Two identical capacitors and of equal capacitance are connected as shown in the circuit. Terminals and of the key are connected to charge capacitor using a battery of emf volt. Now disconnecting and , the terminals and are connected. Due to this

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Solution

Q.
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 a 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 [2019]

1 75%

2 0%

3 50%

4 25%

Ans.
(3)

As we know that, loss of electrostatic energy,

$E_{loss} = \frac{1}{2} \frac{C_{1} C_{2}}{(C_{1} + C_{2})} V^{2} = \frac{1}{2} \times \frac{C^{2}}{2 C} V^{2}$

$= \frac{1}{2} (\frac{1}{2} C V^{2}) = \frac{1}{2} E$ $[∵ C_{1} = C_{2} = C]$

$∴$ $Percentage of loss of energy = \frac{\frac{1}{2} E}{E} \times 100 %$

$= \frac{1}{2} \times 100 % = 50 %$

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