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.

Explanation

(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`