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Derive the equations of stored energy for series or parallel connection of many capacitors. |
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Answer» Solution :For series connection : Charge REMAINS same Q on each capacitor `:.` TOTAL energy stored, `U =(Q^(2))/(2).(1)/(C)` `=(Q^(2))/(2)[(1)/(C_(1))+(1)/(C_(2))+CDOTS+(1)/(C_(n))]` `=(Q^(2))/(2C_(1))+(Q^(2))/(2C_(2))+cdots+(Q^(2))/(2C_(n))` `:. U = U_(1)+U_(2)+cdots+U_(n)` For parallel connection : Potential difference V remains same for each capacitor `:.` Total energy stored , `U=(1)/(2) CV^(2)` `=(1)/(2)[C_(1)+C_(2)+cdots+C_(n)]V^(2)` `=(1)/(2) C_(1)V^(2)+(1)/(2)C_(2)V^(2)+cdots+(1)/(2)C_(n)V^(2)` `:. U_(1)+U_(2)+cdots+U_(n)` So the total energy stored in both the connection is the sum of energy stored in each capacitor. |
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