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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 reverse polarity. The potential difference across the terminal of A is

A
2εn
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B
ε(11n)
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C
2ε(11n)
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D
ε(n2n)
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Solution

The correct option is C 2ε(11n)
If polarity of m cells out of n cells in series combination is reversed, then equivalent emf, εeq=(n2m)ε, while the equivalent resistance is still nr+R where R is the external resistance, so current in R will be

i=(n2m)εnr+R

Here, m=1, R=0, n=n

i=(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 the potential difference across the terminals of A is,

V=ε+ir (From KVL)

V=ε+(n2)εnrr=ε[1+n2n]

V=(2n2)εn=2ε(11n)

Hence, option (c) is correct.

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