Two batteries A and B each of emf 2 V are connected in series to an external resistance R = 1 Ω. If the internal resistance of battery A is 1.9 Ω and that B is 0.9 Ω. What is the potential difference between the terminals of battery ?
Answer & explanation
Correct answer: option 3
i = $\frac{2+2}{1+1.9+0.9}=\frac{4}{3.8}$A
For cell A,
E = V + ir
V = E - ir = $2 - \frac{4}{3.8} \times 1.9$
V = 0 (zero)