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Question

# A parallel plate condenser of capacity C is connected to a battery and is charged to potential V. Another condenser of capacity 2C is connected to another battery and is charged to potential 2V. The charging batteries are removed and now the condensers are connected in such a way that the positive plate of one is connected to negative plate of another. The final energy of this system is :

A
Zero
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B
25CV26
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C
3CV22
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D
9CV22
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Solution

## The correct option is D 3CV22 When the condensers are connected in such a way that the positive plate of one is connected to negative plate of another then the total charge Q=4CV−CV=3CV Now, let it is distributed as shown in figure below, potential across the capacitors is same So,q2C=3CV−qC⇒q=2CV therefore , charge of one capacitor (C) is Q1=3CV−2CV=CV, charge of other capacitor (2C) is Q2=2CV Total potential energy =Q212C1+Q222C2=C2V22C+4C2V22×2C=3CV22

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