A bulb rated `60 W` at `220 V` is connected across a household supply of alternating voltage of `220 V`. Calculate the maximum instantaneous current through the filament.

Explanation

Here `P= 60W`, `V=220 V=E`
`R=(V^2)/P=(220 xx 220)/(360)`
`=806.67`
`in_0 = sqrt2 E = 1.414 xx 220`
`=311.08`
`i_0=(in_0)/R = 806.67/311.08`
`0.385=0.39 A`