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

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Correct answer: D. 1.1 H

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+ω^2 L^2 ...(i) Here, Z=(200)/(0.5)=400 Now, X_L^2=Z^2-R^2 =(400)^2-(200)^2 (2 π f L)^2=12 × 10^4 L= 2 3 × 100 2 π × 50 = 2 3 π =1.1 H

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

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