Q: An electric bulb has a rated power of 50 W at 100 V. If it used on an AC source of 200 V, 50 Hz, a choke has to be used in series with it. This choke should have an inductance of

Sol: Here, P = 50 W, V = 100 Volt

I = P/V = 50/100 = 0.5 A, R = V/I = 100/0.5 = 200 Ω

Let L be the inductance of the choke coil .

$\large I_v = \frac{E_v}{Z} $

$\large Z = \frac{E_v}{I_v} = \frac{200}{0.5} $

Z = 400 Ω

Now , $\large X_L = \sqrt{Z^2 – R^2}$

$\large X_L = \sqrt{400^2 – 200^2} = 200\sqrt{3} $

$\large \omega L = 200\sqrt{3} $

$\large L = \frac{200\sqrt{3}}{\omega} = \frac{200\sqrt{3}}{2 \pi f} $

$\large L = \frac{200\sqrt{3}}{100 \pi} $

L = 1.1 H