An LCR circuit contains resistance of 110Ω and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 45°. If on the other hand, only inductor is removed the current leads by 45° with the applied voltage. The rms current flowing in the circuit will be :

Solution

Physics sir · 4 steps

Each 45° phase angle tells you a reactance equals R, and equal reactances mean resonance.

1. Remove the capacitor. Only R and L remain, so tan φ = X_L/R. A lag of 45° gives tan 45° = 1, so X_L = R = 110 Ω.
2. Remove the inductor. Only R and C remain, so tan φ = X_C/R. A lead of 45° gives X_C = R = 110 Ω.
3. Full circuit. With all three in place, X_L = X_C, so the reactances cancel and the circuit is at resonance. The impedance is just the resistance.

   Z = √(R² + (X_L - X_C)²) = √(110² + 0) = 110 Ω

4. So. Iᵣₘₛ = Vᵣₘₛ/Z = 220/110 = 2 A, which is option (c).

The idea to keep. A 45° phase angle always means the reactance equals the resistance. When X_L = X_C, the LCR circuit is at resonance and Z = R.

Answer from the NTA final answer key. Solution by PrepWiser.