A bulb of 40 W is connected across a source of 220 V . Find the resistance of the bulb.

Explanation

Data : P=40 W, V=220 V , R=?
P=VI=`V((V)/(R)) =(V^(2))/(R)`
`therefore R=(V^(2))/(P)=((220V)^(2))/(40W)=(220xx220)/(40)Omega=(220xx220)/(2xx20)Omega`
`=11xx110 Omega=1210 Omega`
The resistance of the bulb =`1210Omega`.