A charged capacitor $(C_{1} = 3 μ F)$ is getting discharged in the circuit shown. When the current I was observed to be 2.5A, switch ‘S’ was opened. Determine the amount of heat $(μ J)$ that will be liberated in the circuit after switch ‘S’ is opened.

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answer is 60.

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Detailed Solution

Initially $V_{R} = V_{C_{1}} = 2.5 \times 4 = 10 V$
$Q_{1} = C V = 3 \times 10 = 30 μ C U_{1} = \frac{1}{2} \times 3 \times 10^{2} = 150 μ C$

Finally $V_{C_{1}} = V_{C_{2}} = \frac{Q_{1}}{C_{1} + C_{2}} = \frac{30}{5} = 6 V$
$U_{2} = \frac{1}{2} (C_{1} + C_{2}) \times V^{2} = \frac{1}{2} \times 5 \times 36 = 90 μ J$

heat liberated $= 150 - 90 = 60 μ J$