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An inductance of 100 mH and a resistance...

An inductance of 100 mH and a resistance of 100 `Omega` are connected in series and an alternating emf of peak value 100 V, 50 Hz is applied across the combination. The power factor of the circuit is

A

0.954

B

9.54

C

95.4

D

0.845

Text Solution

Verified by Experts

The correct Answer is:
A
`X_(L)=2pifL=2xx3.14xx50xx0.1`
=31.4`Omega`
Power factor`=(R)/(Z)=(R)/(sqrt(R^(2)+X_(L)^(2)))`
`=(100)/(sqrt(100^(2)+(31.4)^(2)))=0.954`
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