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A capacitor of capacitance $3 μF$ is charged to a potential of 6 volt. Now the charging battery is removed and the capacitor is connected in parallel with another capacitor of capacitance $9 μF$ . Then the loss of energy is

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$45.2 μJ$

$38 μJ$

$40.5 μJ$

$48.5 μJ$

answer is A.

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Detailed Solution

$U_{i} = \frac{1}{2} x 3 x 6^{2} μJ = 54 μJ$

Initial charge on capacitor = $3 \times 6 μ = 18 μC$

Final p.d. across the combination $V_{f} = \frac{q_{tot}}{C_{tot}} = \frac{18}{3 + 9} V = \frac{3}{2} V$

$∴ U_{f} = \frac{1}{2} x (3 + 9) x {(\frac{3}{2})}^{2} μJ = 13.5 μJ$

$∴$ Lost energy $= U_{i} - U_{f} = (54 - 13.5) μJ = 40.5 μJ$

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