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Q.

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

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a

20 mJ

b

40mJ

c

10mJ

d

80mJ

answer is B.

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Detailed Solution

Energy stored in capacitor when it is charged upto $2 V = \frac{1}{2} 102^{2} = 20 μJ = u_{1}$ (suppose) Energy stored in capacitor when it is charged upto $4 V = \frac{1}{2} 104^{2} = 80 μJ = u_{2}$ (suppose)
$Increase in charge = 40 - 20 = 20 μc$
Energy drawn from cell = $= 20 \times 4 = 80 μJ = u$ (suppose)
$\begin{aligned} Heat produced = u_{1} + u - u_{2} \\ = 20 + 80 - 80 \\ = 20 μJ \end{aligned}$

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