Two sources of current of equal emf are connected in series and have different internal resistances $r_{1}$ and $r_{2} (r_{2} > r_{1})$ Find the external resistance $R$ at which the potential difference across the terminals of one of the sources becomes equal to zero.

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

$R = r_{2} + r_{1}$

$R = r_{2} - r_{1}$

$R = r_{1} + r_{2}$

$R = r_{1} - r_{2}$

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

$V = E - i r$

$E$ and $i$ for both the sources are equal. Therefore, potential difference ( $V$ will be zero for a source having greater internal resistance, i.e. $r_{z}$

$∴$ $0 = E - i r_{2}$

or $E = i r_{2} = (\frac{2 E}{R + r_{1} + r_{2}}) \cdot r_{2}$

$2 r_{2} = R + r_{1} + r_{2}$

$R = r_{2} - r_{1}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!