Question

n identical cells are joined in series with its two cells A and B in the loop with reversed polarities.
EMF of each shell is E and internal resistance r. Potential difference across cell A or B is:

• A. $\frac{2E}{n}$
• B. $2E|1-\frac{1}{n}|$
• C. $\frac{4E}{n}$
• D. $2E|1-\frac{2}{n}|$

Answer 1

The correct option is C $\frac{4E}{n}$

Number of cells $=n$

EMF of each cell $=E$

Internal resistance of each cell, $=r$

All the cells are in series.

Thus,

Total internal resistance, $R=nr$

Total EMF, $e=nE$

As the two cells are connected in reverse polarity, these two cells will cancel-out the contribution of other two cells in the loop.

Then, the net EMF across the circuit will be

$E_{eq}=nE-4E$

The current $I$ in the circuit $=\frac{nE-4E}{nr}$

Voltage across the opposite connected batteries, $V=E-Ir$

$V=E-\frac{E(n-4)}{nr}\times r$

$V=\frac{4E}{n}$

Therefore, the voltage across the battery A or B will be $\frac{4E}{n}$.