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.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{E_{2}}{E_{1}}$ 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.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE