Alternating Current: PYQ Walkthrough (3)

Question

A series LCR circuit has $L = 0.1$ H, $C = 100 \mu$F, $R = 10$ Ω, connected to an AC source $V = 200 \sin(100t)$ V. Find the current amplitude, the phase angle, and the average power dissipated. (JEE Main 2023)

Solution — Step by Step

Final answers: $I_0 \approx 2.21$ A, $\phi \approx -83.7°$ (leading), $P \approx 24.4$ W.

Why This Works

LCR circuits resolve into a single impedance $Z$ once you compute the reactances. The phase tells you whether current leads or lags the voltage — sign of $X_L - X_C$ decides. Negative means capacitive ($X_C > X_L$), so current leads.

Power dissipates only in the resistor; the inductor and capacitor swap energy with the source without net dissipation. That’s why $P = I_{\text{rms}}^2 R$ also works as a cross-check.

Alternative Method

Use $P = I_{\text{rms}}^2 R$ directly. $I_{\text{rms}} = I_0/\sqrt{2} = 1.56$ A. So $P = (1.56)^2 \times 10 \approx 24.3$ W. Matches the previous answer (small rounding).

Common Mistake

Forgetting the $\frac{1}{2}$ factor when converting amplitudes to averages. $P = V_0 I_0 \cos\phi$ would double-count; the correct formula is $P = \frac{V_0 I_0}{2} \cos\phi = V_{\text{rms}} I_{\text{rms}} \cos\phi$.