An AC circuit drawing power from a source of angular frequency 50 rad s - 1 has a power factor of 0.6 . In this condition, a resistance of 100 Ω is present and the current is lagging behind the voltage. If a capacitor is connected in series, then the required capacitance that will result in a power factor of unity is

Step-by-step solution

As the current lags the voltage, the circuit must contain a combination of resistance and inductance. Power factor cos ⁡ ϕ = 0.6 ⇒ R R 2 + X L 2 = 0.6 ⇒ R 2 + X L 2 = R 0.6 2 ⇒ X L 2 = R 2 0.6 2 - R 2 ⇒ X L 2 = R 2 × 0.64 0.36 ∴ X L = 0 .8 R 0 .6 = 4 R 3 Now we want the power factor cos ⁡ ϕ = 1 This is the condition of resonance in which X L = X C ∴ X C = 4 R 3 ⇒ 1 ωC = 4 R 3 ⇒ C = 3 4 × 50 × 100 = 150 μF