A battery of emf $10 V$ is connected to a group of resistances as shown in figure. The potential difference $V_A - V_B$ between the points A and B is

Resistance between upper branch and lower branch in parallel part is same, so, equal amount of current flows through them. Let main current is i.

$i=\frac{V}{R_{eq}}$

Equivalent resistance of circuit,

$$\begin{gathered} R_{eq}=3+2 \\ =5\Omega \\ \Rightarrow i=\frac{10}{2} \\ i= 2 A \end{gathered}$$

So, current in each branch $=1 A$

Now, $$\begin{gathered} V_C-V_A=1\times 1 \\ =1 V ..... \left(i\right) \end{gathered}$$
Also  $$\begin{gathered} V_C-V_B=1\times 3 \\ =3 V ..... \left(ii\right) \end{gathered}$$
Solving Equations  (i) and (ii), we have
$$\begin{gathered} V_A -V_B \\ = 3 -1 \\ = 2 V \end{gathered}$$