Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals $a$ and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using battery of emf $V$ volt. Now disconnecting $a$ and $b$ , the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy?

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Two identical capacitors

C_{1}

and

C_{2}

of equal capacitance are connected as shown in the circuit. Terminals

a

and

b

of the key

k

are connected to charge capacitor

C_{1}

using battery of emf

V

volt. Now disconnecting

a

and

b

, the terminals

b

and

c

are connected. Due to this, what will be the percentage loss of energy?

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Question Text	Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals $a$ and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using battery of emf $V$ volt. Now disconnecting $a$ and $b$ , the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy?
Topic	Electrostats
Subject	Physics
Class	Class 12

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