A series LCR circuit is connected to an ac source of 220 V, 50 Hz. The circuit contain a resistance R=100 Ω and an inductor of inductive reactance X_L=79.6 Ω. The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be _________ μ F.

Solution

Physics sir · 3 steps

Maximum average power means resonance, so make the capacitor's reactance equal X_L = 79.6 Ω.

1. When power peaks. The average power is P = Iᵣₘₛ² R. With R fixed, P is largest when the current is largest, and that happens when the impedance Z = √(R² + (X_L - X_C)²) is smallest. That is at resonance, X_C = X_L.
2. Set the reactance. So the capacitor must have X_C = 79.6 Ω at f = 50 Hz. Use X_C = 1/(2π f C) and solve for C.

   C = 1/(2π f X_C) = 1/(2π × 50 × 79.6) = 1/(25 007) F

3. So. C ≈ 4.0 × 10⁻⁵ F, which is 40 μF. The 220 V and the 100 Ω only fix how large the peak power is, not where it occurs.

The common slip: 251. Using f = 50 in place of ω = 2π f gives C = 1/(50 × 79.6) ≈ 251 μF, not 40.

The idea to keep. In a series LCR circuit with fixed R, maximum average power, maximum current and minimum impedance all mean the same thing: resonance, X_L = X_C.

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