When `100V` dc is applied across a coil a current of `1` A flows through it When `100V` ac of `50Hz` is applied across the same coil only `0.5A` flows The resistance and inductance of the coil are (`take pi^(2) =10)` .

Options

A. `50 Omega 0.3H`
B. `50 Omega sqrt((0.3)H)`
C. `100 Omega 0.3 H`
D. `100 Omega sqrt(0.3)H`

Answer

D

Explanation

In dc circuit resistance offered by inductor is zero in steady state
` I = (V)/R implies 1 =(100)/R implies R 100 Omega`
`I' = (1)/(2)A`
`I' = (V)/(sqrt(R^(2) + X_(L)^(2))) implies (1)/(2) = (100)/(sqrt(R^(2) + X_(L)^(2))`
`R^(2) + X_(L)^(2) = (200)^(2) implies X_(2)^(L) = (200)^(2) - (100)^(2)`
`X_(L)^(2) = 300 xx 100`
`(2pi fL)^(2) = 300 xx 100`
`4pi^(2) xx (50)^(2) xx L^(2) = 300 xx 100`
`L^(2) = 0.3 implies L = sqrt(0.3) H`

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