In a series L R circuit X_L=R and power factor of the circuit is P₁. When capacitor with capacitance C such that X_L=X_C is put in series, the power factor becomes P₂. The ratio P₁/P₂ is:

Solution

Physics sir · 4 steps

Get the power factor from cos φ = R/Z; adding the capacitor brings the circuit to resonance.

1. Power factor. In a series AC circuit the power factor is cos φ = R/Z, where Z = √(R² + (X_L - X_C)²).
2. LR circuit. With X_L = R, the impedance is Z = √(R² + R²) = √2 R. So P₁ = R/(√2 R) = 1/√2.
3. Add the capacitor. With X_C = X_L, the reactances cancel, so Z = R. This is resonance, and P₂ = R/R = 1.

   P₁/P₂ = (1/√2)/1 = 1/√2

4. So. The ratio is 1/√2, which is option (b).

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

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