In a series L R 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 a

Question

In a series L R 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 ( 3 π n ​ ) μF , then value of n is

TestHub · Accepted Answer

P = V m ​ . in cos ϕ 400 = 250 × 1 m × 0.8 i r m s ​ = 2 A ( 1 m ) 2 . R = P 4 × R = 400 ⇒ R = 100Ω . cos ϕ = R 2 + X L 2 ​ ​ R ​ 10 0 2 + X L 2 ​ = ( 0.8 100 ​ ) 2 10 0 2 + X L 2 ​ = ( 0.8 100 ​ ) 2 X L ​ = 75Ω Power factor is unity X C ​ = X L ​ = 75 ω 1 ​ = 75 ⇒ C = 75 × 2 H × 50 1 ​ = 7500 π 1 ​ F 3 π × 2500 = 3 π 1 ​ × 4 × 1 0 2 mF = 3 π 400 ​ μF N = 400

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