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A $2 μ F$ capacitor is charged to a potential of $15 V$ and a $3 μ F$ is charged to a potential of $10 V$ and the capacitor are connected such that positive plate of one is connected to the negative plate of the other capacitor and negative plate of one is connected to the positive plate of the other capacitor. Select the correct statement(s) about the final circuit.

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Total loss of energy is 375 $μ J$

Total loss of energy is 225 $μ J$

Final total electrical energy of the capacitor will be non zero

Final charge on each capacitor is zero.

answer is A, C.

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Detailed Solution

The common potential is given by

$V = \frac{C_{1} V_{1} - C_{2} V_{2}}{C_{1} + C_{2}} = \frac{30 - 30}{C_{1} + C_{2}} = 0$

Final charge on the first capacitor is

$Q_{1}^{'} = C_{1} V = 0$

Final charge on the second capacitor is

$Q_{2}^{'} = C_{2} V = 0$

The final energy stored in the arrangements is zero.

So total loss of energy = Total initial energy stored in the system = $(\frac{1}{2} \times 2 \times 15^{2} + \frac{1}{2} \times 3 \times 10^{2}) μ J = 375 μ J$

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