Two cells of the same emf e but difference internal resistances r1 and r2 are connected in series with an external resistance R. The potential drop across the first cell is found to be zero. The external resistance R is
A
r1−r2
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B
r1/r2
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C
r1r2
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D
r1+r2
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Solution
The correct option is Ar1−r2 i=2er1+r2+R
Potential across first cell =ΔV=e−ir1=e−(2er1+r2+R)=e[r1+r2+R−2r1r1+r2+R]=e[r2+R−r1r1+r2+R]