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Dear Student,

Here i am considering the internal resistance as r.

Please find below the solution to the asked query:

Let n cells are connected wrongly, then the terminal voltage of the battery is,

$V=\left(12-2n\right)\epsilon -12r$

Given that the two more cells are connected in opposite direction, the current is 2A. Therefore,

$$\begin{gathered} V-2\epsilon -2Ir=0\Rightarrow \left(12-2n\right)\epsilon -12Ir-2\epsilon -2Ir=0 \\ \Rightarrow \left(10-2n\right)\epsilon -14\times 2\times r=0 \\ \Rightarrow \left(10-2n\right)\epsilon =28r\Rightarrow 10-2n=\frac{28r}{\epsilon } \\ \Rightarrow n=5-\frac{14r}{\epsilon } \\ \Rightarrow n=\frac{5\epsilon -14r}{\epsilon } \end{gathered}$$

If you know the value of internal resistance of the cell, then you can find the value of n from the above equation.

Hope this information will clear your doubts about the topic.

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