Q: 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.
Sol: In case of a soil, i.e, L – R circuit,
$\large Z = \sqrt{R^2 + X_L^2}$
$\large Z = \sqrt{R^2 + ( \omega L)^2}$
So when dc is applied, ω = 0, so Z = R
and hence I = V/R, i.e, R = V/I = 100/1 = 100 Ω
and when ac of 50 Hz is applied.
I = V/Z , i.e. Z = V/I = 100/0.5 = 200 Ω
$\large Z = \sqrt{R^2 + ( \omega L)^2}$
$\large \omega^2 L^2 = Z^2 – R^2$
$\large (2 \pi f L)^2 = 200^2 – 100^2 = 3 \times 10^4 $
$\large L = \frac{\sqrt{3} \times 10^2}{2 \pi \times 50} = \frac{\sqrt{3}}{\pi} H$