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Question

n identical cells, each of emf ε and internal resistance r, are joined in series to form a close circuit. One cell A is joined with reversed polarity. The potential difference across each cell, except A is

A
2εn
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B
n1nε
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C
n2nε
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D
2nn2ε
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Solution

The correct option is A 2εn
As one of the cell is connected in reverse polarity, so there will be (n1) cells having same polarity and 1 having reverse polarity so net emf will be

εnet=(n1)εε=(n2)ε

And equivalent internal resistance of the circuit will be

req=nr (series combination)

So current in the circuit will be

i=εnetreq=(n2)εnr

Current in all other cells will be flowing out of the positive terminal (discharging) and in cell A it will be flowing into the positive terminal (charging).

So, potential difference across any cell except A,

V=εir (From KVL)

V=ε(n2)εnrr=ε[1n2n]

V=2εn

Hence, option (a) is correct.
Why this question ?
Concept: If polarity of m cells out of n cells in series combination is reversed, then equivalent emf, Eeq=(n2m)E, while the equivalent resistance is still nr+R where R is the external resistance, so current in R will be
i=(n2m)Enr+R

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