Question

A series LCR circuit consisting of $R = 10\Omega$, $|X_L| = 20\Omega$ and $|X_C| = 20\Omega$, is connected across an a.c. supply of $200\,\text{V}_{\text{rms}}$. The rms voltage across the capacitor is}

• A. $200 \angle -90^\circ \text{ V}$
• B. $200 \angle +90^\circ \text{ V}$
• C. $400 \angle +90^\circ \text{ V}$
• D. $400 \angle -90^\circ \text{ V}$

Hint:

At resonance in a series LCR circuit, the current is maximum and voltages across L and C can be much larger than the supply voltage.

Answer 1

The Correct Option is D

Step 1: Identify circuit condition.
Given that $|X_L| = |X_C| = 20\Omega$, the inductive and capacitive reactances cancel each other. Hence, the circuit operates at resonance.
Step 2: Calculate total impedance.
At resonance, the net reactance is zero and the impedance is purely resistive:
\[ Z = R = 10\Omega \]
Step 3: Calculate circuit current.
\[ I = \frac{V_{\text{rms}}}{Z} = \frac{200}{10} = 20\text{ A} \]
Step 4: Voltage across the capacitor.
\[ V_C = I \times X_C = 20 \times 20 = 400\text{ V} \]
Step 5: Phase angle of capacitor voltage.
The voltage across a capacitor lags the current by $90^\circ$. Hence,
\[ V_C = 400 \angle -90^\circ \text{ V} \]
Step 6: Final conclusion.
Therefore, the rms voltage across the capacitor is $400 \angle -90^\circ$ V.