P = VIcos φ. Why purely inductive/capacitive circuits dissipate no power.

Power Factor in AC Circuits — cos φ

Question

A series LCR circuit has resistance $R = 100\,\Omega$ , inductive reactance $X_L = 200\,\Omega$ , and capacitive reactance $X_C = 100\,\Omega$ . The circuit is connected to a 220 V (rms) AC source. Find the power factor of the circuit and the average power dissipated.

Solution — Step by Step

The net reactance is $X = X_L - X_C = 200 - 100 = 100\,\Omega$ .

We subtract $X_C$ from $X_L$ because they oppose each other — inductive reactance drives current to lag, capacitive reactance drives it to lead.

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + 100^2} = \sqrt{20000} = 100\sqrt{2}\,\Omega

Impedance is the “total opposition” — it’s what limits current in an AC circuit, just like resistance does in DC.

\cos\phi = \frac{R}{Z} = \frac{100}{100\sqrt{2}} = \frac{1}{\sqrt{2}} \approx 0.707

The angle $\phi = 45°$ . This means voltage leads current by 45° — the circuit is more inductive than capacitive.

I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{100\sqrt{2}} = \frac{22}{10\sqrt{2}} = 1.1\sqrt{2} \approx 1.56\,\text{A}

P = V_{rms}\,I_{rms}\cos\phi = 220 \times 1.1\sqrt{2} \times \frac{1}{\sqrt{2}}

P = 220 \times 1.1 = \mathbf{242\,W}

Why This Works

In AC circuits, voltage and current are sinusoidal but generally out of phase. When we write $P = VI\cos\phi$ , the $\cos\phi$ factor accounts for the fact that only the component of current in phase with voltage does actual work.

Think of it this way: a purely inductive or purely capacitive element stores energy in one half-cycle and returns it in the next. Over a full cycle, the net energy transfer is zero — no heat is generated. Only the resistive part of the circuit actually dissipates power.

This is why $\cos\phi = R/Z$ makes physical sense. If $R = 0$ (pure L or pure C), then $\cos\phi = 0$ and $P = 0$ . The entire circuit is just “cycling” energy back and forth with the source.

\cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}}

P_{avg} = V_{rms}\,I_{rms}\cos\phi = I_{rms}^2\,R

For pure resistor: $\cos\phi = 1$ , $P = VI$

For pure L or pure C: $\cos\phi = 0$ , $P = 0$

Alternative Method

We can get average power directly without finding $I_{rms}$ first:

P = \frac{V_{rms}^2 \cdot R}{Z^2} = \frac{220^2 \times 100}{(100\sqrt{2})^2} = \frac{48400 \times 100}{20000} = \frac{4840000}{20000} = 242\,\text{W}

This formula $P = V^2 R / Z^2$ comes from substituting $I = V/Z$ into $P = I^2 R$ . It’s often faster in MCQs when the question gives $V_{rms}$ , $R$ , and $Z$ directly — saves one calculation step.

In JEE Main MCQs, if $X_L = X_C$ (resonance condition), then $Z = R$ , $\cos\phi = 1$ , and $P = V_{rms}^2/R$ . This is the maximum power the circuit can draw — a common PYQ scenario from JEE Main 2023 and 2022.

Common Mistake

Using peak values instead of rms values in the power formula.

Students write $P = V_0 I_0 \cos\phi$ and get double the correct answer. The formula $P = VI\cos\phi$ uses rms values only. If the question gives peak voltage $V_0 = 220\sqrt{2}\,\text{V}$ , convert first: $V_{rms} = V_0/\sqrt{2} = 220\,\text{V}$ .

The other version of the mistake: computing $P = I^2 Z$ instead of $P = I^2 R$ . Power is only dissipated in the resistor — impedance $Z$ is not a resistor, it includes reactive components that store and return energy.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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