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A certain capacitor is charged to a potential V. By what percentage the potential to be increased, so that energy increases by 10%? |
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Answer» Solution :Energy `alpha (p.d)^(2)` `U alpha V^(2) and U. alpha (V.)^(2)` `(U.)/(U)=(110)/(100) =1.1, U.=1.1U` `(U.)/(U)=((V.)^(2))/(V^(2)) therefore (V.)^(2)=(1.1)^(2)` `V.=(sqrt(1.1))V=1.049V` % increase in potential `=((V.-V)/(V)) XX 100 =((1.049V-V)/(V)) xx 100=4.9%` |
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