Q: An electric bulb has a rated power of 50 W at 100 V. If it is 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

(a) 1 mH

(b) 0.1 mH

(c) 0.1 H

(d) 1.1 H

Ans: (d)

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

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

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

$ \displaystyle Z = \frac{200}{0.5}$

Z = 400 Ω

$ \displaystyle X_L = \sqrt{Z^2 – R^2}$

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

$ \displaystyle X_L = 200\sqrt{3}$

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

$ \displaystyle L = \frac{200\sqrt{3}}{\omega} $

$ \displaystyle L = \frac{200\sqrt{3}}{2 \pi \nu} $

$ \displaystyle L = \frac{200\sqrt{3}}{2 \pi \times 50} $

L = 1.1 H