An electric bulb has a rated power of 50W at 100 V. if it used on an a.c. source

An electric bulb has a rated power of 50W at 100 V. if it used on an a.c. source 200V, 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

Correct Answer:

1.1 H

Explanation:

No explanation available. Be the first to write the explanation for this question by commenting below.

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  • maximum current that can flow =p/v = 50/100 =0.5
    R= V²/p= 100*100/50 =200
    vrms=Irma * Z
    220 = 0.5 √(R²+X²)
    X comes out to be 346.14
    Now,
    X = 2πfL
    L = 346.4/(2*3.14*50)

    Hope it helps😃

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