A series LCR circuit with R = 250/11 Ω and X_L = 70/11 Ω is connected across a 220 V, 50 Hz supply. The value of capacitance needed to maximize the average power of the circuit will be _________ μF. (Take : π = 22/7)

Solution

Physics sir · 3 steps

Average power peaks at resonance, so pick C that makes X_C exactly equal to X_L.

1. Why resonance. Average power is P = Iᵣₘₛ² R, and Iᵣₘₛ = Vᵣₘₛ/Z. With R fixed, P is largest when the impedance Z = √(R² + (X_L - X_C)²) is smallest, which happens when X_C = X_L.
2. Set the reactances equal. So we need X_C = 70/11 Ω. The angular frequency is ω = 2π f = 2 × 22/7 × 50 = 2200/7 rad s⁻¹.

   C = 1/(ω X_C) = 1/(2200/7 × 70/11) = 1/2000 F

3. So. C = 5 × 10⁻⁴ F = 500 μF. The value of R is not needed for this question.

The idea to keep. For a fixed R, maximum average power in a series LCR circuit means resonance: X_L = X_C. Use ω = 2π f, not f, inside X_C = 1/(ω C).

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