Source: High school physics (Chinese)
Problem
Two identical batteries are connected in series with a load resistor $R$, forming a single loop. Each battery has EMF $\varepsilon = 2 \text{ V}$ and internal resistance $r = 0.1 \text{ Ω}$. The load resistance is $R = 4.8 \text{ Ω}$. Point $a$ is at the negative terminal of the first battery, and point $b$ is at the junction between the two batteries (the positive terminal of the first battery).
What is the potential difference $U_a - U_b$ between points $a$ and $b$?
A. $1.92 \text{ V}$
B. $2.0 \text{ V}$
C. $-2.0 \text{ V}$
D. $-1.92 \text{ V}$
Total EMF in the loop: $\varepsilon_{\text{tot}} = 2\varepsilon = 4 \text{ V}$.
Total resistance: $R_{\text{tot}} = R + 2r = 4.8 + 0.2 = 5.0 \text{ Ω}$.
Loop current: $I = \dfrac{\varepsilon_{\text{tot}}}{R_{\text{tot}}} = \dfrac{4}{5} = 0.8 \text{ A}$.
The terminal voltage of the first battery (during discharge) equals its EMF minus the internal $Ir$ drop:
$$U_b - U_a = \varepsilon - I r = 2 - 0.8 \times 0.1 = 1.92 \text{ V}$$Therefore $U_a - U_b = -1.92 \text{ V}$. Answer: D.