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

Step-by-step solution

Resistance of bulb R= V^2 P = (100)^2 50 =200 Current through bulb (I)= V R = 100 200 =0.5 ~A In a circuit containing inductive reactance (X_L ) and resistance (R) , impedance (Z) of the circuit is Z= R^2+ ^2 L^2 ...(i) Here, Z= 200 0.5 =400 Now, X_L^2=Z^2-R^2=(400)^2-(200)^2(2 f L)^2=12 10^4L= 2 3 100 2 50 = 2 3 =1.1 H