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(D) Total number of cells is $n$. Two cells are connected with reversed polarity. These two cells will cancel the effect of two other cells in the ser

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

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(D) Total number of cells is $n$. Two cells are connected with reversed polarity. These two cells will cancel the effect of two other cells in the series. Therefore,the effective $EMF$ of the circuit is $E_{net} = (n - 2 - 2)E = (n - 4)E$. The total resistance of the circuit is $R_{total} = nr$. The current in the circuit is $I = \frac{E_{net}}{R_{total}} = \frac{(n - 4)E}{nr}$. Since cells $A$ and $B$ are connected with reversed polarity,they are being charged. The potential difference across a cell being charged is $V = E + Ir$. Substituting the values: $V = E + \left( \frac{(n - 4)E}{nr} \right)r = E + \frac{(n - 4)E}{n} = E \left( 1 + \frac{n - 4}{n} \right) = E \left( \frac{n + n - 4}{n} \right) = E \left( \frac{2n - 4}{n} \right) = 2E \left( \frac{n - 2}{n} \right) = 2E \left( 1 - \frac{2}{n} \right)$.

$n$ identical cells are joined in series,with two cells $A$ and $B$ in the loop having reversed polarities. The $EMF$ of each cell is $E$ and the internal resistance is $r$. The potential difference across cell $A$ or $B$ is (where $n > 4$):

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