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

Explanation

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