Alternating Current: Application Problems (6)

Question

A series LCR circuit has $R = 100\,\Omega$, $L = 0.5\text{ H}$, $C = 50\,\mu\text{F}$, connected to a $V_{\text{rms}} = 220\text{ V}$, $f = 50\text{ Hz}$ source. Find the rms current and the average power dissipated.

Solution — Step by Step

Why This Works

In AC circuits, inductors and capacitors store energy reversibly — they take energy from the source and return it. Only resistors actually dissipate. The phase angle $\phi = \tan^{-1}((X_L - X_C)/R)$ tells us how out-of-phase current is with voltage; $\cos\phi$ is the power factor.

When $X_L = X_C$, the circuit is at resonance: $Z = R$ (minimum), current is maximum, and power factor is unity. This is why tuners in radios work — they pick a frequency by matching $X_L = X_C$.

Alternative Method

Use phasor diagrams. The resistor’s voltage is in phase with current; inductor’s voltage leads by $90°$; capacitor’s lags by $90°$. The vector sum of voltage phasors equals the source voltage. The phase angle tells us power factor directly.

Common Mistake

Confusing $\omega = 2\pi f$ with $f$ alone. Forgetting the $2\pi$ underestimates $X_L$ and overestimates $X_C$ — both wrong, in opposite directions, so the answer is way off.