Resistance, Capacitance & 3-Phase Circuits

Resistance in Series

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When resistors are connected in series, the same current flows through each one. The total resistance is simply the sum of all individual resistances.

R_T = R₁ + R₂ + R₃
The total resistance is always greater than the largest individual resistor.

Key rules in series: Current is the same throughout. The supply voltage divides across each resistor: V_T = V₁ + V₂ + V₃.

Textbook Example

Three 6 Ω resistors in series across a 12 V battery.

R_T = 6 + 6 + 6 = 18 Ω

I = V/R = 12/18 = 0.67 A

V across each: V = IR = 0.67 × 6 = 4 V each (and 4+4+4 = 12 V ✅)

Extra Example

R₁ = 10 Ω, R₂ = 22 Ω, R₃ = 47 Ω across 24 V.

R_T = 10 + 22 + 47 = 79 Ω

I = 24/79 = 0.304 A

V₁ = 0.304 × 10 = 3.04 V, V₂ = 0.304 × 22 = 6.69 V, V₃ = 0.304 × 47 = 14.29 V

Check: 3.04 + 6.69 + 14.29 = 24.02 V ≈ 24 V ✅

Resistance in Parallel

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When resistors are in parallel, the same voltage appears across each one. The total current divides between the branches.

1/R_T = 1/R₁ + 1/R₂ + 1/R₃
The total resistance is always less than the smallest individual resistor.

Two resistors shortcut: R_T = (R₁ × R₂) / (R₁ + R₂) — "product over sum"

Textbook Example

Three 6 Ω resistors in parallel across 12 V.

1/R_T = 1/6 + 1/6 + 1/6 = 3/6 = 0.5

R_T = 1/0.5 = 2 Ω

I_T = V/R_T = 12/2 = 6 A

Each branch: I = 12/6 = 2 A (and 2+2+2 = 6 A ✅)

Extra Example — Product Over Sum

R₁ = 20 Ω and R₂ = 30 Ω in parallel.

R_T = (20 × 30) / (20 + 30) = 600 / 50 = 12 Ω

Notice: 12 Ω is less than the smallest resistor (20 Ω) ✅

Combined Series-Parallel Resistance

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Real circuits often mix series and parallel. The trick is to simplify step by step — solve the parallel groups first, then add them in series.

R₁ in series with (R₂ ∥ R₃), then in series with R₄

Worked Example

R₁ = 4 Ω in series with R₂ (12 Ω) ∥ R₃ (6 Ω), then R₄ = 2 Ω in series. Supply = 24 V.

Parallel first: R₂∥R₃ = (12×6)/(12+6) = 72/18 = 4 Ω

Then series: R_T = 4 + 4 + 2 = 10 Ω

I = 24/10 = 2.4 A

V across R₁ = 2.4 × 4 = 9.6 V; V across parallel = 2.4 × 4 = 9.6 V; V across R₄ = 2.4 × 2 = 4.8 V

Check: 9.6 + 9.6 + 4.8 = 24 V ✅

Capacitance in Series

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⚠ Key Difference from Resistors

Capacitors in series use the reciprocal formula (like resistors in parallel)!
Capacitors in parallel simply add up (like resistors in series)!

1/C_T = 1/C₁ + 1/C₂ + 1/C₃
Total capacitance in series is always less than the smallest capacitor.

Worked Example

C₁ = 10 μF, C₂ = 20 μF, C₃ = 30 μF in series.

1/C_T = 1/10 + 1/20 + 1/30 = 6/60 + 3/60 + 2/60 = 11/60

C_T = 60/11 = 5.45 μF (less than the smallest ✅)

Capacitance in Parallel

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C_T = C₁ + C₂ + C₃
Simply add them up! Total is always greater than the largest capacitor.

Worked Example

C₁ = 10 μF, C₂ = 47 μF, C₃ = 100 μF in parallel.

C_T = 10 + 47 + 100 = 157 μF

⚡ Memory Trick

Capacitors are the opposite of resistors: series uses reciprocal (like R parallel), parallel just adds (like R series).

Combined Series-Parallel Capacitance

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Just like resistors, simplify step by step: solve parallel groups first (add them), then combine in series (reciprocal).

Worked Example

C₁ = 20 μF in series with (C₂ = 30 μF ∥ C₃ = 10 μF).

Parallel first: C₂∥C₃ = 30 + 10 = 40 μF

Then series: 1/C_T = 1/20 + 1/40 = 2/40 + 1/40 = 3/40

C_T = 40/3 = 13.33 μF

Star & Delta Connections

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Three-phase generators produce three separate AC voltages 120° apart. These windings can be connected in two ways:

Property	Star (Y)	Delta (Δ)
Line voltage	V_L = √3 × V_P	V_L = V_P
Line current	I_L = I_P	I_L = √3 × I_P
Neutral	Available (4th wire)	Not available (3-wire)
Voltages available	230 V & 400 V	400 V only
Typical use	Distribution (homes/commercial)	Motors, transmission

√3 = 1.732 — this is the constant for all three-phase calculations.

Three-Phase Calculations

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Three-phase power: P = √3 × V_L × I_L × cos φ
This works for BOTH star and delta connections.

Textbook Example 1 — Star

Balanced star load, 10 Ω/phase, 400 V supply, unity PF.

V_P = V_L / √3 = 400 / 1.732 = 230.9 V

I_P = V_P / R = 230.9 / 10 = 23.09 A (= I_L in star)

P = √3 × 400 × 23.09 × 1 = 16 kW

Textbook Example 2 — Delta

20 kW balanced delta, 400 V, PF = 0.8.

I_L = P / (√3 × V_L × cos φ) = 20000 / (1.732 × 400 × 0.8) = 36.08 A

I_P = I_L / √3 = 36.08 / 1.732 = 20.83 A

Textbook Example 3 — Star vs Delta with Impedance

Three loads: R = 30 Ω, X_L = 40 Ω each, 400 V supply.

Z = √(30² + 40²) = √2500 = 50 Ω

Star: V_P = 400/1.732 = 230.9 V. I_P = 230.9/50 = 4.62 A = I_L

Delta: V_P = 400 V. I_P = 400/50 = 8 A. I_L = 1.732 × 8 = 13.86 A

Why Balanced Loads Matter

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A balanced three-phase load means every phase carries the same current. When balanced:

⚡ Benefits of Balanced Loads

1. The neutral current is zero — the three currents cancel by phasor addition
2. All line conductors can be the same size
3. Maximum efficiency from generators and transformers
4. No voltage imbalance at the loads
5. Reduced losses in cables and switchgear

⚡ Real-World Importance

Distribution boards should have single-phase loads spread evenly across L1, L2 and L3. If one phase carries significantly more current, the neutral carries the difference, cables overheat, and voltages become unequal. BS 7671 requires that every effort is made to balance loads.

Unbalanced Loads & Neutral Current

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When the three phase currents are not equal, the system is unbalanced and a current flows in the neutral conductor. This is found using the formula:

I_N = √(I_A² + I_B² + I_C² − I_AI_B − I_BI_C − I_AI_C)

Where I_A, I_B, I_C are the three phase currents.

Textbook Example

I_A = 20 A, I_B = 15 A, I_C = 10 A. Find I_N.

I_A² = 400, I_B² = 225, I_C² = 100

I_AI_B = 300, I_BI_C = 150, I_AI_C = 200

I_N = √(400 + 225 + 100 − 300 − 150 − 200)

I_N = √(725 − 650) = √75 = 8.66 A

Extra Example — Heavily Unbalanced

I_A = 30 A, I_B = 10 A, I_C = 25 A.

Squares: 900 + 100 + 625 = 1625

Products: (30×10) + (10×25) + (30×25) = 300 + 250 + 750 = 1300

I_N = √(1625 − 1300) = √325 = 18.03 A

Extra Example — Balanced Check

I_A = I_B = I_C = 15 A (balanced).

Squares: 225 + 225 + 225 = 675

Products: 225 + 225 + 225 = 675

I_N = √(675 − 675) = √0 = 0 A ✅ (proves balanced = zero neutral current)

⚠ Why This Matters

If the neutral conductor is undersized or disconnected in an unbalanced system, the star point voltage shifts. This causes overvoltage on lightly loaded phases (damaging equipment) and undervoltage on heavily loaded phases. This is why BS 7671 requires the neutral to be the same size as the line conductors in single-phase circuits and properly rated in three-phase systems.

Capacitive Reactance & Inductive Reactance

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In AC circuits, inductors and capacitors oppose the flow of current — but unlike resistance, this opposition depends on frequency. This frequency-dependent opposition is called reactance, measured in ohms (Ω).

Inductive Reactance (X_L)

An inductor opposes changes in current. The faster the current changes (higher frequency), the more it opposes. A coil with more inductance also opposes more.

X_L = 2πfL

Where: X_L = inductive reactance (Ω), f = frequency (Hz), L = inductance (Henrys, H)
2π = 2 × 3.142 = 6.284

Textbook Example

A coil of 0.15 H on a 50 Hz supply. Find X_L.

X_L = 2πfL = 2 × 3.142 × 50 × 0.15

X_L = 6.284 × 50 × 0.15 = 47.1 Ω

Extra Example

A 0.5 H inductor on a 60 Hz supply (US frequency). Find X_L.

X_L = 2 × 3.142 × 60 × 0.5

X_L = 6.284 × 30 = 188.5 Ω

Notice: higher frequency = higher reactance. The inductor opposes more at 60 Hz than it would at 50 Hz.

Capacitive Reactance (X_C)

A capacitor opposes changes in voltage. Unlike an inductor, a capacitor opposes less at higher frequencies (it lets more AC through).

X_C = 1 / (2πfC)

Where: X_C = capacitive reactance (Ω), f = frequency (Hz), C = capacitance (Farads, F)
Remember: C is usually given in μF — convert to Farads first! (divide by 1,000,000 or use × 10⁻⁶)

Textbook Example

A 60 μF capacitor on a 50 Hz supply. Find X_C.

Convert: 60 μF = 60 × 10⁻⁶ F = 0.00006 F

X_C = 1 / (2 × 3.142 × 50 × 0.00006)

X_C = 1 / 0.01885 = 53.05 Ω

Extra Example

An 8 μF power factor correction capacitor on 50 Hz. Find X_C.

C = 8 × 10⁻⁶ F

X_C = 1 / (2 × 3.142 × 50 × 8 × 10⁻⁶)

X_C = 1 / 0.002513 = 397.9 Ω

Small capacitors have very high reactance — they strongly oppose low-frequency AC.

Property	Inductive Reactance X_L	Capacitive Reactance X_C
Formula	X_L = 2πfL	X_C = 1 / (2πfC)
Unit	Ohms (Ω)	Ohms (Ω)
Frequency ↑	X_L increases	X_C decreases
At DC (f=0)	X_L = 0 (short circuit)	X_C = ∞ (open circuit)
Phase	Current lags voltage by 90°	Current leads voltage by 90°

⚡ CIVIL Memory Aid

Capacitor: I before V (current leads voltage)
Inductor (L): V before I (voltage leads current, i.e. current lags)

Impedance, Power Factor & the Impedance Triangle

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Impedance (Z) is the total opposition to current flow in an AC circuit. It combines resistance (R) and reactance (X) — but because R and X are 90° out of phase, we cannot simply add them. We must use Pythagoras' theorem.

Key Terms and Units

Term	Symbol	Unit	What It Is
Resistance	R	Ohms (Ω)	Opposition from resistive elements (heat-producing)
Inductive Reactance	X_L	Ohms (Ω)	Opposition from inductors (frequency-dependent)
Capacitive Reactance	X_C	Ohms (Ω)	Opposition from capacitors (frequency-dependent)
Impedance	Z	Ohms (Ω)	Total AC opposition (combines R and X)
Phase Angle	φ	Degrees (°)	Angle between voltage and current
Power Factor	cos φ	No unit (0 to 1)	Ratio of useful power to total power
True Power	P	Watts (W)	Useful power doing real work
Reactive Power	Q	VAr	Power stored and returned (not useful)
Apparent Power	S	VA	Total power drawn from supply

The Impedance Triangle — Pythagoras Applied

Z = √(R² + X²)

Where X = X_L (inductive) or X_C (capacitive), or X = X_L − X_C if both are present.

The impedance triangle: R along the base, X vertical, Z is the hypotenuse. Pythagoras finds Z; trigonometry finds φ.

Power Factor from the Impedance Triangle

Power Factor = cos φ = R / Z

PF = 1.0 (unity) → purely resistive, all power is useful
PF = 0 → purely reactive, no useful power
PF < 1.0 lagging → inductive load (motors) — current lags voltage
PF < 1.0 leading → capacitive load — current leads voltage

Textbook Example — RL Circuit

R = 50 Ω in series with a coil of X_L = 47.1 Ω, supply = 100 V.

Impedance: Z = √(50² + 47.1²) = √(2500 + 2218) = √4718 = 68.69 Ω

Current: I = V / Z = 100 / 68.69 = 1.46 A

Power factor: cos φ = R / Z = 50 / 68.69 = 0.728 lagging

Phase angle: φ = cos⁻¹(0.728) = 43.3°

Textbook Example — RC Circuit

R = 100 Ω in series with X_C = 53.05 Ω, supply = 230 V.

Z = √(100² + 53.05²) = √(10000 + 2814) = √12814 = 113.2 Ω

I = 230 / 113.2 = 2.03 A

PF = 100 / 113.2 = 0.883 leading

Extra Example — Full Power Calculation

R = 30 Ω, X_L = 40 Ω, supply = 250 V. Find Z, I, PF and all three powers.

Z = √(30² + 40²) = √(900 + 1600) = √2500 = 50 Ω

I = 250 / 50 = 5 A

PF = cos φ = 30 / 50 = 0.6 lagging

φ = cos⁻¹(0.6) = 53.13°

True Power: P = VI cos φ = 250 × 5 × 0.6 = 750 W

Reactive Power: Q = VI sin φ = 250 × 5 × 0.8 = 1000 VAr

Apparent Power: S = VI = 250 × 5 = 1250 VA

Check: S² = P² + Q² → 1250² = 750² + 1000² → 1,562,500 = 1,562,500 ✅

Extra Example — RLC Circuit

R = 12 Ω, X_L = 20 Ω, X_C = 15 Ω, supply = 100 V.

Net reactance: X = X_L − X_C = 20 − 15 = 5 Ω (inductive, because X_L > X_C)

Z = √(12² + 5²) = √(144 + 25) = √169 = 13 Ω

I = 100 / 13 = 7.69 A

PF = 12 / 13 = 0.923 lagging

⚠ Common Exam Mistake

Never add R and X directly! R = 30 Ω and X = 40 Ω does NOT give Z = 70 Ω. You must use Pythagoras because R and X are 90° apart. Z = √(30² + 40²) = 50 Ω.

Resistivity — Why Cable Size Matters

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Every material has a natural opposition to current flow called resistivity. Copper is an excellent conductor with very low resistivity. The resistance of a cable depends on three things: the material, its length, and its cross-sectional area.

R = ρL / A

Where:
R = resistance of the conductor (Ω)
ρ (rho) = resistivity of the material (Ωm — ohm metres)
L = length of the conductor (m)
A = cross-sectional area (m²)

Copper resistivity: ρ = 17.2 × 10⁻⁹ Ωm (or 1.72 × 10⁻⁸ Ωm)

What Affects Conductor Resistance?

Factor	Effect on Resistance	Why
Length ↑	Resistance increases	Electrons travel further, more collisions
Area ↑	Resistance decreases	More room for electrons to flow
Resistivity ↑	Resistance increases	Material opposes current more (e.g. steel vs copper)
Temperature ↑	Resistance increases (conductors)	Atoms vibrate more, impeding electron flow

Worked Example 1 — Finding Resistance

Find the resistance of 100 m of 2.5 mm² copper cable. (ρ = 17.2 × 10⁻⁹ Ωm)

Convert area to m²: 2.5 mm² = 2.5 × 10⁻⁶ m²

R = ρL / A = (17.2 × 10⁻⁹ × 100) / (2.5 × 10⁻⁶)

R = (1.72 × 10⁻⁶) / (2.5 × 10⁻⁶)

R = 0.688 Ω

Understanding the Maths — Moving Powers of 10 Between Numerator and Denominator

In Step 3 above we divided (1.72 × 10⁻⁶) by (2.5 × 10⁻⁶). Many students find dividing with negative powers confusing. Here's the key rule:

When you move a power of 10 from the bottom (denominator) to the top (numerator), the sign of the exponent flips.

10⁻⁶ in the denominator → becomes 10⁺⁶ in the numerator
10⁻⁹ in the denominator → becomes 10⁺⁹ in the numerator

This works the other way too: 10⁺³ moved from top to bottom becomes 10⁻³.

Let's prove it with a simple example:

What is 1 ÷ 10⁻⁶ ?

Dividing by 10⁻⁶ is the same as multiplying by 10⁺⁶ (the sign flips)

So: 1 ÷ 10⁻⁶ = 1 × 10⁶ = 1,000,000 ✅

It's that simple — whenever you divide by a negative power, just multiply by the positive power instead. No long division needed.

Now let's apply this to the resistivity calculation from Step 2 above:

R = (17.2 × 10⁻⁹ × 100) / (2.5 × 10⁻⁶)

Numerator: 17.2 × 10⁻⁹ × 100 = 17.2 × 10⁻⁹ × 10² = 17.2 × 10⁻⁷

Now we have: R = (17.2 × 10⁻⁷) / (2.5 × 10⁻⁶)

Move the 10⁻⁶ from the denominator to the numerator — it becomes 10⁺⁶:

R = (17.2 × 10⁻⁷ × 10⁺⁶) / 2.5

Combine the powers: 10⁻⁷ × 10⁺⁶ = 10⁻⁷⁺⁶ = 10⁻¹

R = (17.2 × 10⁻¹) / 2.5 = 1.72 / 2.5 = 0.688 Ω ✅

⚡ The Quick Rule

Dividing powers of 10: subtract the indices.
10⁻⁷ ÷ 10⁻⁶ = 10^{−7−(−6)} = 10⁻⁷⁺⁶ = 10⁻¹

Remember: subtracting a negative is the same as adding!
−7 − (−6) = −7 + 6 = −1

Worked Example 2 — Finding Required Cable Size

A circuit must have a maximum cable resistance of 0.5 Ω over 50 m. What minimum CSA is needed?

Rearrange: A = ρL / R

A = (17.2 × 10⁻⁹ × 50) / 0.5

A = 8.6 × 10⁻⁷ / 0.5 = 1.72 × 10⁻⁶ m²

Convert: 1.72 × 10⁻⁶ m² = 1.72 mm²

Next standard cable size up = 2.5 mm²

Worked Example 3 — Finding Resistivity

A 200 m cable of 4 mm² has a measured resistance of 0.86 Ω. What is the resistivity?

Rearrange: ρ = RA / L

ρ = (0.86 × 4 × 10⁻⁶) / 200

ρ = 3.44 × 10⁻⁶ / 200 = 1.72 × 10⁻⁸ Ωm

This confirms the cable is copper (ρ = 1.72 × 10⁻⁸ Ωm) ✅

⚡ Why Electricians Need This

Resistivity calculations are essential for volt drop calculations. The longer the cable run or the smaller the CSA, the higher the resistance, and the greater the volt drop. BS 7671 limits volt drop to 3% for lighting and 5% for power circuits. If the calculated volt drop is too high, you must increase the cable size.

Flash Cards — Click to Flip

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Click any card to reveal the answer on the back.

Test Yourself — 20 Questions

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Resistance in Series

Resistance in Parallel

Combined Series-Parallel Resistance

Capacitance in Series

Capacitance in Parallel

Combined Series-Parallel Capacitance

Star & Delta Connections

Three-Phase Calculations

Why Balanced Loads Matter

Unbalanced Loads & Neutral Current

Capacitive Reactance & Inductive Reactance

Inductive Reactance (XL)

Capacitive Reactance (XC)

Impedance, Power Factor & the Impedance Triangle

Key Terms and Units

The Impedance Triangle — Pythagoras Applied

Power Factor from the Impedance Triangle

Resistivity — Why Cable Size Matters

What Affects Conductor Resistance?

Flash Cards — Click to Flip

Test Yourself — 20 Questions

Inductive Reactance (X_L)

Capacitive Reactance (X_C)