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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. |
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Answer» Correct Answer - `30 kJ, (5)/(23)A` 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` |
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