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A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another unchanged 600pF capacitor. How much electrostatic energy is lost in the process. |
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Answer» Solution :Here `C_(1) 600pE ,C_(2) = 600 pF, V_(1)=200 V , V_(2) = 0` V From GIVEN formula `:.` Loss in electrostatic ENERGY `DeltaU = (C_(1)C_(2)(V_(1)-V_(2))^2)/(2(C_(1)+C_(2)))` `=(600xx10^(-12)xx600xx10^(-12)(200-0))/(2(600xx10^(12)600xx10^(-12)))` `=(36xx10^(-20)xx200)/(2xx1200xx10^(-12))` `= 6xx10^(-6)` J Loss in energy. |
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