Question

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`

Answer 1

Correct Answer - 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)]`