Question

When `100" volt" dc` is applied across a coil, a current of `1 amp` flows through it, when `100 V` ac of `50 Hz` is applied to the same coil, only 0.5 amp flows. Calculate the resistance and inductance of the coil.

Answer 1

In case of a coil, i.e. `L-R` circuit,
`I = (V)/(Z)` with `Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(R^(2) + (omega L)^(2))`
So when dc in applied, `omega = 0`, so `Z = R`
and hence `I = (V)/(R ), i.e, 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`
but `Z = sqrt(R^(2) + omega^(2) L^(2))`, i.e., `omega^(2) L^(2) = Z^(2) - R^(2)`
i.e., `(2 pi f L)^(2) = 200^(2) - 100^(2) = 3 xx 10^(4)`
`L = (sqrt(3)xx 10^(2))/(2 pi xx 50) =(sqrt(3))/(pi) H = 0.55 H`