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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:
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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😃