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 2εnAs one of the cell is connected in reverse polarity, so there will be (n−1) cells having same polarity and 1 having reverse polarity so net emf will be εnet=(n−1)ε−ε=(n−2)ε And equivalent internal resistance of the circuit will be req=nr (∵series combination) So current in the circuit will be i=εnetreq=(n−2)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=ε−ir (From KVL) ⇒V=ε−(n−2)εnrr=ε[1−n−2n] ∴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=(n−2m)E, while the equivalent resistance is still nr+R where R is the external resistance, so current in R will be i=(n−2m)Enr+R

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