$12$ cells each having the same emf are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery is connected in series with an ammeter and two cells identical with each other and also identical with the previous cells. The current is $3\, A$ when the external cells aid this battery and is $2\,A$ when the cells oppose the battery. How many cells in the battery are wrongly connected :-
Diffcult
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Assume $\mathrm{M}$ cells are connected correct and $\mathrm{N}$ cells connected wrong.
$\mathrm{M}+\mathrm{N}=12$ ............$(1)$
$(M+2) E-N E=3 R \Rightarrow M-N+2=\frac{3 R}{E}$ ..........$(2)$
$\mathrm{ME}-(\mathrm{N}+2) \mathrm{E}=2 \mathrm{R} \quad \Rightarrow \mathrm{M}-\mathrm{N}-2=\frac{2 \mathrm{R}}{\mathrm{E}}$ ...........$(3)$
from eq $(1)$ and $(2)$
$-\mathrm{M}+\mathrm{N}+10=0 \quad \Rightarrow \mathrm{M}-\mathrm{N}=10$ .........$(4)$
$\mathrm{M}=11, \mathrm{N}=1$
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