate (kg/s), c_p = specific heat of water (4.18 kJ/kg·K), ΔT = temperature rise. For a comfortable shower (8 litres/min = 0.133 kg/s, ΔT = 32°C from 8°C inlet to 40°C outlet):
\[P = 0.133 \times 4.18 \times 32 = 17.8\text{ kW}\]At this flow rate, 17.8 kW would be needed. Practical showers compromise on flow rate (typically 6–8 litres/min) or temperature rise — which is why electric showers feel less powerful than gas-heated showers. A 10.8 kW shower provides a better flow rate than an 8.5 kW shower from the same cold inlet temperature.
A 9.5 kW electric shower at 230 V. Current = 9500/230 = 41.3 A. Required circuit:
- 45 A double-pole isolating switch or shower switch adjacent to the shower enclosure (outside Zone 1)
- 40 A or 45 A MCB (Type B or C) at the consumer unit
- 6 mm² or 10 mm² cable (depending on cable installation method and length)
- 30 mA RCD protection mandatory (bathroom — BS 7671 Section 701)
- The shower unit must be bonded to the supplementary equipotential bonding network in the bathroom if it is not already effectively bonded via the pipework
| Feature | Storage (immersion heater) | Instantaneous (electric shower) |
|---|---|---|
| Supply current | ~13 A (3 kW) | 41 A (9.5 kW) |
| Hot water quantity | Large (full cylinder — 120–200 litres) | Unlimited (continuous flow) |
| Flow rate | High — fills bath or multiple taps simultaneously | Low — one shower point only |
| Running cost | Higher (heats full tank; heat losses) | Lower (only heats water actually used) |
| Off-peak tariff | Yes — charge at cheap rate overnight | No — operates at peak rate whenever used |
| Response time | 2–3 hours to heat from cold | Instant — immediate hot water |
| Failure mode | All hot water lost; re-heat takes hours | Loss of power = no shower immediately |
Temperature Control — Types of Thermostat
All heating appliances require temperature control to prevent overheating, save energy, and maintain safe operating conditions. The most common method is the thermostat — a device that automatically switches the heating element off when the set temperature is reached and on again when it falls below the set point.
Bimetal Strip Thermostat
A bimetal strip consists of two different metals (typically invar — a nickel-iron alloy with very low thermal expansion — and brass with high thermal expansion) bonded together along their full length. When heated, the brass expands more than the invar, forcing the combined strip to curve. This bending movement operates a set of electrical contacts:
- Below set temperature: strip is relatively straight, contacts closed, heater energised
- At set temperature: strip has bent sufficiently to open the contacts, heater de-energised
- Below set temperature again: strip straightens, contacts reclose, heater re-energises
The set temperature is adjusted by changing the initial contact gap — the user's thermostat dial varies this gap. Bimetal thermostats are used in: convector heaters, fan heaters, storage heater input thermostats, room thermostats (air temperature sensing), and frost stats.
A room convector heater has its bimetal thermostat set to 21°C. The room is cold (15°C): the bimetal strip is straight, contacts closed, element energised. Room warms to 21°C: strip has bent, contacts open, element off. Room slowly cools to ~20°C (the hysteresis — the thermostat doesn't re-close at exactly 21°C but slightly below): contacts reclose, cycle repeats. The temperature "floats" between approximately 20–22°C — the hysteresis bandwidth that prevents rapid on-off cycling.
Rod Thermostat (Immersion Heater Thermostat)
The rod thermostat is specifically designed for immersion heaters. It consists of:
- An outer brass tube (high thermal expansion) inserted into the hot water cylinder alongside (or around) the immersion heater element
- An inner invar rod (very low thermal expansion) fixed at the tip of the brass tube only
- A contact mechanism at the open end of the tube, operated by the differential movement of the rod relative to the tube
As water temperature rises, the brass tube expands and elongates faster than the invar rod — effectively pulling the invar rod toward the closed end. This relative movement operates the contacts. When the set temperature is reached, contacts open, element de-energises. When water cools, brass contracts, rod relatively moves back, contacts reclose.
Rod thermostats typically include a separate overheat safety cut-out (non-resettable thermal fuse or manual reset button) that permanently disconnects the element if the thermostat fails and temperature reaches a dangerous level (~85–95°C for standard thermostats). This prevents the water reaching boiling point (and potentially scalding from discharge, or creating steam pressure in vented cylinders).
Fault: immersion heater not heating water. Test: disconnect supply; measure continuity of immersion heater element (a 3 kW, 230 V element has resistance = \(V^2/P = 52900/3000 = 17.6\text{ Ω}\) — should measure approximately this). If element is continuous, suspect the thermostat. Press the manual reset button on the thermostat (if it has tripped on overheat). If this restores operation, investigate why it overheated (thermostat set too high, element failure causing local hot spot, limescale insulating element). Replace thermostat if defective — always replace with correct type and temperature rating.
Thermocouple
A thermocouple consists of two dissimilar metal wires joined at one end (the hot junction). When this junction is heated, a small emf is generated (the Seebeck effect) — typically in the range of 10–50 µV per degree Celsius, depending on the metal combination. Common thermocouple types and their typical ranges:
| Type | Metals | Range | Sensitivity | Application |
|---|---|---|---|---|
| K | Chromel / Alumel | −200°C to +1260°C | ~41 µV/°C | Most common industrial type; kilns, furnaces, motors |
| J | Iron / Constantan | −40°C to +750°C | ~51 µV/°C | Older industrial processes; boilers, furnaces |
| T | Copper / Constantan | −200°C to +350°C | ~43 µV/°C | Refrigeration, cryogenic, food processing |
The small thermocouple output (millivolts) is amplified by an electronic circuit and compared with a reference to determine temperature. With appropriate control electronics, the thermocouple output can operate relay contacts to control a heating load. Modern boiler gas valves use thermocouples as pilot flame safety devices — the pilot flame heats the thermocouple, generating an emf that holds the gas valve open; if the pilot goes out, the thermocouple cools, the emf falls to zero, and the valve closes (fail-safe gas shutoff).
Modern thermostatic electric showers use a thermistor (a semiconductor device whose resistance changes with temperature) or a thermocouple to sense the outlet water temperature. The electronic controller compares this to the user's set temperature and adjusts the power to the element (via a triac) to maintain the outlet temperature regardless of changes in cold water inlet temperature. If the cold mains temperature drops in winter, the controller increases power; if a user opens a cold tap nearby (reducing mains pressure), the controller responds to the resulting temperature change and adjusts flow restriction or power accordingly.
Central heating (CH) systems use a boiler (gas, oil, LPG, or heat pump) to heat water which is pumped around a circuit of radiators and through a coil inside the domestic hot water (DHW) cylinder. An array of electrical input, control, and output devices manages the system to maintain desired temperatures in each zone, minimise energy waste, and comply with Building Regulations Part L (Energy Efficiency). Electricians are responsible for wiring the entire control system — understanding every device is essential for correct installation, commissioning, and fault-finding.
Input Devices — What They Sense
| Device | What it senses | Type | Typical setting | Wiring notes |
|---|---|---|---|---|
| Room thermostat | Air temperature in the heated space | Bimetal (older) or electronic thermistor (modern); wireless types send RF signal to a receiver wired into the controller | 21°C living room, 18°C bedroom — user adjustable | Two-wire (switched live) or three-wire (live, neutral, switched live) connection to controller; position away from draughts, direct sunlight, and heat sources (radiators, TVs) to avoid false readings |
| Cylinder thermostat | Temperature of stored domestic hot water | Bimetal or electronic; strapped to the side of the DHW cylinder at approximately one-third height from the bottom | 60°C (Legionella prevention minimum) | Two-wire or three-wire; the strap position determines which portion of the cylinder is sensed — position correctly per manufacturer's instructions |
| Frost thermostat (frost stat) | Outside air temperature or internal boiler house temperature | Bimetal; wide hysteresis (typically 0°C set point, operates when temperature drops to 0°C, resets at 6°C) | 2–5°C | Overrides programmer; calls for heating regardless of time schedule; wired in parallel with room thermostat or directly to controller override input |
| Pipe thermostat (pipe stat) | Return water temperature to the boiler | Bimetal clip-on type; fits around the return pipe | ~55°C — boiler shuts off if return water is still warm enough, preventing unnecessary re-firing | Used in conjunction with frost stat — prevents boiler operating unnecessarily when pipes are already warm. Connected in series with the boiler control circuit |
| Programmer / timer | Time of day and day of week | Electronic with digital display; wireless types available; some can be programmed via smartphone app (smart thermostats — Nest, Hive, etc.) | Multiple heating periods per day; separate CH and DHW schedules | Three-wire or four-wire connection to controller; provides separate switched live outputs for CH demand and DHW demand; mains supply input |
Legionella pneumophila bacteria can multiply rapidly in water at 20–45°C. Above 60°C they are killed within 2 minutes; above 70°C, instantly. The HSE HSG274 guidance (Legionnaires' Disease — The Control of Legionella Bacteria in Water Systems) requires domestic hot water to be stored at ≥60°C and distributed at ≥50°C at all outlets within 1 minute. Setting the cylinder thermostat below 60°C creates a potential health risk. The electrician must set the thermostat correctly at commissioning and advise the client never to reduce it below 60°C even to reduce energy costs. Thermostatic mixing valves (TMVs) set to 43°C should be fitted at showers and baths to prevent scalding risk from the 60°C stored water.
Control Devices — How the System Is Managed
- Main controller (wiring centre): the electrical hub of the system that receives all input device signals and operates the output devices. Receives mains supply, programmer outputs, thermostat signals; outputs to pump, boiler, and motorised valves. Available as proprietary wiring centres (e.g. Honeywell, Drayton) with clearly labelled terminals or as individual components wired together.
- Motorised zone valves: electrically operated valves that direct the pump flow to the appropriate circuits:
- Two-port zone valve: simply opens (allows flow) or closes (stops flow) on command from the controller. Used in S-plan systems — one valve for CH circuit, one for DHW cylinder coil circuit. The valve motor drives the valve open when its circuit calls for heat; a microswitch in the fully-open position sends a signal to the controller to start the pump and boiler.
- Three-port mid-position valve (Y-plan): has three positions — port A open only (DHW), port B open only (CH), or mid-position (both open). Motor-driven to each position as required by thermostat signals. If only DHW calls for heat, all flow goes to cylinder. If only CH calls, all flow goes to radiators. If both call simultaneously, mid-position allows flow to both.
- Boiler interlock: the boiler must only fire when the pump is running and when either the room or cylinder thermostat is calling for heat. Wiring the boiler via a volt-free contact from the motorised valve's end-switch (or via the room/cylinder thermostat) provides this interlock — preventing the boiler from cycling unnecessarily when all thermostats are satisfied (a requirement of BS 7671 and Building Regulations Part L).
- Automatic bypass valve: when TRVs on most radiators close (all rooms at set temperature), the circuit resistance increases, potentially causing the pump to cavitate and the boiler to overheat. An automatic bypass valve (pressure-differential valve) opens when the pressure across it exceeds the set differential, allowing flow to bypass through a short circuit back to the boiler — maintaining minimum boiler flow. Required by Part L on all new CH systems with TRVs.
S-plan system sequence for CH only calling:
- Room thermostat satisfied (above set temperature) or programmer in OFF period → CH valve remains closed, boiler and pump off
- Room temperature falls below set point AND programmer in ON period → room thermostat closes its contact → CH zone valve motor energised → valve drives to open position
- CH zone valve reaches fully open position → end-switch closes → boiler live and pump live are energised
- Pump circulates hot water through radiators → rooms warm up → room thermostat opens its contact
- CH zone valve motor de-energised → valve driven to closed position by return spring → end-switch opens → boiler and pump de-energised
The DHW zone valve operates identically but is controlled by the cylinder thermostat. If both CH and DHW call simultaneously, both zone valves open, and both boiler and pump run together — the pump flow divides between the two circuits.
Output Devices — What Does the Work
- Circulating pump: a centrifugal pump with an electric motor (typically 25–75 W, three-speed). Continuously circulates hot water from the boiler through radiators and cylinder coil. Most modern pumps are variable-speed (auto-adapt), automatically adjusting speed to match system resistance — saving energy and reducing noise. The pump runs whenever the boiler fires (via the interlock).
- Boiler: raises water temperature by burning fuel (gas, oil) or using heat pump technology. The boiler receives a call for heat from the controller and fires its burner. Condensing boilers (now mandatory for new installations under Part L) recover latent heat from the flue gases by condensing water vapour — achieving efficiencies of 90–99% (compared to 75–80% for non-condensing). The resulting condensate (weakly acidic water) must be drained to the soil stack or a soak-away.
- Thermostatic Radiator Valves (TRVs): fitted to individual radiators, TRVs sense the room air temperature via a wax-filled sensor capsule. As the room warms, the wax expands and progressively closes the valve, reducing radiator output. This prevents radiators overheating rooms beyond their individual set temperatures. TRVs should not be fitted to the radiator in the room containing the room thermostat (conflict of control).
Without TRVs: the room thermostat controls the boiler but all radiators run at full output — some rooms overheat. Occupants open windows to cool down, wasting the heat. With TRVs but no room thermostat: TRVs control individual rooms but there is no overall "satisfied" signal to shut down the boiler — the boiler cycles against the TRVs. With both TRVs AND a room thermostat (in the hallway, say): TRVs control room temperatures individually; the room thermostat provides the overall system setback when the reference room is satisfied; the boiler interlock prevents unnecessary cycling. This combination is required by Building Regulations Part L Appendix D for new CH systems.
Central Heating System Plans
The two most widely used wiring plans for fully-pumped domestic central heating systems are:
| Plan | Valve type | Advantages | Disadvantages |
|---|---|---|---|
| S-plan | Two separate 2-port motorised zone valves (one CH, one DHW) | Simple to wire; easy to fault-find; CH and DHW fully independent; adding extra zones straightforward | Two valves (higher component count); slightly higher cost than Y-plan |
| Y-plan (Sundial) | One 3-port mid-position motorised diverter valve | Single valve; lower component count; traditional in UK domestic CH | More complex valve mechanism; mid-position means CH and DHW always share flow when both demand simultaneously — DHW heating may be slower; wiring more complex to understand |
Both plans comply with Building Regulations Part L when combined with a programmer, room thermostat, cylinder thermostat, and TRVs on radiators.
- Condensing boiler (minimum SEDBUK efficiency 92%)
- Full programmer with separate CH and DHW timing
- Room thermostat (or equivalent zone control)
- Cylinder thermostat (for all unvented and indirect cylinders)
- Thermostatic radiator valves (TRVs) on all radiators except in the room containing the room thermostat
- Automatic bypass valve
- Boiler interlock (boiler must not run when no heat is demanded)
🃏 Chapter 11 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Covers space heating, water heating, thermostats and central heating controls.
📝 Chapter 11 — Multiple Choice Questions
Select your answer and click Submit. You have two attempts per question. First-attempt score shown at the end.
1A vehicle workshop has a high ceiling, poor insulation, and roller doors that are frequently opened. Which type of electric heater is most efficient and appropriate, and why?
2A storage heater installed in a new-build dwelling must include which feature to comply with Building Regulations Part L?
3Underfloor heating cables must NOT be installed under fixed equipment such as baths, WC pans, or kitchen units. What is the technical reason?
4A 3 kW immersion heater at 230 V requires a dedicated circuit. Which combination correctly meets BS 7671 requirements?
5Why must a cylinder thermostat on a domestic hot water cylinder be set to a minimum of 60°C?
6A bimetal strip thermostat uses two different metals bonded together. How does it open the contacts when temperature rises?
7In an S-plan central heating system using two-port motorised zone valves, what must happen before the boiler and pump are energised?
8Why is an automatic bypass valve required in a central heating system fitted with thermostatic radiator valves (TRVs) on all radiators?
9A 9.5 kW electric shower requires a high power rating. What physical reason makes this necessary?
10A rod thermostat used in an immersion heater works by differential expansion of an outer brass tube and an inner invar rod. Which metal expands MORE when heated, and what does this do?
11Which wiring plan for a domestic central heating system uses a single three-port mid-position motorised valve to serve both central heating and domestic hot water circuits?
12A thermocouple works by generating a small EMF when its hot junction is heated (Seebeck effect). In a gas boiler pilot flame safety circuit, what happens if the pilot flame goes out?
Sends your Chapter 11 MCQ score summary from your registered email to your tutor.
Electronic Components
- State the basic operating principles of different types of electronic components
- Identify typical applications of different types of electronic components
- Describe the function and application of electronic components in electrotechnical systems
- Identify the limitations of electronic components used in different electrotechnical systems
Active vs Passive Components
All electronic components are classified as either active or passive, depending on whether they can control or amplify the current flowing through them:
| Category | Definition | Energy behaviour | Examples |
|---|---|---|---|
| Passive | Cannot amplify or control current — they only respond to the voltage or current applied to them | Consume or store energy; never deliver more energy than they receive | Resistors, capacitors, inductors, transformers, diodes (technically passive — no gain) |
| Active | Can control or amplify current — a small input signal can control a much larger output current | Require an external power supply to provide the additional energy; can deliver more power to the load than the input signal provides | Transistors (BJT, MOSFET), thyristors (SCRs), triacs, integrated circuits (ICs), operational amplifiers |
The key distinction in practical terms: a transistor acting as an amplifier uses a microamp base current to control a milliamp collector current — the additional power comes from the power supply, not from the input signal. A resistor simply limits current — no external power supply involvement.
Fixed Resistors — Construction and Power Ratings
A resistor opposes current flow and converts electrical energy to heat (P = I²R). The resistor must be physically large enough to dissipate this heat without overheating — the power rating specifies the maximum continuous power the resistor can safely dissipate. Exceeding the power rating causes overheating, drift in resistance value, and ultimately open-circuit failure.
| Type | Construction | Typical power rating | Application |
|---|---|---|---|
| Carbon composition | Carbon granules and binder in a ceramic rod | 0.125–2 W | General purpose signal circuits, old radio and TV equipment |
| Carbon film | Carbon film deposited on a ceramic substrate | 0.125–1 W | General purpose; low noise; stable |
| Metal film | Nichrome or similar alloy film on ceramic | 0.1–3 W | Precision circuits; low temperature coefficient; tighter tolerances (±1% or better) |
| Wirewound | Resistance wire (nichrome) wound on a ceramic former; ceramic or aluminium cladding | 5–250 W | High-power loads; motor starters; braking resistors for VFDs; load banks |
| Metal oxide film | Tin oxide or similar on ceramic | 0.5–10 W | High stability; wide temperature range; flameproof types for safety-critical applications |
A 470 Ω resistor is connected across a 24 V dc supply. Power dissipated:
\[P = \frac{V^2}{R} = \frac{24^2}{470} = \frac{576}{470} = 1.23\text{ W}\]Standard carbon film resistors are available at 0.125 W, 0.25 W, 0.5 W, 1 W, 2 W. A 1 W resistor would be operating at 123% of its rating — it would overheat. Select a 2 W metal film resistor (allowing a 1.6× safety margin — good practice is to derate to 50% of rated power in continuous service).
Resistor Applications
Potential divider (voltage divider): Two resistors in series divide the supply voltage in the ratio of their resistances. The output voltage is taken from the junction of the two resistors:
\[V_{\text{out}} = V_{\text{in}} \times \frac{R_2}{R_1 + R_2}\]R₁ = 7 kΩ, R₂ = 5 kΩ in series across 12 V. Output voltage at the junction:
\[V_{\text{out}} = 12 \times \frac{5000}{7000 + 5000} = 12 \times \frac{5}{12} = 5\text{ V}\]This 5 V reference is used to bias the base of a transistor or feed a logic circuit. The potential divider principle also appears in the Wheatstone bridge (used in sensor circuits with thermistors and LDRs).
Current-limiting resistor (in series with LED): LEDs require a series resistor to limit current to their rated value (typically 10–20 mA). The resistor drops the difference between the supply voltage and the LED forward voltage (~2 V for red, ~3.2 V for blue):
Supply = 12 V dc. Red LED (V_F = 2.0 V, I_F = 15 mA). Required series resistance:
\[R = \frac{V_{\text{supply}} - V_F}{I_F} = \frac{12 - 2}{0.015} = \frac{10}{0.015} = 667\text{ Ω}\]Select the next standard value: 680 Ω. This is the most frequently used resistor application in LED indicator circuits found in switchgear, fire alarm panels, and control equipment.
Variable Resistors (Potentiometers and Rheostats)
A variable resistor has three terminals: two ends of a resistive track and a sliding wiper contact. When all three terminals are used (wiper + both track ends), it is a potentiometer — providing a variable output voltage. When only one track end and the wiper are used, it functions as a rheostat — providing a variable resistance in series with a circuit.
Track characteristics:
- Linear (log A): resistance changes proportionally with slider position — used for precision adjustment, speed controls, dimmer knobs
- Logarithmic (log B): resistance changes logarithmically — used for audio volume controls because human hearing perceives volume logarithmically
A 12 V, 100 mA dc motor needs speed control from 0 to full speed. At full speed, motor resistance ≈ 120 Ω (R = V/I = 12/0.1). A wirewound rheostat of 0–120 Ω is connected in series. At full resistance: voltage across motor ≈ 6 V; speed ≈ 50%. At zero resistance: full 12 V; full speed. This simple method wastes energy as heat in the rheostat — modern VFDs or PWM controllers are more efficient for larger motors.
Thermistors
Thermistors are temperature-sensitive resistors made from sintered semiconductor oxides (manganese, nickel, cobalt). Unlike metallic conductors (which have a small, linear positive temperature coefficient), thermistors have a large, non-linear temperature coefficient that makes them extremely sensitive temperature sensors.
| Type | Behaviour | Temperature range | Applications |
|---|---|---|---|
| NTC (negative temperature coefficient) | Resistance decreases as temperature increases — typically 10× decrease per 50°C rise | −50°C to +150°C | Temperature measurement (replacing bimetallic strips); temperature compensation in electronic circuits; in-rush current limiting (thermistor cools = high resistance limits initial current; warms = low resistance, normal operation); thermostats in boilers (Chapter 11) |
| PTC (positive temperature coefficient) | Resistance increases with temperature — very sharply at a critical switching temperature (may increase 10,000× over a narrow range) | −20°C to +120°C (typical) | Motor winding protection — PTC thermistors embedded in each phase winding; when winding temperature exceeds critical point, resistance rises dramatically, triggering the protection relay and shutting the motor off |
A three-phase induction motor has three PTC thermistors embedded in the stator windings (one per phase). Each has a nominal resistance of 100 Ω at 20°C, rising to >4000 Ω at 130°C (the critical protection temperature). The three thermistors are connected in series (total resistance ≈ 300 Ω normally) and connected to an electronic relay module. When any winding reaches 130°C (e.g. due to overload, stall, or cooling fan failure), the total resistance exceeds the relay's trip threshold (~5000–10000 Ω), the relay de-energises, and the motor contactor opens. The motor stops before the winding insulation is damaged. This is more reliable than a simple thermal overload relay which measures current — the PTC directly senses the winding temperature.
Light-Dependent Resistors (LDRs)
An LDR (also called a photoresistor or photocell) is a passive component whose resistance decreases as the intensity of light falling on it increases. The semiconductor material (most commonly cadmium sulphide, CdS) generates additional free electrons when photons are absorbed — reducing resistance. In darkness, resistance can be 1 MΩ or more; in bright light, resistance falls to 100–1000 Ω.
An LDR is connected in series with a fixed resistor (100 kΩ) across a 12 V supply, forming a potential divider. The junction voltage feeds the input of an electronic switch (transistor or IC). In daylight (LDR resistance = 5 kΩ): \(V_{\text{junction}} = 12 \times 5/(100+5) = 0.57\text{ V}\) — below the switching threshold, street light OFF. At dusk (LDR resistance = 100 kΩ): \(V_{\text{junction}} = 12 \times 100/(100+100) = 6\text{ V}\) — above threshold, electronic switch activates relay, street light ON. This circuit demonstrates how a passive sensor (LDR) combined with a potential divider and active component (transistor) creates an automatic control system.
Colour Code (4-Band System)
Most resistors are too small to have their value printed legibly on the body. Instead, coloured bands are used. For the most common 4-band system, the bands are read from left to right (the first band is the one closest to the lead end):
- Band 1: first significant digit
- Band 2: second significant digit
- Band 3: multiplier (power of 10 to multiply the two-digit number by)
- Band 4: tolerance (how accurate the stated value is)
Memory aid for the colour sequence (0–9): Black Brown Red Orange Yellow Green Blue Violet Grey White — "Bad Boys Race Our Young Girls But Violet Generally Wins."
| Colour | Digit value (Bands 1 & 2) | Multiplier (Band 3) | Tolerance (Band 4) |
|---|---|---|---|
| Black | 0 | ×1 | — |
| Brown | 1 | ×10 | ±1% |
| Red | 2 | ×100 | ±2% |
| Orange | 3 | ×1,000 | — |
| Yellow | 4 | ×10,000 | — |
| Green | 5 | ×100,000 | ±0.5% |
| Blue | 6 | ×1,000,000 | ±0.25% |
| Violet | 7 | ×10,000,000 | ±0.1% |
| Grey | 8 | ×100,000,000 | ±0.05% |
| White | 9 | ×1,000,000,000 | — |
| Gold | — | ×0.1 | ±5% |
| Silver | — | ×0.01 | ±10% |
| None | — | — | ±20% |
Band 1 = Yellow = 4; Band 2 = Violet = 7; Band 3 = Orange = ×1,000; Band 4 = Gold = ±5%.
Value: 47 × 1,000 = 47,000 Ω = 47 kΩ ±5%. Acceptable range: 44.65 kΩ to 49.35 kΩ.
Band 1 = Red = 2; Band 2 = Yellow = 4; Band 3 = Orange = ×1,000; Band 4 = Gold = ±5%.
Value: 24 × 1,000 = 24,000 Ω = 24 kΩ ±5% ✓ (matches self-assessment answer).
Band 1 = Brown = 1; Band 2 = Black = 0; Band 3 = Red = ×100; Band 4 = Brown = ±1%.
Value: 10 × 100 = 1,000 Ω = 1 kΩ ±1%. Acceptable range: 990 Ω to 1,010 Ω.
This is a precision resistor (1% tolerance) — likely a metal film type used in a precision voltage divider or measurement circuit.
Band 1 = Green = 5; Band 2 = Blue = 6; Band 3 = Gold = ×0.1; Band 4 = Silver = ±10%.
Value: 56 × 0.1 = 5.6 Ω ±10%. Sub-ohm resistors (Gold or Silver multiplier) are used as current-sensing resistors in power supplies and motor drives.
5-Band Resistors (Precision)
Precision resistors (±1% and better) use a 5-band system with an additional significant digit:
- Band 1: first significant digit
- Band 2: second significant digit
- Band 3: third significant digit
- Band 4: multiplier
- Band 5: tolerance (usually brown for ±1%, red for ±2%)
Digits = 1, 0, 0; Multiplier = ×100; Tolerance = ±1%.
Value: 100 × 100 = 10,000 Ω = 10 kΩ ±1%.
Number-Letter Code (BS 1852)
The alphanumeric code (BS 1852) is used on PCB component references and in parts lists. The letter (R, K, or M) acts as the decimal point and indicates the unit:
- R = ohms (×1)
- K = kilohms (×1,000)
- M = megohms (×1,000,000)
| Code | Value | Reading method |
|---|---|---|
| R56 | 0.56 Ω | R = decimal point: 0.56 Ω |
| 1R0 | 1.0 Ω | R = decimal point: 1.0 Ω |
| 27R | 27 Ω | R at end = 27 Ω (no decimal) |
| 6K8 | 6.8 kΩ | K = decimal point: 6.8 kΩ |
| 470K | 470 kΩ | K at end = 470 kΩ |
| 3M3 | 3.3 MΩ | M = decimal point: 3.3 MΩ |
| 2M2 | 2.2 MΩ | M = decimal point: 2.2 MΩ |
Tolerance letters (appended after the value code): F = ±1%, G = ±2%, J = ±5%, K = ±10%, M = ±20%.
PCB label 22KJ: 22 kΩ (K = 22 × 1000), ±5% tolerance (J) = 22 kΩ ±5%.
PCB label 4R7K: 4.7 Ω (R = decimal), ±10% tolerance (K) = 4.7 Ω ±10%.
PCB label 100RJ: 100 Ω (R at end), ±5% tolerance (J) = 100 Ω ±5%.
This coding system is used on bill-of-materials sheets and in fault-finding guides for PCB-level repair of fire alarm control panels, VFD drives, and building management systems.
What a Capacitor Does
A capacitor stores electrical energy in an electric field between two conducting plates separated by an insulating material called the dielectric. The amount of charge stored for a given voltage is the capacitance (C), measured in farads (F). One farad is a very large unit — practical capacitors range from picofarads (pF, 10⁻¹² F) in radio circuits to thousands of microfarads (µF) in power supply smoothing. The energy stored by a charged capacitor is:
\[E = \frac{1}{2} CV^2 \quad (\text{joules})\]This stored energy makes capacitors both useful (power supply smoothing, energy storage) and dangerous — large electrolytic capacitors can retain charge for minutes or hours after a circuit is switched off, delivering a potentially fatal shock if accidentally discharged through a person.
Capacitors in DC Circuits
When a dc voltage is first applied to an uncharged capacitor, current flows as the plates charge up and the electric field builds. As the voltage across the capacitor rises toward the supply voltage, the charging current progressively falls to zero. Once fully charged, no further dc current flows — the capacitor blocks dc. The charge stored is:
\[Q = C \times V \quad (\text{coulombs})\]When the dc supply is disconnected, the capacitor retains its charge (the electric field is maintained) — it acts as a temporary voltage source. On discharge through a load, the voltage and current both decay exponentially with time constant \(\tau = RC\) (seconds).
A 100 µF capacitor charges through a 10 kΩ resistor from a 12 V dc supply. Time constant:
\[\tau = R \times C = 10000 \times 100 \times 10^{-6} = 1\text{ s}\]After 1 time constant (1 s), the capacitor has charged to 63.2% of 12 V = 7.6 V. After 5 time constants (5 s), it is 99.3% charged ≈ 11.9 V. This RC time constant is used to create time delays in control circuits — for example, a delayed-release relay that holds on for a set period after the input signal is removed.
Capacitors in AC Circuits
On one half-cycle, the ac supply charges the capacitor to one polarity. On the next half-cycle, the supply reverses and charges the capacitor to the opposite polarity — the capacitor repeatedly charges and discharges. This continuous charge reversal creates an apparent current flowing through the capacitor circuit. The opposition a capacitor offers to ac is the capacitive reactance X_C (in ohms):
\[X_C = \frac{1}{2\pi fC}\]X_C decreases with increasing frequency — at very high frequencies a capacitor is essentially a short circuit to ac. At dc (f = 0), X_C is infinite — a complete open circuit. This is the basis of the rule: capacitors block dc but pass ac. In practice they pass more ac as frequency increases.
A 10 µF capacitor:
\[X_C\text{ at 50 Hz} = \frac{1}{2\pi \times 50 \times 10 \times 10^{-6}} = \frac{1}{0.00314} = 318\text{ Ω}\] \[X_C\text{ at 1 kHz} = \frac{1}{2\pi \times 1000 \times 10 \times 10^{-6}} = \frac{1}{0.0628} = 15.9\text{ Ω}\]At 1 kHz the capacitor passes current 20× more easily than at 50 Hz — demonstrating the frequency-selective behaviour used in filter circuits (ADSL telephone line splitters, audio crossovers, VFD output filters).
Equivalent Capacitance — Series and Parallel
Capacitors combine by rules opposite to resistors — parallel capacitors add (more plates in parallel = more charge storage); series capacitors combine like parallel resistors (effective plate separation increases):
Series: \(\dfrac{1}{C_T} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} + \ldots\) (capacitance decreases — same as resistors in parallel)
Two capacitors in series: \(C_T = \dfrac{C_1 \times C_2}{C_1 + C_2}\)
10 µF and 22 µF in parallel: \(C_T = 10 + 22 = 32\text{ \mu F}\)
Step 1: two 50 µF in series: \(C_{50\text{s}} = \frac{50 \times 50}{50+50} = \frac{2500}{100} = 25\text{ \mu F}\)
Step 2: 25 µF in parallel with 100 µF: \(C_T = 25 + 100 = 125\text{ \mu F}\) ✓ (matches self-assessment answer)
A 4700 µF electrolytic capacitor in a power supply is charged to 325 V peak (the peak of a 230 V rms supply). Energy stored:
\[E = \frac{1}{2} CV^2 = \frac{1}{2} \times 4700 \times 10^{-6} \times 325^2 = 0.5 \times 4700 \times 10^{-6} \times 105625 = 248\text{ J}\]248 joules is enough energy to cause a serious or fatal electric shock. This capacitor must be fully discharged before working on the power supply — using a bleed resistor (permanently connected across the capacitor) or a discharge probe. Never assume large electrolytic capacitors are safe simply because the mains is disconnected.
Types of Capacitor
| Type | Dielectric | Value range | Polarized? | Max voltage | Key application |
|---|---|---|---|---|---|
| Mica | Mica sheets | 1 pF – 10 nF | No | Up to 500 V | High-frequency filters, precision tuning |
| Polystyrene (PS) | Polystyrene film | 100 pF – 10 nF | No | 100 V | Precision, low-loss; timing circuits |
| Polyester film (MKT) | Polyester film | 1 nF – 10 µF | No | 100–600 V | General purpose; decoupling; motor run caps |
| Polypropylene (MKP) | Polypropylene film | 1 nF – 100 µF | No | Up to 2000 V | Power factor correction; motor run; snubbers |
| Ceramic disc | Barium titanate ceramic | 1 pF – 100 nF | No | 50–500 V | Decoupling; bypass; high-frequency use |
| Aluminium electrolytic | Aluminium oxide (formed electrolytically) | 1 µF – 100,000 µF | Yes | 6.3–450 V | Power supply smoothing; large bypass; audio |
| Tantalum electrolytic | Tantalum oxide | 0.1 µF – 1000 µF | Yes | 4–50 V | Compact precision timing; signal coupling; low-voltage circuits |
A factory installs a 50 kVAr, 400 V three-phase PF correction capacitor bank. Each capacitor in the bank must: handle 400 V ac continuously (so specify at least 450 V rated polypropylene capacitors); tolerate the inrush current when first energised; operate in both half-cycles (non-polarized — electrolytic would be destroyed instantly on ac). Polypropylene film capacitors are specified for this application because they are non-polarized, have low loss (high efficiency), and tolerate the harmonic voltages present in industrial supplies. Electrolytic capacitors must never be used on ac supplies — the dielectric would break down on the reverse half-cycle.
Capacitor Identification and Safety
A capacitor's markings specify: capacitance value (in µF, nF, or pF), maximum working voltage (WVDC or VAC), type, and polarity markers for polarized types.
Polarity identification for electrolytics:
- The negative lead is marked with a stripe down the side of the capacitor body (containing minus signs)
- The positive lead is longer than the negative lead in new (uncut) components
- The positive terminal may also be marked with a + symbol on the case
- In metal-can (TO-3) electrolytics, the case itself is typically the negative terminal
- Never exceed the working voltage — the dielectric breaks down, creating a short circuit and potentially an explosion (especially electrolytics)
- Never reverse-connect a polarized (electrolytic or tantalum) capacitor — reverse voltage breaks down the dielectric; the capacitor may vent gas, catch fire, or explode violently
- Never connect a polarized capacitor to an ac supply
- Always discharge large capacitors before working on a circuit — use a discharge resistor (1–10 kΩ, rated for the stored energy) connected across the terminals
- In VFD drives and UPS systems, large dc link capacitors may remain charged to 600–800 V for minutes after mains disconnection — always wait for the "DC LINK DISCHARGED" indicator before opening the enclosure
Semiconductor Fundamentals
Semiconductors have electrical conductivity between conductors and insulators, and their conductivity can be precisely controlled by adding impurity atoms (doping). Silicon (Si) is the most widely used semiconductor material:
- P-type silicon: doped with trivalent atoms (e.g. boron, aluminium) which have one fewer electron than silicon — creating "holes" (positive charge carriers). Conventional current in P-type flows from + to − (holes move toward the negative terminal).
- N-type silicon: doped with pentavalent atoms (e.g. phosphorus, arsenic) which have one extra electron — creating free electrons (negative charge carriers). Current flows from − to + (electrons move toward the positive terminal).
By joining P-type and N-type silicon at a junction (PN junction), a diode is created. More complex arrangements (PNP or NPN) create transistors; four-layer structures (PNPN) create thyristors.
PN Junction Diode — Operating Principle
A diode has two terminals: the anode (P-type end) and the cathode (N-type end). Its fundamental property is that it conducts current readily in one direction but blocks it in the other:
| Bias condition | Connection | Behaviour | Voltage across diode |
|---|---|---|---|
| Forward biased | Positive to anode (P), negative to cathode (N) | Conducts — acts as a closed switch with small voltage drop | ~0.6–0.7 V (silicon), ~0.3 V (germanium), ~3 V (blue LED) |
| Reverse biased | Negative to anode (P), positive to cathode (N) | Blocks — acts as an open switch; only tiny leakage current flows | Full supply voltage appears across the diode |
| Reverse breakdown | Reverse bias exceeds breakdown voltage (V_BR) | Diode conducts in reverse — destroys ordinary diodes; deliberate in Zener diodes | Equal to the reverse breakdown voltage |
A relay coil connected to a 24 V dc supply. When the transistor switch opens, the coil's collapsing magnetic field induces a back-emf that would damage the transistor. A flyback diode is connected in reverse-bias across the coil (anode to negative supply, cathode to positive supply). During normal operation: diode reverse-biased — passes no current. When transistor opens: back-emf reverses the coil voltage — the diode becomes forward-biased and provides a path for the induced current, clamping the voltage spike to approximately 0.7 V. The transistor is protected from voltages exceeding its maximum rating.
Diode Types and Applications
| Type | Symbol identifier | Key characteristic | Application |
|---|---|---|---|
| Signal diode | e.g. 1N4148 | Small (100 mA max), fast switching (4 ns), low forward voltage | High-speed switching, signal detection, protection in logic circuits |
| Rectifier (power) diode | e.g. 1N4007 (1 A, 1000 V) | Higher current capacity (1–100 A), higher reverse voltage rating | Mains rectification in power supplies, battery chargers, UPS systems |
| Zener diode | e.g. BZX79C5V1 (5.1 V Zener) | Designed to operate in reverse breakdown — maintains constant reference voltage across its terminals | Voltage reference, voltage regulation, over-voltage protection |
| Schottky diode | e.g. 1N5817 | Metal-semiconductor junction; very low forward voltage (0.2–0.4 V); extremely fast switching | Power supply rectification where efficiency is critical; fast switching in VFD snubber circuits |
| Photodiode | — | Reverse-biased PN junction that generates current proportional to incident light intensity | Optical sensors, fibre-optic receivers, smoke detectors, light barriers |
| LED | — | Forward-biased junction emits photons; colour determined by bandgap energy (Chapter 10) | Indicator lights, displays, illumination (Chapter 10), optocouplers |
A 5.1 V Zener diode provides a stable 5.1 V reference from an unstabilised 12 V supply. A series resistor R limits the Zener current. If load current is 20 mA and minimum Zener current (for regulation) is 5 mA, total current = 25 mA. Resistor value:
\[R = \frac{V_{\text{supply}} - V_Z}{I_T} = \frac{12 - 5.1}{0.025} = \frac{6.9}{0.025} = 276\text{ Ω}\]Select 270 Ω (next standard value). Power in resistor: \(P = I^2 R = 0.025^2 \times 276 = 0.173\text{ W}\) — a 0.5 W resistor is adequate. This simple 5.1 V reference circuit is used to power microcontroller logic from a higher voltage supply in fire alarm panels and building automation systems.
Rectifier Circuits
Rectifier circuits convert ac to dc using diodes. The dc output always has some ripple (variation) that must be smoothed for most electronic circuits:
| Circuit | Diodes needed | Output | Ripple frequency | Average dc output |
|---|---|---|---|---|
| Half-wave | 1 | Only positive (or negative) half-cycles; half the ac is "missing" | 50 Hz (= supply frequency) | 0.318 × V_peak = 0.45 × V_rms |
| Full-wave bridge | 4 (in bridge) | Both half-cycles rectified; both contribute to dc output | 100 Hz (= 2 × supply frequency) | 0.637 × V_peak = 0.9 × V_rms |
A bridge rectifier is supplied from a 12 V rms transformer secondary. Average dc output voltage (before smoothing) = 0.9 × 12 = 10.8 V. Peak voltage = 12 × 1.414 = 16.97 V. After smoothing with a large capacitor, output rises toward the peak value minus the two diode forward voltage drops (2 × 0.7 = 1.4 V): smoothed output ≈ 16.97 − 1.4 = 15.6 V dc. This is why a "12 V transformer" combined with a bridge rectifier and smoothing capacitor gives approximately 15–16 V dc — important to know when specifying power supply components.
Transistors — Bipolar Junction Transistors (BJTs)
A BJT is a three-terminal device comprising two PN junctions — PNP or NPN. The three terminals are the base (B), collector (C), and emitter (E). The small base current controls the much larger collector current — making it a current amplifier.
The current gain (β or h_FE) is the ratio of collector current to base current:
\[\beta = \frac{I_C}{I_B} \qquad \text{typically 50–500 for small-signal transistors}\]Two key operating modes:
- Switch mode (saturation/cut-off): base current either absent (transistor fully OFF — high collector-emitter voltage, zero collector current) or sufficient to saturate the transistor (fully ON — very low collector-emitter voltage, maximum collector current). Used to switch relay coils, LED arrays, motor drivers.
- Amplifier mode (active region): base current at an intermediate level; collector current proportional to base current × gain. Used in audio amplifiers, sensor signal conditioning, and analogue control circuits.
A fire alarm panel PLC output provides 5 V at 2 mA — insufficient to drive a 24 V relay coil (requiring 50 mA). An NPN transistor (e.g. BC547, β = 200) is used as a switch:
Base current needed to saturate transistor: \(I_B = I_C / \beta = 50 / 200 = 0.25\text{ mA}\). A base resistor of \(R_B = (5 - 0.7) / 0.25 = 17.2\text{ kΩ}\) — use 15 kΩ (closest standard value) to ensure saturation. The PLC drives 2 mA through 15 kΩ to the base; the transistor delivers 50 mA to the relay coil from the 24 V supply. Current amplification ratio = 50/2 = 25:1. A flyback diode across the relay coil protects the transistor from the inductive back-emf.
Modern VFD inverters use IGBTs (Insulated Gate Bipolar Transistors) rather than BJTs. IGBTs switch much faster (up to 100 kHz), handle higher voltages (600–1700 V), and are voltage-controlled (gate, not current-controlled base) — requiring virtually zero gate drive power. A 75 kW VFD contains six IGBTs arranged in a three-phase bridge inverter, each switching on and off ~10,000 times per second to synthesise the variable-frequency output. This is why VFDs can control motor speed smoothly and efficiently — the speed of solid-state switching with the power handling of large thyristors.
Thyristor (Silicon-Controlled Rectifier — SCR)
A thyristor is a four-layer (PNPN) semiconductor device with three terminals: anode (A), cathode (K), and gate (G). It behaves like a diode with a controlled switching threshold — it will not conduct even when forward-biased until a trigger pulse is applied to the gate. Once triggered into conduction, it latches ON and continues conducting even if the gate pulse is removed. It can only be switched OFF by reducing the current through it below a minimum "holding current" (which occurs naturally at the zero-crossing of an ac supply).
Operating states:
- Blocking state: forward biased but no gate pulse — acts as open circuit. No current flows.
- Triggered (ON) state: gate pulse applied — switches to conducting state, voltage drop falls to ~1–2 V, current limited only by external circuit resistance.
- Reverse blocking: reverse biased — acts as open circuit (like a reverse-biased diode).
Phase-angle control on an ac supply: The thyristor is triggered at a controlled point in each positive half-cycle. Triggering at α = 0° (very start of positive half-cycle) allows maximum conduction — full power. Triggering at α = 90° allows only the second quarter of the half-cycle — reduced power. The average power to the load is thus controlled by varying the firing angle α from 0° (full power) to 180° (zero power).
A thyristor dc motor drive (Ward-Leonard drive) uses a thyristor bridge to convert ac to variable dc for the motor armature. At firing angle α = 0°, average dc output ≈ 0.9 × V_rms (full speed). At α = 60°, average dc ≈ 0.45 × V_rms (half speed). At α = 90°, output ≈ 0 V (motor stopped). The firing angle is controlled by the speed reference signal from a potentiometer or PLC — giving smooth, stepless speed control of large dc motors (e.g. rolling mills, mine hoists, paper mills) from 0 to full speed.
Main limitation: a single thyristor only conducts during positive half-cycles — it controls only 50% of the available ac cycle. For full ac power control, two thyristors back-to-back (one per half-cycle) or a Triac (the equivalent in a single package) are needed.
Triac — Bi-directional Thyristor
A Triac is equivalent to two thyristors connected in antiparallel (back-to-back) in a single semiconductor device — it can switch current in both directions. Terminals: Main Terminal 1 (MT1), Main Terminal 2 (MT2), and Gate (G). There is no anode or cathode — either MT can be the "anode" depending on the polarity.
A trigger pulse to the gate switches the Triac ON in either the positive or negative half-cycle, making it ideal for ac power control. The firing angle determines what fraction of each half-cycle the Triac conducts:
| Firing angle (α) | Fraction of cycle conducted | Power to load (% of full) |
|---|---|---|
| 0° | Full half-cycle (180°) | 100% — maximum power |
| 45° | 135° of each half-cycle | ~85% |
| 90° | 90° of each half-cycle | ~50% |
| 135° | 45° of each half-cycle | ~15% |
| 180° | 0° — Triac never fires | 0% — zero power |
A 250 W incandescent lamp is controlled by a trailing-edge triac dimmer. The user turns the dial — this varies a resistor in an RC timing circuit which controls the firing angle of the Triac. At full brightness: Triac fires at ~10° → conducts for 170° of each half-cycle → 100% power. At half brightness: Triac fires at ~90° → 50% power. At minimum: Triac fires at ~170° → ~3% power (just enough to sustain the arc but lamp appears very dim).
Important: LED and CFL lamps require specific "LED-compatible" dimmers. Standard Triacs may not fire correctly at the very low currents drawn by LEDs, causing flicker or failure to dim smoothly. Modern LED dimmers use more sophisticated firing algorithms or MOSFET-based circuits to handle LED drivers correctly.
Diac — Bilateral Trigger Device
A Diac (DIode AC switch) is a two-terminal semiconductor device that blocks current in both directions until the voltage across it exceeds its breakover voltage (typically 28–32 V). At breakover, it conducts and presents a low impedance. Unlike a diode, it conducts symmetrically in both directions (bidirectional). After conduction the voltage across it drops to a lower holding voltage.
The Diac is almost always used in combination with a Triac as a triggering device. The Diac and an RC network set the firing angle of the Triac precisely and symmetrically for both half-cycles — providing more stable, reproducible control than a direct gate resistor. The RC circuit charges to the Diac breakover voltage, at which point the Diac fires and delivers a sharp current pulse to the Triac gate.
In a standard phase-control dimmer: (1) At the start of each half-cycle, the mains voltage charges capacitor C through resistor R. (2) When C reaches the Diac breakover voltage (~30 V), the Diac fires. (3) The Diac delivers a sharp current pulse to the Triac gate. (4) The Triac turns on and conducts for the remainder of the half-cycle. (5) At the zero-crossing, the Triac turns off automatically. (6) Cycle repeats. Varying R changes the time constant, changing when C reaches breakover — adjusting the firing angle from 0° to 180°.
Inverters — DC to AC Conversion
An inverter converts direct current (dc) to alternating current (ac). It is the essential final stage of variable frequency drives (Chapter 8, Part 10), solar PV systems (Chapter 6), UPS systems, and battery-to-mains power converters. The inverter does not require a transformer for operation (unlike older vibrator-based inverters) — it uses high-frequency semiconductor switching to synthesise an ac output waveform from a dc input.
How a modern IGBT inverter works:
- Six IGBT switches are arranged in three pairs (one pair per phase for a three-phase output)
- Each pair switches rapidly (typically 2–16 kHz) between the positive and negative dc bus rails
- A PWM (Pulse Width Modulation) algorithm controls the on/off time of each IGBT to synthesise a sinusoidal average voltage at the desired frequency and amplitude
- The motor's inductance filters the PWM switching — the motor sees an approximately sinusoidal current even though the voltage is a series of pulses
- By varying the PWM frequency and modulation depth, any output frequency (1–400 Hz) and any voltage (up to the dc bus voltage) can be generated
Output waveform types:
- Modified sine wave (quasi-sine): simple staircase approximation. Suitable for resistive loads (heaters, incandescent lamps) and some motors. Not suitable for sensitive electronics, medical equipment, or active power factor correction loads.
- Pure sine wave: very high-quality PWM filtering gives a low-distortion sine wave output indistinguishable from the mains. Required for sensitive electronics, VFD-driven motors (minimises harmonic heating), medical equipment, and active loads. More complex and expensive.
A 4 kWp solar array produces dc (typically 300–800 V depending on string configuration). A grid-tie inverter converts this to 230 V ac at exactly 50.00 Hz, synchronised to the grid phase. The inverter constantly monitors grid frequency and voltage, adjusting its output to match. It must disconnect automatically within 5 seconds if the grid fails (anti-islanding protection — to prevent the inverter energising a dead section of grid while engineers work on it). Connection to the public supply requires DNO approval under Engineering Recommendation G98/G99.
A server room UPS must supply 8 kVA for 15 minutes from a 96 V battery bank. The inverter output: 230 V ac, 50 Hz, 8 kVA. Battery current during discharge: \(I = P/(V \times \eta) = 8000/(96 \times 0.90) = 92.6\text{ A}\). Battery capacity required: \(92.6 \times 0.25 = 23.2\text{ Ah}\) (15 min = 0.25 h). Specify 30 Ah battery bank (margin for ageing and temperature derating). The inverter is rated for 8 kVA continuous output at unity power factor — server power supplies typically draw close to unity PF.
| Inverter application | Input dc | Output ac | Key requirement |
|---|---|---|---|
| VFD (motor drive) | ~565 V dc (from 400 V mains rectifier) | Variable frequency, variable voltage (0–400 V, 0–120 Hz) | PWM accuracy; heat sinking; EMC filtering (Chapter 8) |
| Solar grid-tie | Variable (200–800 V from PV array) | 230 V ac, 50 Hz (single-phase) or 400 V, 50 Hz (three-phase) | Grid synchronisation; anti-islanding; G98/G99 compliance |
| UPS | Battery voltage (24–480 V depending on system) | 230 V ac, 50 Hz pure sine wave | Transfer time <10 ms; pure sine wave output; battery monitoring |
| EV traction | Battery pack (300–800 V) | Three-phase ac to traction motor | Very high efficiency; regenerative braking (motor becomes generator) |
| 12 V to 230 V (portable) | 12 V battery | 230 V ac, 50 Hz (modified or pure sine) | High current from 12 V (e.g. 1000 W = 83 A at 12 V) |
Electronic components — individually or integrated into ICs (integrated circuits) on PCBs (printed circuit boards) — now form the heart of virtually every electrical installation system: fire alarms, access control, emergency lighting, motor drives, building management systems, and renewable energy systems all depend on the electronic principles covered in this chapter. Understanding the components allows electricians to understand system behaviour, interpret fault conditions, and correctly specify and interface with electronic control equipment.
All electronic control circuits require a suitable dc power supply, typically derived from the mains ac supply using a bridge rectifier, smoothing capacitors, and a voltage regulator (Zener diode or IC regulator) to provide stable 5 V, 12 V, or 24 V dc.
Dimmer Switches
Modern phase-angle dimmer switches use a Triac triggered via a Diac (the Diac-Triac circuit described in Part 5) to vary the fraction of each ac half-cycle that reaches the lamp. The user's control knob varies a resistor in the RC network, shifting the firing angle from near 0° (bright) to near 180° (dim).
Why Triacs, not thyristors: A thyristor only conducts during positive half-cycles — using one would produce a pulsating half-wave output that causes visible 50 Hz flicker and produces dc current which would saturate transformer-type LED drivers. A Triac conducts on both half-cycles, producing a balanced output and correct waveform.
A standard leading-edge Triac dimmer was designed for resistive and inductive loads (incandescent and halogen lamps). LED lamps use an electronic driver that may not conduct at the very start of each half-cycle — the Triac needs a minimum "hold-on" current to remain conducting, and an LED driver's current waveform may drop below this level, causing the Triac to drop out early. This produces flicker, buzzing, and incomplete dimming. The solution: use a trailing-edge (MOSFET-based) dimmer specifically rated for LED loads, or an LED driver specified as "dimmable" and rated for the chosen dimmer type. Always match the dimmer and driver type at the design stage.
Heating Control Systems
Thyristors and Triacs control electrical heating loads (kilowatt-scale) using two distinct methods, each with different advantages:
| Method | Mechanism | Advantages | Best for |
|---|---|---|---|
| Phase-angle control | Triac fires at a controlled angle in each half-cycle — the later it fires, the shorter the conduction period and the less power delivered | Smooth, continuous control; responds quickly to temperature changes; inexpensive control circuit | Resistive heating elements that tolerate partial-cycle current (fan heaters, radiant panels); motor speed control (universal motors) |
| Burst (zero-crossing) control | Triac is fully ON for a fixed number of complete ac cycles, then fully OFF for a fixed number — the ratio of ON cycles to total cycles determines average power | Switching only at zero-crossings eliminates EMI; load sees full sine wave during ON periods; less stress on components; lower harmonic content | Slower-response thermal masses (furnaces, immersion heaters, large ovens); any load where EMI from phase-angle control is problematic |
A proportional temperature controller uses burst firing to maintain water at exactly 60°C. At 50% power demand: Triac fires for 5 complete 50 Hz cycles (100 ms), then off for 5 cycles (100 ms) — 50% average power = 1.5 kW heating. Each switch-on occurs at a voltage zero-crossing — no di/dt switching spike, no EMI. The element dissipates full 3 kW for 100 ms bursts, which the thermal mass of the water easily absorbs. This is more efficient than resistive phase control, which would require additional energy to charge/discharge the element inductance on each cycle.
Motor Temperature Protection
PTC thermistors embedded in motor windings protect against overheating from overload, stalling, loss of a phase, blocked ventilation, or ambient temperature extremes. Three PTC thermistors are embedded in the stator windings — one per phase, in the hottest slot — and connected in series to an electronic relay module:
A 15 kW four-pole cage induction motor with embedded PTC thermistors (critical temperature 130°C) is connected to a PTC relay module (e.g. Siemens 3RN1010-1CB00). The relay measures the series resistance of the three PTC thermistors and trips if resistance exceeds 3.5 kΩ (indicating temperatures above 130°C). During commissioning, check: (1) PTC thermistor circuit continuity (expect 200–400 Ω at 20°C for standard PTC types); (2) relay module supply voltage and trip threshold setting; (3) relay output wired to motor contactor coil so trip de-energises the contactor. The thermistor protection is independent of the thermal overload relay in the motor starter — both are required for full BS 7671 Regulation 552.1 motor protection compliance.
Motor Speed Control (Universal Motors)
A Triac in series with the supply provides phase-angle speed control for universal (AC/DC) motors found in power tools, vacuum cleaners, and kitchen equipment. A variable resistor sets the RC time constant which determines the Triac firing angle. At full resistance (late firing): low average voltage → slow speed. At minimum resistance (early firing): high average voltage → fast speed. A tachogenerator (small generator on the motor shaft) or back-emf sensing circuit can provide closed-loop speed feedback — automatically correcting the firing angle if load changes cause speed variation, maintaining constant speed under varying loads.
Security and Access Control Systems
Electronic components underpin all modern security systems:
- Passive infrared (PIR) motion detectors: a pyroelectric sensor (ferroelectric crystal) detects changes in infrared radiation from a warm body moving across its field of view. The sensor output is amplified by an IC op-amp, compared with a reference, and drives a transistor switch that operates the alarm relay.
- Active infrared beam sensors: an infrared LED transmitter emits a modulated beam (to reject ambient light interference). A photodiode receiver converts the beam to a dc signal processed by a comparator IC. Breaking the beam removes the receiver signal → transistor de-energises → relay activates alarm. Used for perimeter protection (door frames, windows, safes).
- Access control keypads: microcontroller ICs (containing millions of transistors) compare entered codes with stored values, control MOSFET or relay outputs to operate electric strikes or magnetic door locks.
Central Heating Boiler Control
Modern condensing boilers use multiple thermistors for precise temperature management:
- Flow temperature sensor (NTC thermistor): measures the temperature of water leaving the boiler heat exchanger. The boiler's electronic controller (microprocessor-based) adjusts the gas valve modulation to maintain the target flow temperature (typically 55–80°C for radiator systems, 35–45°C for underfloor heating).
- Return temperature sensor: measures the return water temperature; the difference between flow and return indicates heat transfer rate. Modern "weather compensation" controllers reduce flow temperature when outside temperature is mild — reducing condensing boiler gas consumption by 10–15%.
- Flue temperature sensor: monitors exhaust temperature; excessively high flue temperature may indicate poor combustion or heat exchanger fouling — triggers a fault code.
- Frost protection sensor: dedicated NTC thermistor on the boiler body; triggers the frost protection circuit independently of the programmer.
Fire Alarm Control Panels
A fire alarm control panel (FACP) is a sophisticated electronic system that integrates many of the components in this chapter:
- Microprocessor (IC): processes inputs from all loop devices; makes detection decisions based on programmed algorithms
- Transistor switch arrays: drive sounder circuits and relay outputs for door releases, fire damper controls, and lift recall
- Zener diodes: voltage references for detection threshold circuits
- NTC thermistors: in some heat detector types for rate-of-rise detection
- Bridge rectifier + electrolytic capacitors: the ac/dc power supply providing 24 V dc to the panel and detection loop
- Lead-acid battery + charger circuit: standby supply, monitored by the processor for fault condition (BS 5839 requires 24-hour standby)
A fire alarm panel displays a "PSU Fault" warning. Possible causes: (1) Mains supply failure (check 230 V ac input, check MCB in distribution board); (2) Failed bridge rectifier — measure dc bus voltage; should be ~27–28 V dc. If absent, one or more bridge diodes may be open-circuit. (3) Failed smoothing capacitor — if dc bus voltage is present but very low or heavily rippled, the large smoothing electrolytic may have failed (common failure mode in electrolytics older than 5 years). (4) Battery disconnect — check battery fuse and battery terminal voltage (should be 24–27.6 V). (5) Battery overvoltage — Zener reference failure in charger circuit causing overcharge. In each case, understanding which electronic component is responsible for each function enables systematic fault isolation.
ADSL Telephone Line Filters (Splitters)
ADSL internet uses the same copper pair as telephone service — but at much higher frequencies (25 kHz–1.1 MHz for ADSL2+) than voice (300 Hz–3.4 kHz). Without a filter, the broadband and voice signals would interfere. A microfilter (splitter) installed at each telephone socket separates the signals using LC filters:
- Low-pass filter (inductor + capacitor): passes voice frequencies (below ~4 kHz) to the telephone — blocks the high-frequency internet signal
- High-pass filter (capacitor + resistor): passes broadband frequencies (above ~25 kHz) to the modem/router — blocks the low-frequency voice signal
Understanding LC filter principles (capacitors block dc and pass high-frequency ac; inductors pass dc and low-frequency ac but block high frequency) is directly applicable to other filtering applications: EMC input filters on VFDs, power supply noise suppression, and audio speaker crossover networks.
🃏 Chapter 12 Flashcards — Test Your Knowledge
Click any card to reveal the answer. Use these to revise all six sections.
📝 Chapter 12 — Multiple Choice Questions
Select your answer and click Submit. You have two attempts per question. First-attempt score shown at the end.
1What is the key distinction between an active electronic component and a passive one?
2A 470 Ω resistor is connected across a 24 V dc supply. It dissipates 1.23 W continuously. Which power rating should be selected, applying good engineering practice of 50% derating?
3A resistor is colour-coded: Yellow, Violet, Orange, Gold. What is its value and tolerance?
4A PTC thermistor is embedded in the stator winding of a three-phase induction motor. What happens when the winding temperature reaches the critical protection threshold?
5Two 50 µF capacitors are connected in series, and that combination is then connected in parallel with a 100 µF capacitor. What is the total equivalent capacitance?
6Why must an electrolytic capacitor never be connected to an ac supply or with reversed polarity?
7A silicon diode in a circuit has its anode connected to +12 V and its cathode to +5 V. What is the approximate voltage across the diode and its state?
8A full-wave bridge rectifier is fed from a 12 V rms transformer secondary. What is the approximate average dc output voltage before smoothing?
9An NPN transistor (β = 200) is used as a switch to drive a 24 V relay coil requiring 50 mA. What base current is needed to saturate the transistor, and why is a flyback diode essential?
10A thyristor (SCR) is triggered into conduction by a gate pulse during the positive half-cycle of an ac supply. The gate pulse is then removed. What happens next?
11In a domestic light dimmer, a Triac is fired at a 90° phase angle. Approximately what percentage of full power reaches the lamp?
12A VFD drive contains large dc link electrolytic capacitors charged to approximately 565 V when the mains supply is connected. What is the correct safety procedure before opening the drive enclosure for maintenance?
Sends your Chapter 12 MCQ score summary from your registered email to your tutor.
Mock Tests
- Three cumulative quizzes drawn from every chapter (Ch 1–12)
- Each quiz contains 48 questions — 4 per chapter in rotation
- One attempt per question — choose carefully!
- Finish & Submit to reveal your score and grade
- Review answers with detailed explanations after submitting
- Fail (<50%) · Pass (50–64%) · Merit (65–79%) · Distinction (≥80%)
📝 Mock Test 1
Answer all 48 questions — one attempt each. Click Finish & Submit when done.
1What is the value of ¾ + ⅙?
2Transpose V = IR to make I the subject.
3Express 4,700,000 Ω in standard form.
4A right-angled impedance triangle has R = 8 Ω and XL = 6 Ω. What is the impedance Z and the power factor (cos θ)?
5What are the seven SI base units? Which one directly measures electric current?
6The cross-sectional area of a 75 mm × 75 mm trunking is:
7Conductor 1: diameter 4 mm, length 2 m. Conductor 2: same material, diameter 2 mm. What length of conductor 2 gives the same resistance?
8Resistance measurements with an ohmmeter must be taken with the supply disconnected, and the ohmmeter uses its own internal supply. Which statement is correct?
9A bundle of conduit has a mass of 200 kg. What force is required to lift it?
10A lever has a load of 300 N at 0.2 m from the pivot. What effort at 0.6 m from the pivot maintains equilibrium?
11A driving gear has 30 teeth and meshes with a driven gear of 90 teeth. If the driver rotates at 1500 rpm, what is the driven gear speed?
12A hoist lifts a mass of 20 kg through 15 m in 40 s. What is the output power?
13How many electrons does a copper atom have in its neutral state?
14A conductor has resistivity ρ = 1.72 × 10⁻⁸ Ωm, length 100 m, CSA 2.5 mm². What is its resistance?
15Three resistors of 6 Ω, 12 Ω and 4 Ω are connected in parallel. What is the total resistance?
16BS 7671 limits volt drop on a lighting circuit to 3% of the 230 V nominal supply. What is the maximum permitted volt drop?
17The unit of magnetic flux density is the:
18A conductor 0.4 m long carries 5 A in a field of 0.8 T perpendicular to the conductor. What force acts on it?
19The output from a rotating coil of a single-phase alternator is taken by:
20A coil of 200 turns has its flux change by 0.05 Wb in 0.1 s. What is the induced EMF?
21A CCGT (Combined Cycle Gas Turbine) power station has an overall efficiency of approximately:
22The UK supply voltage tolerance is 230 V +10% / −6%. What is the minimum permitted supply voltage at the consumer's terminals?
23Six 2 V lead-acid cells are connected in series. What is the terminal voltage and what happens to the capacity (Ah)?
24A transformer has copper losses of 15 kW and iron losses of 10 kW at full load. Full-load output is 900 kW. What is the efficiency?
25A coil has an inductance of 0.2 H. What is its inductive reactance XL at a frequency of 50 Hz?
26In a phasor diagram for a series R-L-C circuit, which quantity is used as the reference phasor (drawn horizontally at 0°)?
27A single-phase motor draws 10 A from a 230 V supply and consumes 1840 W of true power. What is the power factor?
28In a balanced three-phase star (Y) connected system, the line voltage is 400 V. What is the phase voltage?
29A 230 V dc motor has an armature resistance of 0.5 Ω and a back-EMF of 218 V at full speed. What is the armature current at full speed?
30Which type of dc motor maintains an almost constant speed regardless of load and is best suited to machine tools and centrifugal pumps?
31A motor connected to a 50 Hz supply runs at approximately 960 rev/min at full load. How many poles does it have?
32During the star phase of a star-delta starter, the voltage across each motor winding and the starting current are each reduced to what fraction of their direct-on-line values?
33What is the fundamental advantage of using a relay to switch a load rather than connecting the control signal directly to the load circuit?
34Which fuse type can safely interrupt fault currents up to 80–100 kA by limiting peak let-through current before it reaches its prospective maximum — protecting downstream equipment from full fault energy?
35Which is a genuine advantage of an MCB over a semi-enclosed (rewireable) fuse?
36What is the key advantage of fitting individual RCBOs to each final circuit rather than one shared RCD protecting all circuits?
37A photometer measures 450 lux at a point on a work surface. What does this value represent?
38Using the lumen method, how many lamps are required to illuminate a 20 m × 15 m factory to 400 lux, if each lamp delivers 5,000 lm, UF = 0.5 and MF = 0.8?
39A stroboscopic effect occurs in fluorescent-lit machine shops because the discharge arc extinguishes 100 times per second (at 50 Hz). Which solution completely eliminates this hazard?
40White LED lamps are most commonly produced using which method?
41A vehicle workshop has a high ceiling, poor insulation, and roller doors that are frequently opened. Which type of electric heater is most efficient and appropriate, and why?
42A 3 kW immersion heater at 230 V requires a dedicated circuit. Which combination correctly meets BS 7671 requirements?
43In an S-plan central heating system using two-port motorised zone valves, what must happen before the boiler and pump are energised?
44A rod thermostat used in an immersion heater works by differential expansion of an outer brass tube and an inner invar rod. Which metal expands MORE when heated, and what does this do?
45What is the key distinction between an active electronic component and a passive one?
46A PTC thermistor is embedded in the stator winding of a three-phase induction motor. What happens when the winding temperature reaches the critical protection threshold?
47A silicon diode in a circuit has its anode connected to +12 V and its cathode to +5 V. What is the approximate voltage across the diode and its state?
48A thyristor (SCR) is triggered into conduction by a gate pulse during the positive half-cycle of an ac supply. The gate pulse is then removed. What happens next?
Sends your Mock Test 1 score from your registered email to your tutor.
📝 Mock Test 2
Answer all 48 questions — one attempt each. Click Finish & Submit when done.
1A cable carries a current of 20 A and the allowable volt drop is 5%. The supply voltage is 230 V. What is the maximum permitted volt drop in volts?
2Evaluate using BODMAS: 10 + 3 × 2² − (12 ÷ 4)
3Five daily energy readings (kWh) are: 45, 50, 45, 60, 55. What is the mean daily energy consumption?
4Using trigonometry, an ac circuit has Z = 25 Ω and R = 20 Ω. What is the phase angle θ?
5Convert 85°C to Kelvin.
6A room measures 8 m × 5 m × 3 m. What is its volume?
7What is the difference between EMF and potential difference (pd)?
8In an AC circuit a wattmeter indicates the:
9A transformer of mass 800 kg has a base area of 0.8 m². What pressure does it exert on the floor?
10The effort required to lift a 40 kg load using a four-pulley system is:
11How much kinetic energy is possessed by a 2 kg mass moving at 4.5 m/s?
12A motor delivers 5 kW of mechanical output from a 7 kW electrical input. What is the efficiency?
13Conventional current flows from _____ to _____ in an external circuit.
14A 110 V supply feeds two 22 Ω resistors in series. What is the circuit current?
15A 230 V circuit has a measured load current of 10 A. What is the power consumed?
16Which of the following correctly describes the thermal effect of electric current?
17A solenoid coil has 500 turns and carries a current of 2 A. What is the magnetomotive force (MMF)?
18State Lenz's Law. What does it tell us about the direction of an induced EMF?
19The RMS voltage of a sinusoidal supply is 200 V. What is the peak voltage?
20State three advantages of three-phase AC generation over single-phase. Which of the following is NOT an advantage?
21Why is electricity transmitted at very high voltage (400 kV) rather than at low voltage?
22Which earthing system uses a combined neutral and protective earth (PEN) conductor in the supply cable, split at the consumer's earth block?
23A transformer has 1000 primary turns and 250 secondary turns supplied at 250 V. What is the secondary voltage?
24Why is an isolating transformer used in a bathroom shaver socket (BS 7671 Section 701)?
25A 100 µF capacitor is connected to a 50 Hz supply. What is its capacitive reactance XC? (Use 2π = 6.283)
26In a purely inductive ac circuit, what is the phase relationship between voltage and current?
27An installation has a true power of 15 kW and a reactive power of 20 kVAr. What is the apparent power in kVA?
28A balanced three-phase load draws a line current of 20 A at a line voltage of 400 V with a power factor of 0.85. What is the total true power consumed?
29What is the function of the commutator in a dc motor?
30In a three-phase cage-rotor induction motor, how is the rotor driven to rotate?
31A capacitor-start induction-run motor is used instead of a basic split-phase motor for a compressor because:
32A VFD maintains a constant V/f ratio when reducing motor speed. On a 400 V, 50 Hz motor running at half speed (25 Hz), the VFD output voltage should be:
33A machine guarding circuit requires the motor to run only when both guard switch A AND guard switch B are closed. Using relay N/O contacts, this AND logic is achieved by connecting the two contacts:
34A motor circuit fuse must tolerate a starting current surge of 7× full-load current for several seconds without blowing, while still protecting against sustained overload and faults. Which HBC fuse category is correct?
35Why will a standard 32 A MCB NOT protect a person receiving a 50 mA electric shock through a line-to-earth fault?
36A fault on a table lamp (plugged into a ring circuit socket) draws 30 A — seen by both a 13 A plug fuse AND the 32 A Type B ring circuit MCB. What demonstrates correct discrimination?
37A lamp of 600 cd is suspended 3 m directly above a workbench. What is the illuminance at the point directly below it?
38A lighting maintenance factor (MF) of 0.65 is appropriate for which type of environment?
39A high-pressure mercury vapour (HPMV/MBF) lamp fails to restart immediately after a brief supply interruption. What is the technical reason?
40An inductive fluorescent circuit draws 5 A at a poor power factor. The switch controlling it is not inductive-rated. According to BS 7671 Regulation 559.6.1.1, what minimum current rating must the switch have?
41A storage heater installed in a new-build dwelling must include which feature to comply with Building Regulations Part L?
42Why must a cylinder thermostat on a domestic hot water cylinder be set to a minimum of 60°C?
43Why is an automatic bypass valve required in a central heating system fitted with thermostatic radiator valves (TRVs) on all radiators?
44Which wiring plan for a domestic central heating system uses a single three-port mid-position motorised valve to serve both central heating and domestic hot water circuits?
45A 470 Ω resistor is connected across a 24 V dc supply. It dissipates 1.23 W continuously. Which power rating should be selected, applying good engineering practice of 50% derating?
46Two 50 µF capacitors are connected in series, and that combination is then connected in parallel with a 100 µF capacitor. What is the total equivalent capacitance?
47A full-wave bridge rectifier is fed from a 12 V rms transformer secondary. What is the approximate average dc output voltage before smoothing?
48In a domestic light dimmer, a Triac is fired at a 90° phase angle. Approximately what percentage of full power reaches the lamp?
Sends your Mock Test 2 score from your registered email to your tutor.
📝 Mock Test 3
Answer all 48 questions — one attempt each. Click Finish & Submit when done.
1Solve for x: 5x + 3 = 28
2Simplify: 2³ × 2⁻¹ ÷ 2²
3In a pie chart of building energy use, lighting accounts for 90 kWh out of a total 450 kWh. What is the sector angle for lighting?
4The decimal 0.000 001 is equal to which SI prefix?
5Velocity is measured in which unit?
6A copper busbar is 1 m long with a 40 mm × 5 mm cross-section. Density of copper = 8900 kg/m³. What is its mass?
7The UK mains frequency is 50 Hz. What is the periodic time of one complete cycle?
8A test instrument is rated CAT III. Which location is it suitable for?
9Which of the following is a vector quantity?
10A compound pulley system has:
11A cable drum of mass 50 kg is lifted 3 m. How much gravitational potential energy is stored?
12A motor (efficiency 85%) drives a gearbox (efficiency 92%). What is the overall system efficiency?
13Which cable insulating material has the highest temperature rating?
14Two 12 V 50 W lamps are connected in series across a 12 V supply. What is the total power dissipated?
15A 24 V supply circuit has a diode with a 0.6 V forward volt drop. What is the voltage across the load?
16What is the chemical effect of electric current called when current passes through an electrolyte causing material to deposit or dissolve?
17When applying Fleming's left-hand rule, the thumb indicates:
18A conductor of length 0.5 m moves at 4 m/s through a field of 1.5 T perpendicular to both. What EMF is induced?
19The maximum value of a sinusoidal current is 25 A. What is its average value?
20The UK mains supply has a frequency of 50 Hz. What is the angular frequency ω in radians per second?
21In the UK voltage hierarchy, at what voltage does the 11 kV/LV distribution transformer deliver power to final consumers?
22A wind turbine doubles its wind speed. By what factor does its power output change?
23Why are transformers rated in kVA rather than kW?
24What is the critical safety rule for a current transformer (CT) secondary circuit?
25A series R-L circuit has R = 5 Ω and XL = 12 Ω. What is the impedance Z?
26In a series R-L-C circuit at resonance, which statement is correct?
27What is the most common method of power factor correction for industrial installations?
28A three-phase installation has large numbers of single-phase computer loads. Even when the fundamental phase currents are balanced, the neutral conductor may carry significant current because:
29A dc series motor must never be run on no-load because:
30A 4-pole induction motor is connected to a 50 Hz supply and runs at 1440 rev/min at full load. What is the percentage slip?
31A double-cage rotor motor resolves a fundamental design conflict in single-cage motors by:
32A centrifugal fan motor rated at 40 kW is run at 80% of full speed by a VFD. Approximately what power does the fan consume at this reduced speed?
33A contactor is used to start and stop a standard three-phase cage-rotor induction motor (DOL). Which BS EN 60947-4 utilisation category must it be rated for?
34A Type D MCB has an instantaneous magnetic trip threshold of 10–20 × In. In which situation would it be specified in preference to Type B or Type C?
35A socket outlet circuit supplies a washing machine. Which RCD type does BS 7671 require, and why?
36A flyback (freewheeling) diode is connected in reverse-bias across a dc relay coil driven by a transistor. What is its purpose?
37A 1000 cd luminaire is mounted 3 m above a working plane. Find the illuminance at a point 4 m horizontally from directly below the lamp (use the cosine law).
38Which lamp type produces light by heating a tungsten filament to approximately 2500°C until it glows, and has a Colour Rendering Index (CRI) of 100?
39Low-pressure sodium (SOX) lamps have the highest efficacy of any lamp type (up to 160 lm/W). Why are they being replaced by LED for street lighting despite this efficiency advantage?
40An LED luminaire is installed in a ceiling with thermal insulation directly above it. What installation risk does this create, and what is required to avoid it?
41Underfloor heating cables must NOT be installed under fixed equipment such as baths, WC pans, or kitchen units. What is the technical reason?
42A bimetal strip thermostat uses two different metals bonded together. How does it open the contacts when temperature rises?
43A 9.5 kW electric shower requires a high power rating. What physical reason makes this necessary?
44A thermocouple works by generating a small EMF when its hot junction is heated (Seebeck effect). In a gas boiler pilot flame safety circuit, what happens if the pilot flame goes out?
45A resistor is colour-coded: Yellow, Violet, Orange, Gold. What is its value and tolerance?
46Why must an electrolytic capacitor never be connected to an ac supply or with reversed polarity?
47An NPN transistor (β = 200) is used as a switch to drive a 24 V relay coil requiring 50 mA. What base current is needed to saturate the transistor, and why is a flyback diode essential?
48A VFD drive contains large dc link electrolytic capacitors charged to approximately 565 V when the mains supply is connected. What is the correct safety procedure before opening the drive enclosure for maintenance?
Sends your Mock Test 3 score from your registered email to your tutor.