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

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(B) Let $n$ be the number of wrongly connected cells.Number of cells helping = $(12 - n)$.Number of cells opposing = $n$.Resultant $emf$ of the $12$ c

Vedclass · Accepted Answer

$1$

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(B) Let $n$ be the number of wrongly connected cells.Number of cells helping = $(12 - n)$.Number of cells opposing = $n$.Resultant $emf$ of the $12$ cells = $(12 - n)E - nE = (12 - 2n)E$.Total resistance of the $12$ cells = $12r$.When the two additional cells aid the battery,the total $emf$ is $(12 - 2n)E + 2E = (14 - 2n)E$ and total resistance is $14r$.Given current $I_1 = 3 \, A$,so $\frac{(14 - 2n)E}{14r} = 3$  --- $(i)$.When the two additional cells oppose the battery,the total $emf$ is $(12 - 2n)E - 2E = (10 - 2n)E$ and total resistance is $14r$.Given current $I_2 = 2 \, A$,so $\frac{(10 - 2n)E}{14r} = 2$  --- $(ii)$.Dividing $(i)$ by $(ii)$:$\frac{14 - 2n}{10 - 2n} = \frac{3}{2}$$28 - 4n = 30 - 6n$$2n = 2 \implies n = 1$.Thus,the number of wrongly connected cells is $1$.

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