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

Options

A. `0.1` mH
B. `1` mH
C. `0.1` H
D. `1.1` H

Answer

D

Explanation

Resistance of bulb (R) = `(V^(2))/(P) = (100^(2))/(50) = 200 Omega`
Current through bulb ` I = (V)/(R) = (100)/(200) = 0.5`A.
Choke of inductive reactance `X_(L)` and resistance of the bulb `200 Omega` are in series.
`therefore X_(L)^(2) + R^(2) = Z^(2)`, where `Z = (200 ("volt"))/(0.5 ("amp")) = 400Omega`
or `X_(L)^(2) = Z^(2) - R^(2) = (400)^(2) - (200)^(2)`
or `(2pifL)^(2) = 12 xx 10^(4)`
or `L = (2sqrt3 XX100)/(2pi xx 50) = (2sqrt3)/(pi) = (2 xx 1.73)/(3.14) = 1.10` henry
`therefore` Inductance (L) of choke = 1.1 H.