Two cells, having the same emf, are connected in series through an external resistance `R`. Cells have internal resistance `r_(1)` and `r_(2) (r_(1) gt r_(2))` respectively. When the circuit is closed, the potentail difference across the first cell is zero the value of `R` is
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Refer Fig, Current in circuit, `I=(E+E)/(r_(1)+r_(2)+R) = (2E)/(r_(1)+r_(2)+R)` As per question, let for the given value of `R`, terminal potential difference across first cell be zero, i.e., `E-Ir_(1) =0` or `E = Ir_(1) = (2E)/((r_(1)+r_(2)+R)) . r_(1)` or `I=(2r_(1))/((r_(1)+r_(2)+R))` or `r_(1)+r_(2)+R = 2r_(1)` or `R = r_(1) - r_(2)`
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