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The two batteries of emf E₁ and E₂ having internal resistances r₁ and r₂ respectively are connected in series to an external resistor R. Both the batteries are getting discharged. The above described combination of these two batteries has to produce a weaker current then when any one of the battery is connected to same resistor. For this requirement to be fulfilled.

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$\frac{E_{2}}{E_{1}}$ must not lie between $\frac{r_{2}}{r_{1} + R} and \frac{r_{1}}{r_{2} + R}$

$\frac{E_{2}}{E_{1}}$ must not lie between $\frac{r_{2}}{r_{1} + R} and \frac{r_{2} + R}{r_{1}}$

$\frac{E_{2}}{E_{1}}$ must lie between $\frac{r_{2}}{r_{1} + R} and \frac{r_{1}}{r_{2} + R}$

$\frac{E_{2}}{E_{1}}$ must lie between $\frac{r_{2}}{r_{1} + R} and \frac{r_{2} + R}{r_{1}}$

answer is A.

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Detailed Solution

When two cells are connected in series,

$I = \frac{(E_{1} + E_{2})}{(r_{1} + r_{2} + R)}$

When resistor is connected across battery of emf $E_{1},$

$I_{1} = \frac{E_{1}}{(r_{1} + R)}$

When resistor is connected across battery of emf $E_{2},$

$I_{2} = \frac{E_{2}}{(r_{2} + R)}$

For required condition to be fulfilled:

$I < I_{1} and I < I_{2}$

$\Rightarrow \frac{E_{2}}{E_{1}} < \frac{r_{2}}{(r_{1} + R)}$

and $\frac{E_{2}}{E_{1}} > \frac{(r_{2} + R)}{r_{1}}$

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