Q.A current of 3 A flows through a 3 Ω resistor when connected across a battery. The same battery supplies a current 0.75 A when connected across a 15 Ω resistor. The internal resistance of the battery 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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