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# 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√r1r2r1−r2r1+r2r1+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+iR−E+ir2=0⇒i=2E(R+r1+r2) Potential across first cell, VAB=ir1−E=2Er1(R+r1+r2)−EFor VAB=0, we get: 2Er1(R+r1+r2)−E=0⇒ 2r1(R+r1+r2)=1 ⇒2r1=R+r1+r2 ⇒R=r1−r2

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