Q.Two cells each of emf e but internal resistances r1 and r2 are connected in series through an external resistance R. If the potential difference across the first cell is zero while current flows, the relation of R in terms of r1 and r2 is

Writing the EMF equation for both cases and solving the two equations simultaneously gives r = 1.0 Ω.

Let the battery have EMF $E$ and internal resistance $r$.

Case 1: $I_1 = 3\ A$ through $R_1 = 3\ \Omega$:

$E = I_1(R_1 + r) = 3(3 + r) = 9 + 3r \quad \text{...(i)}$

Case 2: $I_2 = 0.75\ A$ through $R_2 = 15\ \Omega$: …

Setting the potential difference across the first cell to zero and using the series-circuit current gives R = r₁ − r₂.

Two cells, each of EMF e, with internal resistances r1 and r2, connected in series with external resistance R. Current in the circuit: I = 2e/(R + r1 + r2).

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