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

Step-by-step solution

Here, P=50 ~W , V=100 ~V I= P V = 50 100 =0.5 ~A , R= V I = 100 0.5 =200 Let L be the inductance of the choke coil aligned & I_v= E_v Z or Z= E_v I_v = 200 0.5 =400 & =100 12 & aligned Now, X_L= Z^2-R^2 = 400^2-200^2 =100 12 or, L=100 2 3 aligned L= 200 3 = 200 3 2 v = 200 3 100 & = 2 1.732 3.14 & =1.1 H aligned