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Q.

 A cell E1 of emf 6V and internal resistance 2Ω is connected with another cell E2 of emf 4 V and internal resistance 8Ω (as shown in the figure). The potential difference across points X and Y is : 
 

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a

5.6 V

b

10.0 V

c

2.0 V

d

3.6 V

answer is C.

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Detailed Solution

 Current I=6−410=15A
 

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As the 4 v cell is in charging, P.D across x and y is 4 + (8x1/5) =5.6 V

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