The power factor of the circuit in below figure is 1√2. The capacitance of the circuit is equal to
A
400μF
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B
300μF
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C
500μF
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D
200μF
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Solution
The correct option is C500μF Given that V=2sin100t Comparing this with general form V=V0sinωt we get V0=2ω=100rad/s R=10ΩL=0.1H Power factor cosϕ=1√2 We know that cosϕ=RZ=R√R2+(XL−XC)2 ......(i) So XL=ωL=100×0.1=10ΩXC=1ωC=1100C Putting all value in eq(i) and squaring both sides we get 12=102102+(10−1100C)2 102+(10−1100C)2=200 (10−1100C)2=102 10−1100C=±10 (cannot be +10 because that would give indefinite C) 10−1100C=−10 C=500μF