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. 0.55 H
B. 5.5 mH
C. 0.55mH
D. 55mH.

Answer

A

Explanation

For dc, V = lR (R is resistance of inductor )
100 = `1 xx R implies R = 100 Omega`
For ac , `Z = sqrt(R^(2) + X_(L)^(2))`
`because V = IZ`
`implies 100 = 0.5 sqrt(R^(2) + X_(L)^(2))`
`implies 100 = 0.5 sqrt((100)^(2) + X_(L)^(2))`
`implies 100 = (5)/(10) sqrt((100)^(2) + X_(L)^(2))`
`X_(L)^(2) = (200)^(2) - (100)^(2) = 300 xx 100`
`implies X_(L) = sqrt(3 xx (100)^(2)) = 100 sqrt3`
`implies omega L = 100 sqrt3 implies 2 pi v L = 100 sqrt3`
`implies 2 xx pi xx 50 xx L = 100 sqrt3`
`therefore L = (sqrt3)/(pi) = (sqrt3)/(3.14) implies L = 0.55 H`.