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3. current electricity — Physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 SciencePhysics3. current electricity1 Mark
MCQ
$12$ 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
    $4$
  • ✓
    $1$
  • C
    $3$
  • D
    $2$
✓

Answer

Correct option: B.
$1$
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 $(ii)$, $n = 1$

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