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A light bulb is rated at 100W for a 220 ...

A light bulb is rated at 100W for a 220 V supply. Find (a) the resistance of the bulb, (b) the peak voltage of the source, and (c) the rms current through the bulb.

Text Solution

Verified by Experts

(a) We are given P = 100 W and V = 220 V. The resistance of the bulb is
`R=(V^(2))/(P)=((220V)^(2))/(100W)=484Omega`
(b) The peak voltage of the source is
`v_(m)=sqrt(2)V=311V`
(c) Since, P = I V
`I=(P)/(V)=(100W)/(220V)=0.454A`
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