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A cell $E_{1}$ of emf 6V and internal resistance 2 $Ω$ is connected with another cell $E_{2}$ of emf 4V and internal resistance 8 $Ω$ (as shown in the figure). The potential difference across points X and Y is :

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10.0V

3.6V

5.6V

2.0V

answer is C.

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Detailed Solution

Current $I = \frac{6 - 4}{10} = \frac{1}{5} A$

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As the 4 V cell is in charging, P.D across X and Y is 4+ (8 x 1/5) = 5.6v

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