Answer to Question #211043 in Physics for Festus

Question

Two batteries A and B each of emf 2 volts are connected in series to external resistance R = 1ohms. Internal resistance of A is 1.9 ohms and that of B is 0.9 ohms, what is the potential difference between the terminals of battery A

Answer 1

Find the current:

$$I=\frac{V_A+V_B}{r_A+r_B+R}.$$

Find the voltage drop across the terminals of A:

$$V_A=Ir_A=\frac{r_A(V_A+V_B)}{r_A+r_B+R}=2\text{V}.$$