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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

`pi`H

B

`sqrt(3)/piH`

C

`pi/sqrt(3)H`

D

`sqrt(2)/pi`H

Text Solution

Verified by Experts

The correct Answer is:
B
Power factor, `cosphi = 1/2 rArr phi = 60^(@)`
`therefore tan60^(@) = (omegaL)/R rArr sqrt(3) = (2piv xx L)/(100)`
`L= (sqrt(3) xx 100)/(2pi xx 50) = sqrt(3)/pi` H
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