In a series LCR circuit, the inductive reactance (X L ) is 10Ω and the capacitive reactance (X C ) is 4Ω. The resistance (R) in the circuit is 6Ω. The power factor of the circuit is :

Solution

Physics sir · 3 steps

The net reactance is X_L - X_C. Find the impedance, then the power factor is R/Z.

1. Net reactance. In a series circuit the inductor and capacitor voltages are opposite in phase, so their reactances subtract. X = X_L - X_C = 10 - 4 = 6 Ω.
2. Impedance. R and X are at right angles in the phasor diagram, so they add like the sides of a right triangle. Z = √(6² + 6²) = √72 = 6√2 Ω.

   cos φ = R/Z = 6/6√2 = 1/√2

3. So. The power factor is 1/√2, which is option (c). The current lags the voltage by 45° because R and the net reactance are equal.

The idea to keep. Power factor in series LCR is always R/Z, with Z = √(R² + (X_L - X_C)²). When R equals the net reactance, the phase angle is 45°.

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