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Question

Two cells of unequal emfs, E1 and E2, the internal resistance r1 and r2 are joined as shown. VA and VB are the potentials at A and B respectively

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A
one cell with continuously supply energy to the other
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B
the potential difference across both the cells will be equal
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C
the potential difference across one cell will be grater than its emf
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D
VAVB=(E1r2E2r1)(r1+r2)
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Solution

The correct options are
A one cell with continuously supply energy to the other
B the potential difference across both the cells will be equal
C the potential difference across one cell will be grater than its emf
Let i be the current flow in clockwise direction in the circuit .
thus, E2E1=i(r1+r2)
or i=E2E1r1+r2
now, VAVB=E2+ir2=E2+(E2E1)r2r1+r2=E2r1+E2r2+E2r2E1r2r1+r2=E2r1+2E2r2E1r2r1+r2
This is also potential across each cell.
Thus, VAVB>E2
Current flows in the cell of emf E2 from the positive plate to the negative plate inside the cell and hence it absorbs energy.
ans :A,B,C

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