12 cells each having same emf are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and

Q: $12$ cells each having same $emf$ are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and two cells which are in series. Current is $3 \,A$ when cells and battery aid each other and is $2\, A$ when cells and battery oppose each other. The number of cells wrongly connected is

b (b) Let $n$ be the number of wrongly connected cells. Number of cells helping one another $ = (12 - n)$ Total $e.m.f.$ of such cells $ = (12 - n)E$ Total $e.m.f.$ of cells opposing = $nE$ Resultant $e.m.f.$ of battery $ = (12 - n)E - nE$$ = (12 - 2n)E$ Total resistance of cells = $12r$ ( resistance remains same irrespective of connections of cells) With additional cells $(a)$ Total $e.m.f.$ of cells when additional cells help battery =$ (12 -2n) E + 2E$ Total resistance = $12r + 2r = 14r$ $\frac{{(12 - 2n)E + 2E}}{{14r}} = 3$ ......$(i)$ $(b)$ Similarly when additional cells oppose the battery $\frac{{(12 - 2n)E - 2E}}{{14r}} = 2$......$(ii)$ Solving $(i)$ and $(ii)$, $n = 1$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

$12$ cells each having same $emf$ are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and two cells which are in series. Current is $3 \,A$ when cells and battery aid each other and is $2\, A$ when cells and battery oppose each other. The number of cells wrongly connected is

A$4$
B$1$
C$3$
D$2$

Diffcult

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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