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(D) The total number of cells is $n$. Two cells are connected with reversed polarity,so their $emf$ opposes the others. The effective $emf$ of the cir

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

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(D) The total number of cells is $n$. Two cells are connected with reversed polarity,so their $emf$ opposes the others. The effective $emf$ of the circuit is $E_{net} = (n - 2)E - 2E = (n - 4)E$.The total internal resistance of the circuit is $R_{total} = nr$.The current $i$ flowing through the circuit is given by $i = \frac{E_{net}}{R_{total}} = \frac{(n - 4)E}{nr}$.For a cell with reversed polarity,the potential difference $V$ across its terminals is given by $V = E + ir$ (since it is being charged).Substituting the value of $i$: $V = E + \left( \frac{(n - 4)E}{nr} \right)r = E + \frac{(n - 4)E}{n} = E + E - \frac{4E}{n} = 2E - \frac{4E}{n}$.Factoring out $2E$,we get $V = 2E(1 - \frac{2}{n})$.

$n$ identical cells are joined in series with two cells $A$ and $B$ 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: $(n > 4)$

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