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. The potential difference across cell A or B is (Here n > 4)

$\frac{2 E}{n}$

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

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$\frac{4 E}{n}$

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

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Solution

The correct option is D
$2 E (1 - \frac{2}{n})$

Equivalent emf of the circuit, $E_{e q} = (n - 2) E - 2 E = (n - 4) E$
Equivalent resistance, $r_{e q} = n r$
So, current in the circuit, $I = \frac{E_{e q}}{r_{e q}} = \frac{(n - 4) E}{n r}$
Potential difference across A or B, $V = E + I r = E + \frac{(n - 4) E}{n r} \times r = 2 E (1 - \frac{2}{n})$

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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. The potential difference across cell A or B is (Here n > 4)

The correct option is D 2E(1−2n) Equivalent emf of the circuit, Eeq=(n−2)E−2E=(n−4)E Equivalent resistance, req=nr So, current in the circuit, I=Eeqreq=(n−4)Enr Potential difference across A or B, V=E+Ir=E+(n−4)Enr×r=2E(1−2n)