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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. |
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Answer» <P> Solution :(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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