Question

An alternating e.m.f. of 0.2 V is applied across an LCR series circuit having \( R = 4 \, \Omega \), \( C = 80 \, \mu \text{F} \), and \( L = 200 \, \text{mH} \). At resonance, the voltage drop across the inductor is

• A. 1 V
• B. 2.5 V
• C. 5 V
• D. 10 V

Hint:

At resonance in an LCR circuit, the voltage drop across the inductor is equal to the total voltage drop in the circuit.

Answer 1

The Correct Option is B

Step 1: Resonance condition in LCR circuit.
At resonance, the inductive reactance \( X_L = \omega L \) and the capacitive reactance \( X_C = \frac{1}{\omega C} \) are equal, so the total impedance \( Z = R \) at resonance.
Step 2: Voltage drop across inductor.
The voltage drop across the inductor is given by: \[ V_L = I X_L \] At resonance, the current \( I \) is given by: \[ I = \frac{V_{\text{rms}}}{R} \] where \( V_{\text{rms}} = \frac{0.2}{\sqrt{2}} \) is the rms voltage. Substituting the values: \[ V_L = 2.5 \, \text{V} \]
Step 3: Conclusion.
Thus, the correct answer is (B) 2.5 V.