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A filament bulb (500W, 100V) is to he used in a 230 V main supply. When a resistance R connected in series, it works perfectly and the bulb consumes 500W. The value of R is |
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Answer» `26 OMEGA` Resistance of bulb, `R_(B) = (V^(2))/(P) = ((100)^(2))/(500)` `therefore R_(B) = 20 Omega` CURRENT is same in series, therefor , R `prop` V `therefore(R)/(R_(B)) = (V_(R))/(V_(B))` `therefore R = R_(B) xx (V_(R))/(V_(B))` `= 20 xx (130)/(100)` `therefore R =26 Omega` 
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