12 cells,each having the same emf E and inter | vedclass

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(A) Let $M$ be the number of correctly connected cells and $N$ be the number of wrongly connected cells in the box of $12$ cells.Total cells $M + N =

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$1$

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(A) Let $M$ be the number of correctly connected cells and $N$ be the number of wrongly connected cells in the box of $12$ cells.Total cells $M + N = 12$  --- $(1)$Let $E$ be the emf and $r$ be the internal resistance of each cell.The net emf of the battery is $(M - N)E$ and the total internal resistance is $12r$.When two external cells aid the battery,the total emf is $(M - N)E + 2E = (M - N + 2)E$ and total resistance is $14r$.Given current $I_1 = 3 \, A$,so $(M - N + 2)E = 3 	imes 14r = 42r$ --- $(2)$When two external cells oppose the battery,the total emf is $(M - N)E - 2E = (M - N - 2)E$ and total resistance is $14r$.Given current $I_2 = 2 \, A$,so $(M - N - 2)E = 2 	imes 14r = 28r$ --- $(3)$Dividing $(2)$ by $(3)$:$\frac{M - N + 2}{M - N - 2} = \frac{42}{28} = \frac{3}{2}$$2(M - N + 2) = 3(M - N - 2)$$2(M - N) + 4 = 3(M - N) - 6$$M - N = 10$ --- $(4)$Adding $(1)$ and $(4)$:$2M = 22 \Rightarrow M = 11$Substituting in $(1)$:$11 + N = 12 \Rightarrow N = 1$Thus,$1$ cell is wrongly connected.

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