Question

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

• A. $\frac{2\epsilon }{n}$
• B. $\frac{n-1}{n}\epsilon$
• C. $\frac{n-2}{n}\epsilon$
• D. $\frac{2n}{n-2}\epsilon$

Answer 1

The correct option is A $\frac{2\epsilon }{n}$

As 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

${\epsilon }_{net}=(n-1)\epsilon -\epsilon =(n-2)\epsilon$

And equivalent internal resistance of the circuit will be

$r_{eq}=nr(\because \text{series combination})$

So current in the circuit will be

$i=\frac{{\epsilon }_{net}}{r_{eq}}=\frac{(n-2)\epsilon }{nr}$

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

So, potential difference across any cell except $\text{A}$,

$V=\epsilon -ir$ (From KVL)

$\Rightarrow V=\epsilon -\frac{(n-2)\epsilon }{nr}r=\epsilon [1-\frac{n-2}{n}]$

$\therefore V=\frac{2\epsilon }{n}$

Hence, option $\left(a\right)$ is correct.

Why this question ? Concept: If polarity of $m$ cells out of $n$ cells in series combination is reversed, then equivalent emf, $E_{eq}=(n-2m)E$, while the equivalent resistance is still $nr+R$ where $R$ is the external resistance, so current in $R$ will be $i=\frac{(n-2m)E}{nr+R}$