It is possible if E < r/(r + R). If one cell of e.m.f. E₁ and internal resistance r₁ be connected to an external resistance R, then the current passing through the circuit
I₁ = E₁/(r₁ + R).
If a second cell of e.m.f E₂ and internal resistance r₂ is then added in series, the current
I₂ = (E₁ + E₂)/(r₁ + r₂ + R).
Clearly I₂ < I₁ if (E₁ + E₂)/(r₁ + r₂ + R) < E₁/(r₁ + R), hence
E₂/E₁ < r₂/(r₁ + R).
In other words for this inequality to obtain, the internal resistance of the second cell must be sufficiently great.