Under what condition, current passing through the resistance R can be increased by short circuiting the battery of emf E₂? The internal resistances of the two batteries are r₁ and r₂respectively

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

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

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

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

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

answer is B.

(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

Initially $I_{1} = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}} .$

After short circuiting the second cell, $I_{2} = \frac{E_{1}}{R + r_{1}}$

The condition required is $I_{2} > I_{1} \cdot$

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

$E_{1} R + E_{1} r_{1} + E_{1} r_{2} > E_{1} R + E_{1} r_{1} + E_{2} R + E_{2} r_{1}$

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

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE