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:

Options

A. `15 V`
B. `20 V`
C. `25 V`
D. `39 V`

Answer

D

Explanation

`I=E/Z, 0.5 =50/Z,Z=100 Omega`
`Z^(2)=R^(2)+omega^(22)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)]`