Question

A series $LCR$ circuit with $L=0.5 \,H$ and $R=10 \Omega$ is connected to an AC supply with rms voltage and frequency equal to $200 \,V$ and $\frac{150}{\pi} Hz$, respectively. The magnitude of the capacitance is varied so that current amplitude in the circuit becomes maximum.The rms voltage difference across the inductor is

• A. 3000 V
• B. 2500 V
• C. 2000 V
• D. 2600 V

Answer 1

Answer: (A)

Current amplitude in series $LCR$ circuit is maximum, if there is resonance i.e, $\omega L=\frac{1}{\omega C}$.
Given, $L=0.5 H , R=10 \Omega, V_{r m s}=200\, V$ and $f=\frac{150}{\pi} \,Hz$
Angular frequency,
$\omega=2 \pi f=2 \pi \times \frac{150}{\pi}=300 \,rad s ^{-1}$
$\therefore $ For $I_{\max }$ in series $LCR$ circuit,
$\omega^{2} =\frac{1}{L C} $
$\Rightarrow \, C =\frac{1}{L \omega^{2}} $
$\Rightarrow \,C=\frac{1}{0.5 \times 300 \times 300} =22.2 \times 10^{-6} F$
Current in the circuit,
$I_{\max }=\frac{V_{ rms }}{R}=\frac{200}{10}=20 \,A$
$\because$ rms voltage difference across the inductor $V_{L}$ and capacitor $V_{C}$ are equal. Then
$V_{L}=I_{\max } \omega L=20 \times 300 \times 0.5$
$\Rightarrow \, V_{L}=3000 \,V$