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The potential differnce between the two ...

The potential differnce between the two ends of an electric lamp is decrased by 1 % . Neglecting the change in its resistance , calculate the percentage increase or decrease in the power of the lamp .

Text Solution

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we kown power , `P=(V^(2))/(R)`
`therefore"In"P=2"In"V-"In"R`
By differentiating , `(dP)/(P)=2.(dV)/(V)[thereforeR="constant",dR=0]`
According to the problem , `(dV)/(V)=-1%=-(1)/(100)`
`therefore(dP)/(P)=-(2)/(100)=-2%`
i.e , the power of the lamp will be decreased by 2 %
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