An electric bulb is marked to `100 W, 220 V`. If the supply voltage drops to `115 V`, what is the heat end light energy produced by the bulb in `20` minutes. Calculate the current flowing through it.

Answer

`30 kJ, (5)/(23)A`

Explanation

Resistance of the bulb `R = ((230)^(2))/(100) = 529 Omega`
when the voltage drops to `V = 115V` , the total heat and light energy produced by the bulb in `20 min will be
`H = (V^(2))/(R ) l = (115)^(2))/(529) xx (20 xx 60) = 30000 J = 30kJ`
Current `l = (V)/(R ) = (115)/(529) = (5)/(23)A`