A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 kΩ, an inductor of inductive reactance X L = 250 πΩ and an unknown capacitor. The value of capacitance to maximize the average power should be : (Take π² = 10)

Solution

Physics sir · 3 steps

Average power is largest at resonance, so make the capacitor's reactance equal to X_L and solve for C.

1. When power peaks. Average power is P = (V² R)/Z² with Z = √(R² + (X_L - X_C)²). With V and R fixed, P is largest when Z is smallest, which happens when X_C = X_L. That is resonance, and the value of R does not affect the answer.
2. Set reactances equal. X_C = 1/(ω C) and ω = 2π f = 100π rad s⁻¹. Setting X_C = X_L = 250π Ω gives the capacitance directly.

   C = 1/(ω X_L) = 1/(100π × 250π) = 1/(25000 π²) = 1/250000 F

3. So. C = 4 × 10⁻⁶ F = 4 μF, which is option (a).

The idea to keep. Maximum average power in a series LCR circuit means resonance, so X_L = X_C. The resistance and the voltage only set how large that power is, not where the maximum occurs.

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