Alternating Current: Tricky Questions Solved (4)

Question

A series LCR circuit has $L = 0.1 \text{ H}$, $C = 100 \mu\text{F}$, $R = 10 \Omega$, connected to an AC source $V = 220 \sin(100 \pi t) \text{ V}$. Find (a) the impedance, (b) rms current, (c) phase angle, (d) average power dissipated.

Solution — Step by Step

Final answers: $Z \approx \mathbf{10.01 \Omega}$, $I_{rms} \approx \mathbf{15.55 \text{ A}}$, $\phi \approx \mathbf{-2.6°}$, $P \approx \mathbf{2.42 \text{ kW}}$.

Why This Works

This circuit is almost at resonance: $\omega^2 = 1/(LC)$ gives $\omega_0 = 1/\sqrt{0.1 \times 10^{-4}} = 1/0.00316 \approx 316.2 \text{ rad/s}$, very close to $\omega = 314$. So $X_L \approx X_C$, $Z \approx R$, and the circuit looks purely resistive.

At resonance, power dissipation is maximum: $P_{max} = V_{rms}^2/R$. For $V_{rms} = 155.6$ and $R = 10$: $P_{max} = 24,212/10 = 2421 \text{ W}$ — matches our answer.

Alternative Method

Since the circuit is at resonance, skip $Z$ entirely. $I_{rms} = V_{rms}/R = 155.6/10 = 15.56 \text{ A}$. $P = V_{rms} I_{rms} = 2421 \text{ W}$. Three lines instead of ten.

Common Mistake