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 ad

Question

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 ( 3 π n ​ ) μ F , then value of n is ____ .

TestHub · Accepted Answer

P = Z E r m s 2 ​ ​ cos ϕ 400 = Z ( 250 ) 2 × 0.8 ​ Z = 400 ( 250 ) 2 × 0.8 ​ = 400 62500 × 0.8 ​ = 400 50000 ​ = 125 So, Z = 125 Ω . We know that cos ϕ = 0.8 . Since sin 2 ϕ + cos 2 ϕ = 1 , we have: sin ϕ = 1 − cos 2 ϕ ​ = 1 − ( 0.8 ) 2 ​ = 1 − 0.64 ​ = 0.36 ​ = 0.6 X L ​ = Z sin ϕ = 125 × 0.6 = 75 Ω

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