Question

In a series LCR circuit, the voltage across resistance of \( 2000 \, \Omega \) is 200 V and the resonant frequency is \( 400 \, {rad/s} \). The capacitance \( C \) value is 4 μF. At resonance, the voltage across \( L \) is

• A. 3.25 V
• B. 6.25 V
• C. 625 V
• D. 62.5 V

Hint:

At resonance in an LCR circuit, the inductive and capacitive reactances are equal, and the impedance is minimized, being purely resistive.

Answer 1

The Correct Option is D

Given:
Resistance, \( R = 2000 \, \Omega \)
Voltage across resistance, \( V_R = 200 \, {V} \)
Resonant frequency, \( \omega_0 = 400 \, {rad/s} \) Capacitance, \( C = 4 \, \mu{F} = 4 \times 10^{-6} \, {F} \)
Step 1: Calculate the current in the circuit At resonance, the impedance of the circuit is purely resistive, so the current \( I \) is: \[ I = \frac{V_R}{R} = \frac{200}{2000} = 0.1 \, {A} \] Step 2: Calculate the inductive reactance \( X_L \) At resonance, the inductive reactance \( X_L \) is equal to the capacitive reactance \( X_C \): \[ X_L = X_C = \frac{1}{\omega_0 C} = \frac{1}{400 \times 4 \times 10^{-6}} = \frac{1}{1.6 \times 10^{-3}} = 625 \, \Omega \] Step 3: Calculate the voltage across the inductor \( V_L \) The voltage across the inductor is given by: \[ V_L = I \times X_L = 0.1 \times 625 = 62.5 \, {V} \] 
Final Answer: 62.5 V