Two cells of emf 2 E and E with internal resistance r 1 and r 2 respectively are connected in series to an…

Two cells of emf 2E and E with internal resistance r1and r2 respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is

  1. r1+r2
  2. r12-r2
  3. r12+r2
  4. r1-r2

Solution

i=3ER+r1+r2

T.P.D.=2E-ir1=0

2E=ir1

2E=3E×r1R+r1+r2

2R+2r1+2r2=3r1

R=r12-r2

Asked in: JEE Main 2021 (17 Mar Shift 2)

Practice more Current Electricity questions on Aicharya