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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 n3πμF, then value of n is _____

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answer is 0400.00.

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Detailed Solution

cosθ=0.8 ⇒tanθ=34=XLR⇒XL=3R4 And Z=XL2+R2⇒Z=5R4 Power P=V2Z2×R ⇒400=2502×R25R216⇒R=100Ω So, XL=75Ω For power fractor to be 1,XC=XL So, 12πfC=75⇒C=4003πμF

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