A capacitor of capacitance 10 $μF$ is charged up to a potential difference of 2V and then the cell is removed. Now it is connected to a cell of emf 4V and is charged fully. In both cases the polarities of the two cells are in the same directions. Total heat produced in the complete charging process is

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40 $μ$ J

10mJ

20 $μ$ J

80mJ

answer is B.

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Detailed Solution

Energy stored in capacitor when it is charged upto

$2 V = \frac{1}{2} 10 \times 2^{2} = 20 μJ = U_{1}$

Energy stored in capacitor when it is charged upto

$4 V = \frac{1}{2} 10 \times 4^{2} = 80 μJ = U_{2}$

Increase in charge = 40 – 20 = 20 $μ$ C

Energy drawn from cell =charge drawn x voltage of cell =20 × 4 = 80 $μ$ J = $U$ (suppose)

Heat produced $= U - (U_{2} - U_{1}) = 20 μJ$

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