In a series LR circuit with X_L=R, power factor P₁. If a capacitor of capacitance C with X_C=X_L is added to the circuit the power factor becomes P₂. The ratio of P₁ to P₂ will be :

Solution

Physics sir · 3 steps

Power factor is just R/Z: find Z before and after the capacitor, and the ratio drops out.

1. Before the capacitor. With X_L = R, the impedance is Z₁ = √(R² + X_L²) = √2 R. So P₁ = R/Z₁ = 1/√2.
2. After the capacitor. Now Z₂ = √(R² + (X_L - X_C)²). Since X_C = X_L, the reactances cancel and Z₂ = R. So P₂ = R/R = 1; the circuit is at resonance.

   P₁/P₂ = (1/√2)/1 = 1/√2

3. So. P₁ : P₂ = 1 : √2, which is option (a).

The idea to keep. Power factor is cos φ = R/Z. When X_L = X_C, the reactances cancel, Z becomes R and the power factor reaches its maximum value of 1.

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