(a) The average power dissipated
`bar(P)=(i^(2).R)=(i_(m)^(2)R sin^(2) omega t)=i_(m)^(2)R(sin^(2)omega t)`
`therefore" "sin^(2)omega t=(1)/(2)(1-cos 2 omega t)" "=(sin^(2)omega t)=(1)/(2)(1-(cos 2 omega t))=(1)/(2)" "therefore" "(cos 2 omega t = 0)`
`therefore" "bar(P)=(1)/(2)i_(m)^(2)R`.
(b) Powet of the bulb, P = 100 W and voltage, V = 220 V.
The resistance of the bulb is given as `R=(V^(2))/(P)=(220 xx 220)/(100)= 484 Omega`.