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Question

A circuit draws a power of 550 W from a source of 220 V, 50 Hz. The power factor of the circuit is 0.8 and the current lags in phase behind the potential difference. To make the power factor of the circuit as 1.0, what capacitance is needed to be connected in series with it.

A
10 μF
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B
35 μF
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C
75 μF
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D
100 μF
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Solution

The correct option is C 75 μF
Given that current lags behind voltage, so the circuit must contain inductance and resistance initially.

Pavg=E2rmsZ×cosϕ

Z=E2rms×cosϕPavg

Power factor =0.8=cosϕ

Z=(220)2×0.8550

Z=70.4 Ω

As, cosϕ=RZ

R=Zcosϕ=70.4×0.8=56.32 Ω

Initially for LR circuit

Z2=R2+ω2L2

ω2L2=Z2R2

ωL=Z2R2=42.24 Ω

Given that by applying capacitor C to make cosϕ=1

So cosϕ=1RZfinal=1

R=Zfinal

R=R2+(ωL1ωC)2

ωL=1ωC

C=1ω2L=1ω(ωL)=12πf(ωL)

C=12×3.14×50×42.2

C75 μF

Hence, option (C) is correct.

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