let the internal resistance of the cell = r, therefore E = I (R+r) (i) E = 0.4(4+r), (ii) E = 0.2 (10+r) therefore 0.4 (4+r)= 0.2 (10+r) 1.6 + 0.4r = \(2.0 \times 0.2\text{r}\) 0.4r - 0.2r = 2.0 - 1.6 therefore 0.2r = 0.4 => r = \(\frac{0.4}{0.2}\) = 2Ω therefore r = 2Ω