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 resistance is

Explanation

Data: `V_(dc) = 100 V, Id_(dc) = 1A, V_("rms")= 100 V, f=50 Hz, I_("rms")= 0.5`A
(i) The resistance of the coil,
`R=(V_(dc))/(I_(dc)) = 100/1 = 100 Omega`
(ii) The impedance of the coil,
`Z=(V_(rms)/I_(rms)) = 100/0.5 = 200 Omega`
`Z^(2) = R^(2) + X_(L)^(2)`
`therefore X_(L)^(2) = Z^(2)-R^(2)`
`=(200)^(2) - (100)^(2) = 3 xx 10^(4)`
`therefore X_(L) = sqrt(3) xx 10^(2) = 173.2 Omega`
`X_(L) = omegaL = (2pif)L`
`therefore` The self-inductance of the coil, `L= X_(L)/(2pif) = 173.2/(2 xx 3.142 xx 50) = 0.5512` H