Question

$n$ identical cells are joined in series with its two cells $A$ and $B$ in the loop with reversed polarities. Emf of each cell is $E$ and internal resistance $r$. Potential difference across cell $A$ or $B$ is (here $n > 4)$

• A. $\frac{2 E}{n}$
• B. $2 E\left(1-\frac{1}{n}\right)$
• C. $\frac{4 E}{n}$
• D. $2 E\left(1-\frac{2}{n}\right)$

Answer 1

Answer: (D)

The two opposite cells $A$ and $B$ will cancel two more other cells, so net emf will be $n-4$. So current
$I=\frac{(n-4) E}{n r} .$ Now p.d. across $A$ or $B$
$=E+I r$ (as they will be in charging state)
$=E+\frac{(n-4) E}{n}=2 E\left(1-\frac{2}{n}\right)$