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Q.

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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Detailed Solution

Ui=12 x 3 x 62 μJ=54 μJInitial charge on capacitor = 3×6μ=18 μCFinal p.d. across the combination Vf=qtotCtot=183 + 9V=32V∴Uf=12 x (3+9) x 322 μJ=13.5 μJ∴ Lost energy =Ui-Uf=(54-13.5) μJ=40.5 μJ
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