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A coil of inductance 0.1 H is connected ...

A coil of inductance `0.1 H` is connected to `50 V, 100Hz` generator and current is found to be `0.5A`. The potential difference across resistance of coil is:

A

`15 V`

B

`20 V`

C

`25 V`

D

`39 V`

Text Solution

Verified by Experts

The correct Answer is:
4
`I = (E)/(Z) , 0.5 = (50)/(Z) = 100 Omega`
`Z^(2) = R^(2) + omega^(2) L^(2)`, then `R = 78 Omega`
Now `V_(R ) = sqrt(V_(LR)^(2) = V_(L)^(2)) = 39 V , [V_(R)^(2) + V_(L)^(2) = V_(LR)^(2)]`
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