RL, RC and RLC Series Circuits — Power and Power Factor

By the end you should be able to: Analyse RL, RC and RLC series circuits, define active, reactive and apparent power, and explain the significance of power factor and its improvement.

The series circuits

In a series circuit the current is common, so it is the natural reference phasor and every voltage is drawn relative to it.

RL series. $Z = R + jX_L$ . Current lags by $\phi = \tan^{-1}(X_L/R)$ , between 0° and 90°.

RC series. $Z = R - jX_C$ . Current leads by $\phi = \tan^{-1}(X_C/R)$ .

RLC series. $Z = R + j(X_L - X_C)$ , and the net reactance decides.

The voltage triangle. Since $V_R = IR$ is in phase with the current and $V_L$ , $V_C$ are at ±90° to it, the voltages add as phasors:

V = \sqrt{V_R^2 + (V_L - V_C)^2}

Not as scalars. This is why the measured voltages across the individual elements of a series AC circuit can each exceed the supply voltage — near resonance $V_L$ and $V_C$ may be many times $V$ and yet cancel almost exactly. Nothing is violated; they are simply in antiphase. It surprises students and it destroys insulation.

The three powers

Multiply the voltage triangle by the common current $I$ and it becomes the power triangle.

Active (real, true) power — actually converted to heat or work:

P = VI\cos\phi \qquad \text{watts (W)}

Only resistance contributes. This is what the energy meter records and what you pay for.

Reactive power — shuttled back and forth between source and field, never consumed:

Q = VI\sin\phi \qquad \text{volt-amperes reactive (VAR)}

Positive for a lagging (inductive) load, negative for leading.

Apparent power — the product of the RMS values, ignoring phase:

S = VI \qquad \text{volt-amperes (VA)}

S = \sqrt{P^2 + Q^2}, \qquad \bar{S} = P + jQ

Why cables and transformers are rated in VA, not W. Conductor heating depends on $I^2R$ and insulation on $V$ — neither cares about the phase angle. A transformer supplying 100 kW at unity power factor and one supplying 100 kW at 0.5 lagging carry 100 kVA and 200 kVA respectively; the second needs twice the copper for the same useful output.

Power factor

\text{pf} = \cos\phi = \frac{P}{S} = \frac{R}{|Z|}

Always quoted with lagging or leading, since $\cos\phi$ alone cannot distinguish them.

Unity (pf = 1): purely resistive; all supplied power is used.
Lagging: inductive. Induction motors, transformers, fluorescent ballasts — most industrial load.
Leading: capacitive. Uncommon as a natural load.

Why a low power factor is penalised. For a given real power $P$ ,

I = \frac{P}{V\cos\phi}

Current is inversely proportional to power factor. Halve the power factor and the current doubles — so the $I^2R$ losses in the cables quadruple, the voltage drop doubles, and the supply equipment must be twice the size for the same useful output. The utility bears those costs, so it bills large consumers on kVA, or applies a power factor penalty.

Power factor improvement

Most industrial load is inductive, drawing lagging reactive power. Add a capacitor in parallel with the load: it supplies leading reactive power, cancelling part of the lagging demand.

Note in parallel, not in series. The load must keep its own voltage and continue to operate exactly as before; the capacitor only supplies the reactive current locally, so that current no longer has to travel from the generator. $P$ is unchanged, $Q$ falls, $S$ falls, and $\phi$ falls with it.

The required capacitor rating is

Q_C = P(\tan\phi_1 - \tan\phi_2)

Correction to exactly unity is usually avoided — component tolerance or a change in load can tip the circuit into leading, and it is expensive for the last few per cent. Typical practice is to correct to 0.95.