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If the power factor is 1//2 in a series ...

If the power factor is `1//2` in a series `RL` circuit with `R = 100 Omega`. If `AC` mains, `50 Hz` is used then `L` is

A

`sqrt3/pi` Henry

B

`pi` Henry

C

`pi/sqrt3` Henry

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`cos phi=R/Z=100/Z=1/2`
`Z=200= sqrt(R^2 + X_L^2)`
`rArr X_L =sqrt3 xx 100 Omega = omegaL = 100 piL`
`L=sqrt3/piH`
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