When two cells of different emf’s are connected in series to an external resistance the current is 5A. When the poles of one cell are interchanged, the current is 3A. The ratio of emf’s of two cell is

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2 : 1

5 : 1

1 : 3

4 : 1

answer is D.

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Detailed Solution

Let the emfs be $E_{1} a n d E_{2}$ and internal resistances be $r_{1} a n d r_{2}$

External resistance be R.

Case 1 :

Current in circuit is

$i = \frac{E_{1} + E_{2}}{r_{1} + r_{2} + R} = 5 A$

Case 2 :

Current in circuit is

$i = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} = 3 A$

Dividing above equations gives

$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{5}{3}$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{4}{1}$

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