When 100 V dc is applied across a coil, a current of 1A flows through it and when 100 V ac of 50 Hz is applied to the same coil, only 0.5 flows
The inductance of coil is

Options

A. 5.5 H
B. `3//(pi)H`
C. `sqrt(3)//(pi)H`
D. `2.5 H`

Answer

C

Explanation

when dc is applied , `R=V/I =(100)/(1) = 100 Omega`
and when ac of 50 Hz is applied
`I=V/Z i.e., Z=V/I = 100/0.5 = 200 Omega`.
`Z=sqrt(R^(2)+omega^(2)L^(2))`
i.e., `omega^(2)L^(2)=Z^(2)-R^(2)`
i.e., `(2 pi fL)^(2)=(200^2)-(100^2)=3xx10^(4)` (`as o,ega = 2 pi f`)
so,` L=(sqrt(3)xx10^(2))/(2 pi xx 50 ) =(sqrt(3))/(pi)H=0.55H`.