In a series LCR circuit R = 1 kΩ, C = 2 μF and the voltage across R is 100 V at resonant frequency (ω) $200\, rad\, s^{-1}$. At resonant frequency, the voltage across L is

250 V

The correct answer is Option (2) → 250 V

Given:

Series LCR circuit: $R = 1\ \text{k}\Omega = 1000\ \Omega$, $C = 2\ \mu\text{F} = 2 \times 10^{-6}\ \text{F}$

Voltage across resistor at resonance: $V_R = 100\ \text{V}$

Resonant angular frequency: $\omega = 200\ \text{rad/s}$

At resonance, current in the circuit: $I = \frac{V_R}{R} = \frac{100}{1000} = 0.1\ \text{A}$

Voltage across inductor: $V_L = I \cdot X_L = I \cdot \omega L$

Resonance condition: $\omega = \frac{1}{\sqrt{LC}} \Rightarrow L = \frac{1}{\omega^2 C}$

$L = \frac{1}{(200)^2 \cdot 2 \times 10^{-6}} = \frac{1}{8 \times 10^{-2}} = 12.5\ \text{H}$

Inductive reactance: $X_L = \omega L = 200 \cdot 12.5 = 2500\ \Omega$

Voltage across L: $V_L = I X_L = 0.1 \cdot 2500 = 250\ \text{V}$

∴ Voltage across inductor at resonance = 250 V