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 

2 V

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.

∴ $\mathrm{i}=\frac{V}{R_{eq}}$

Equivalent resistance of circuit, $R_{\text {eq }}=3+2=5 \Omega \Rightarrow \mathrm{i}=\frac{10}{5}=2 A$

So, current in each branch $=1 A$

Now, $V_{C}-V_{A}=1 \times 1=1 V$                  (i)

also, $V_{C}-V_{B}=1 \times 3=3 V$                  (ii)

Solving eqs. (i) and (ii), we have

$V_{A}-V_{B}=3-1=2 V$