Question

A series R-L-C circuit is supplied by 100 V, 50 Hz AC. What is the power factor of the circuit at resonance?

(where R = 10 Ω, L = 10 mH, C = 100 μF)
1. Zero
2. Unity
3. 0.6 lagging
4. 0.6 leading

Answer 1

Correct Answer - Option 2 : Unity

Concept:

In a series RLC circuit, the impedance is given by

\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}\)

At resonance, the magnitude of inductive reactance is equal to the magnitude of capacitive reactance.

Magnitude of XL = XC

At this condition, Z = R.

Hence at resonance, the impedance is purely resistive and it is minimum.

Current in the circuit, I = V/Z

As impedance is minimum the current is maximum.

As impedance is purely resistive, the power factor is unity.

Application:

R = 10 Ω, L = 10 mH, C = 100 μF

The power factor at resonance is unity irrespective of the values of R, L, and C.