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Question

The power factor of the circuit shown below is 1/2. The capacitance of the circuit is equal to

A
400 μ F
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B
300 μ F
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C
500 μ F
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D
200 μ F
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Solution

The correct option is C 500 μ F

cos ϕ=12=RZ

Z=2R
Z2=2R2
R2+(XLXc)2=2R2
XLXc=±10 Ω
ω L1ω C=±10Ω
As ω=100
(100×0.11100×C)=±10Ω
(101100C)=±10Ω
1100C=0 or 20
Possible value of C is =500 μ F

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