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)

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) (2 E/n) (B) 2 E(1-(1/n)) (C) (4 E/n) (D) 2 E(1-(2/n))

Tardigrade Admin · Accepted Answer

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

Q. $n$ 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 $n > 4)$

$\frac{2 E}{n}$

$2 E (1 - \frac{1}{n})$

$\frac{4 E}{n}$

$2 E (1 - \frac{2}{n})$

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)
$= E + \frac{( n - 4 ) E}{n} = 2 E (1 - \frac{2}{n})$

Q. 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)

n2E​

2E(1−n1​)

n4E​

2E(1−n2​)