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 reverse polarity. The potential difference across the terminal of $\text{A}$ is

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

Answer 1

The correct option is C $2\epsilon (1-\frac{1}{n})$

If polarity of $m$ cells out of $n$ cells in series combination is reversed, then equivalent emf, ${\epsilon }_{eq}=(n-2m)\epsilon$, while the equivalent resistance is still $nr+R$ where $R$ is the external resistance, so current in $R$ will be

$i=\frac{(n-2m)\epsilon }{nr+R}$

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

$\therefore i=\frac{(n-2)\epsilon }{nr}$

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 the potential difference across the terminals of $\text{A}$ is,

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

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

$\therefore V=\frac{(2n-2)\epsilon }{n}=2\epsilon (1-\frac{1}{n})$

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