Question

A series $LCR$ circuit with $L=0.5H$ and $R=10\Omega$ is connected to an AC supply with rms voltage and frequency equal to $200V$ 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, $L=0.5H,R=10\Omega ,V_{rms}=200V$ and $f=\frac{150}{\pi }Hz$
Angular frequency,
$\omega =2\pi f=2\pi \times \frac{150}{\pi }=300rad{s}^{-1}$
$\therefore$ For $I_{\max }$ in series $LCR$ circuit,
${\omega }^2=\frac{1}{LC}$
$\Rightarrow C=\frac{1}{L{\omega }^2}$
$\Rightarrow C=\frac{1}{0.5\times 300\times 300}=22.2\times 10^{-6}F$
Current in the circuit,
$I_{\max }=\frac{V_{rms}}{R}=\frac{200}{10}=20A$
$\because$ rms voltage difference across the inductor $V_L$ and capacitor $V_C$ are equal. Then
$V_L=I_{\max }\omega L=20\times 300\times 0.5$
$\Rightarrow V_L=3000V$