Question

In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 $K\Omega$ with C = 2$\mu$F. The resonant frequency $\omega$ is 200 rad/s. At resonance the voltage across L is :

• A. $25\times10^{-2} V$
• B. 40 V
• C. 250 V
• D. $4\times10^{-3} V$

Answer 1

Answer: (C)

At resonance $\omega L=\frac{1}{\omega C}$
Current flowing through the circuit,
$I=\frac{V_{R}}{R}=\frac{100}{1000}=0.1 A$
So, voltage across L is given by
$V_{L}=I X_{L}=I\omega L$
$but \, \quad\omega L=\frac{1}{\omega C}$
$\therefore \, \quad V_{L}=\frac{I}{\omega C}=\frac{0.1}{200\times2\times10^{-6}}=250 V$