A series LCR circuit with L=100/π mH, C=10⁻³/π F and R=10 Ω, is connected across an ac source of 220 V, 50 Hz supply. The power factor of the circuit would be ________.

Solution

Physics sir · 4 steps

Work out both reactances first; when they turn out equal, the circuit is at resonance.

1. Angular frequency. ω = 2π f = 2π × 50 = 100π rad s⁻¹. The π in L and C is there to cancel this π.
2. Inductive reactance. Convert to henry first: L = 100/π mH = 0.1/π H. Then X_L = ω L = 100π × 0.1/π = 10 Ω.
3. Capacitive reactance. X_C = 1/(ω C) = 1/(100π × 10⁻³/π) = 1/0.1 = 10 Ω. So X_L = X_C, and the impedance is just Z = R = 10 Ω.

   cos φ = R/Z = R/√(R² + (X_L - X_C)²) = 10/√(10² + 0²) = 1

4. So. The circuit is at resonance, current and voltage are in phase, and the power factor is 1.

The idea to keep. In a series LCR circuit, always compare X_L and X_C first. If they are equal, the circuit is purely resistive and the power factor is 1, whatever the value of R.

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