Question

$n$ identical cells are joined in series with its two cells $\text{A}$ and $\text{B}$ in the loop with reversed polarities. EMF of each cell is $E$ and internal resistance $r$. Potential difference across cell $\text{A}$ or $\text{B}$ is:
(Here $n>4$)

• 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 D $2E(1-\frac{2}{n})$

Circuit can be drawn as:


From question, we can clearly observe that two cells $\text{A}$ and $\text{B}$ of same polarity are connected with $(n-2)$ cells with reversed polarities as shown above.

So, equivalent emf of the circuit will be

$E_{eq}=(n-2)E-2E=(n-4)E$

And equivalent resistance is given by

$r_{eq}=nr$

So, current in the circuit,

$i=\frac{E_{eq}}{r_{eq}}=\frac{(n-4)E}{nr}$


Potential difference across cell $\text{B}$, $V_1-V_2$ can be calculated as:

$V_1-E-ir=V_2$ (From KVL)

$\Rightarrow V_1-V_2=E+ir=E+\frac{(n-4)E}{nr}r$

$\therefore V_1-V_2=2E(1-\frac{2}{n})$

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

Why this question? To obtain net potential of batteries connected in series their individual potential should be added with their respective signs.