wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The power factor of the given circuit is 1/2. The capacitance of the circuit (in μF) is equal to


Open in App
Solution

We know that,
Power Factor =cosθ=RZ=12
where Z is the impedance of the circuit.

In the given circuit,
R=10 Ω,L=0.1 H,ω=100 rad/s

We know,
Z=R2+(XLXC)2=(10)2+(XLXC)2
where
XL=ωL=100×0.1=10 Ω
XC=1ωC=1100C

Therefore,
cosθ=R(R)2+(ωL1ωC)2=12
10(10)2+(101100C)2=12

Cross-multiplying and squaring both sides,
200=100+(101100C)2
100=(101100C)2
±10=101100C
10=101100C
[for +10,C which is not possible]

20=1100C
C=12000=500 μF

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Electrical Power in an AC Circuit
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon