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.

Answer 2

Since \(|X_L|=|X_C|\), the circuit is at resonance, and one useful property of series resonance is that the inductor and capacitor voltages are equal in magnitude and exactly opposite in phase, so they cancel each other in the loop, leaving the entire supply voltage to appear across the resistor alone. This gives a shortcut to the current without first computing the total impedance.

By KVL, \(V_R = V_{rms} = 200\) V (since \(V_L+V_C=0\)). The current is then \[ I = \frac{V_R}{R} = \frac{200}{10} = 20 \text{ A} \] and the voltage across the capacitor is \[ V_C = I \times X_C = 20 \times 20 = 400 \text{ V} \] with the capacitor voltage lagging the current (and hence the in-phase resistor voltage/supply) by \(90^\circ\).

1. \(200\angle -90^\circ\) V: The magnitude here matches the supply voltage, not the capacitor voltage, which is boosted above the supply at resonance due to the current magnification through the reactance.
2. \(200\angle +90^\circ\) V: Both the magnitude and the sign of the angle are wrong; a capacitor's voltage lags its current, it does not lead it.
3. \(400\angle +90^\circ\) V: The magnitude matches, but the sign of the angle is wrong; a leading angle would apply to the inductor's voltage, not the capacitor's.
4. \(400\angle -90^\circ\) V: This matches both the magnitude \(I\times X_C=400\) V and the lagging \(-90^\circ\) phase that a capacitor voltage always carries relative to its current.

Therefore, the correct answer is \(400\angle -90^\circ\) V.