In a series LR circuit, power of 400 W is dissipated from a source of 250 V , 50 Hz . The power factor of the circuit is 0 . 8 . In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R . Taking the value of C as n 3 π μF , then value of n is

Step-by-step solution

P = V m . in cosϕ 400 = 250 × 1 m × 0 .8 i rms = 2 A 1 m 2 . R = P 4 × R = 400 ⇒ R = 100 Ω . cosϕ = R R 2 + X L 2 100 2 + X L 2 = 100 0 .8 2 100 2 + X L 2 = 100 0 .8 2 X L = 75 Ω Power factor is unity X C = X L = 75 1 ω = 75 ⇒ C = 1 75 × 2 H × 50 = 1 7500 π F 3 π × 2500 = 1 3 π × 4 × 10 2   mF = 400 3 π μF N = 400