Understanding AC circuits through graphical representation
EAL Level 3 • Unit ELEC3/008 • AC TheoryIn DC circuits, voltage and current stay constant — simple numbers are enough. But in AC circuits, voltage and current are constantly changing in a wave pattern. We need a way to show both how big a quantity is (its magnitude) and when it happens relative to other quantities (its phase angle). That's exactly what a phasor does.
Think of a phasor like the hand of a clock — its length tells you the size of the quantity, and the angle it points tells you its timing relative to a reference.
The UK mains supply is a sinusoidal (sine wave) alternating current at 50 Hz, meaning the waveform completes 50 full cycles every second. The voltage alternates between +325 V and −325 V, giving an RMS (effective) value of 230 V.
When we compare the timing of voltage and current waveforms in an AC circuit, there are three possible relationships:
In a circuit containing resistance only (heaters, kettles, incandescent lamps), the current and voltage rise and fall together. They are in phase.
In a circuit with inductance (motors, transformers, fluorescent lighting), the current reaches its peak after the voltage. We say current lags voltage. In a pure inductor, the lag is 90°.
In a circuit with capacitance, the current reaches its peak before the voltage. We say current leads voltage. In a pure capacitor, the lead is 90°.
To draw phasor diagrams by hand, you need: a sharp pencil, a ruler, a protractor, and graph paper (optional but helpful).
Draw a phasor representing 100 V at the reference angle (0°).
Draw V = 230 V (reference) and I = 10 A lagging by 30°.
To find the combined effect of two AC quantities, you can't just add the numbers (because they're at different phases). Instead, you use phasor addition: place the tail of the second phasor at the tip of the first, then draw the resultant from the origin to the end of the last phasor.
Current A = 4 A (reference), Current B = 3 A leading A by 90°. Find the resultant.
In series circuits, the current I is the same through all components, so we use current as the reference phasor (drawn horizontal). The voltage phasors are drawn relative to I.
| Component | VR / VL / VC relative to I | Phase angle φ |
|---|---|---|
| Resistor (R) | VR is in phase with I | 0° |
| Inductor (L) | VL leads I by 90° | +90° |
| Capacitor (C) | VC lags I by 90° | −90° |
In an RL circuit, VR is in phase with I, and VL leads I by 90°. The total voltage VT is the phasor sum of VR and VL.
Since VR = IR, VL = IXL, and VT = IZ, we can divide every voltage phasor by the current I. This gives us the impedance triangle — a right-angled triangle showing the relationship between R, X and Z.
Power factor tells you how efficiently a circuit uses the supply. A PF of 1.0 means all power is doing useful work (purely resistive). A PF of 0 means no useful work is done (purely reactive).
| Phase Angle φ | cos φ (PF) | Meaning |
|---|---|---|
| 0° | 1.0 | V & I in phase — purely resistive (ideal) |
| 30° | 0.866 | Small phase shift — mostly resistive |
| 45° | 0.707 | Equal R and X |
| 60° | 0.5 | Large phase shift — mostly reactive |
| 90° | 0.0 | Purely reactive — no useful power |
Power factor can also be found from the impedance triangle: PF = R ÷ Z = VR ÷ VT
The power triangle shows the relationship between the three types of power in an AC circuit. It has the same shape as the impedance triangle.
A coil of 0.15 H is in series with a 50 Ω resistor across a 100 V, 50 Hz supply. Find XL, Z, I, PF and draw the phasor diagram.
Measuring the result: Use your ruler to measure VT on the diagram. At your chosen scale it should equal 100 V. Use your protractor to measure the angle between VT and I — it should be approximately 43°.
A 60 μF capacitor in series with a 100 Ω resistor, 230 V 50 Hz. Find XC, Z, I.
R = 30 Ω, XL = 40 Ω, supply = 250 V. Find Z, I, PF, and all three powers.