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Question

A capacitor of capacitance C=900 pF is charges fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C=900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is


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Solution

Initial stored energy of capacitor is:
Uinitial =12CV2=12C×(100)2=5000C
Energy lost is given as:
Lost ΔU=12C1C2C1+C2(V1V2)2=12C22C(1000)2

=14C(100)2=2500C

Ufinal =UiΔU=5000C2500C=2500C

=2500×900×1012

=225×108

Ufinal =2.25×106 J
Hence the correct option is (a)

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