Question

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

• A. $2.5\times {10}^{-2}\text{V}$
• B. $40\text{V}$
• C. $250\text{V}$
• D. $4\times {10}^{-3}\text{V}$

Answer 1

The correct option is C $250\text{V}$

Current in the circuit is,

$I=\frac{V_R}{R}=\frac{100}{1000}=0.1\text{A}$

At resonance,

$X_L=X_C=\frac{1}{\omega C}=\frac{1}{200\times 2\times {10}^{-6}}$

$\Rightarrow X_L=2500\Omega$

Voltage across $L$ is,

$V_L=I{X}_L=0.1\times 2500=250\text{V}$

Hence, option $\left(\text{C}\right)$ is correct.