Alternating Current: Common Mistakes and Fixes (2)

Question

A series LCR circuit has $L = 0.1$ H, $C = 100$ µF, $R = 10$ Ω, connected to an AC source of peak voltage $V_0 = 200$ V at angular frequency $\omega = 100$ rad/s. Find the impedance and the rms current.

Solution — Step by Step

Why This Works

In a series LCR circuit, the inductor’s voltage leads the current by $90°$, the capacitor’s voltage lags by $90°$, and the resistor’s voltage is in phase. Adding these as phasors gives the impedance triangle: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.

When $X_L > X_C$, the circuit is inductive; when $X_L < X_C$, capacitive. Here $X_C \gg X_L$, so the circuit is strongly capacitive.

Alternative Method

Use phasor diagrams: draw $V_R$ along the current direction, $V_L$ perpendicular up, $V_C$ perpendicular down. The resultant gives the source voltage, and its magnitude divided by $I$ gives $Z$.

Common Mistake

Students sometimes use peak voltage $V_0$ to compute current: $I = V_0 / Z$. That gives the peak current $I_0$, not $I_{\text{rms}}$. Remember: $I_{\text{rms}} = V_{\text{rms}}/Z$, where both rms quantities equal peak divided by $\sqrt{2}$.