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Question

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


A
3.6 V
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B
10.0 V
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C
5.6 V
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D
2.0 V
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Solution

The correct option is C 5.6 V
Given:
E1=6 V
r1=2 Ω

E2=4 V
r2=8 Ω

Now,
Eeff=64=2 V
Req=2+8=10 Ω

So, current in the circuit will be
I=EeffReq
I=210=0.2 A

On going from point X to Y in clockwise, we get,

VX+4+0.2×8=VY

VXVY=5.6 V

Therefore, the potential difference across points X and Y is 5.6 V.

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