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Question

Two cells with the same emf E and different internal resistances r1 and r2 are connected in series to an external resistance R. The value of R so that the potential difference across the first cell be zero, is
  1. r1r2
  2. r1r2
  3. r1+r2
  4. r1+r22

A
r1r2
B
r1+r2
C
r1r2
D
r1+r22
Solution
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The two cells have potential E each and respective internal resistance r1 and r2.
Let the current flowing in the circuit be i.

Using Kirchhoff's voltage law:
E+ir1+iRE+ir2=0
i=2E(R+r1+r2)

Potential across first cell, VAB=ir1E=2Er1(R+r1+r2)E

For VAB=0, we get: 2Er1(R+r1+r2)E=0

2r1(R+r1+r2)=1

2r1=R+r1+r2

R=r1r2

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