An a.c. source is applied to a series LR circuit with X_L = 3R and power factor is X₁ . Now a capacitor with X_c = R is added in series to LR circuit and power factor is X₂ . The ratio X₁ to X₂ is

Step-by-step solution

For the series LR circuit, the impedance is Z₁ = R^2 + X_L^2 . Given X_L = 3R , Z₁ = R^2 + (3R)^2 = 10 R . The power factor X₁ = R Z₁ = R 10 R = 1 10 . When a capacitor with X_C = R is added in series, the new impedance is Z₂ = R^2 + (X_L - X_C)^2 . Substituting the values, Z₂ = R^2 + (3R - R)^2 = R^2 + 4R^2 = 5 R . The new power factor X₂ = R Z₂ = R 5 R = 1 5 . The ratio X₁ X₂ = 1/ 10 1/ 5 = 5 10 = 1 2 . Thus, X₁ : X₂ = 1 : 2 . Answer: 1 : 2