Question

In a series resonant $L-C-R$ circuit, the voltage across $R$ is $100\, V$ and $R = 1\, k \Omega$ with $C = 2\, \mu F$. The resonant frequency $\omega$ is $200\, rad s^{−1}$. At resonance the voltage across $L$ is

• A. $2.5 × 10^{−2}\, V$
• B. $4 × 10^{−8}\, V$
• C. $250 \,V$
• D. $2.5 × 10^{−3}\, V$

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
$V_{L}=I X_{L}=l \omega L$
but $\omega L =\frac{1}{\omega C}$
$\therefore V_{L} =\frac{1}{\omega C}$
$=\frac{0.1}{200 \times 2 \times 10^{-6}}$
$=250\, V$