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= V² 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= R²+ ² L² Here, Z= 200 0.5 =400 Now, X_ L ²=Z²-R²=(400)²-(200)² aligned (2 f L)² &=12 10⁴ L &= 2 3 100 2 50 = 2 3 =1.1 H aligned