Step 1: Read the circuit. A source drives two parallel arms across the same two end points A's side and B's side. The upper arm has $4\,\Omega$ and $2\,\Omega$ in series; the lower arm has $1\,\Omega$ and $5\,\Omega$ in series.
Step 2: Idea of a voltage divider. In a series arm, the voltage splits between resistors in the ratio of their values. We can find the potential at the middle node of each arm.
Step 3: Potential at node A. With a $2$ V source, node A in the upper arm takes \[ V_{A} = 2\times\frac{2}{4+2} = \frac{4}{6} = \frac{2}{3} \text{ V} \] Step 4: Potential at node B. In the lower arm node B takes \[ V_{B} = 2\times\frac{5}{1+5} = \frac{10}{6} = \frac{5}{3} \text{ V} \] Step 5: Take the difference. \[ V_{AB} = \left|\frac{5}{3} - \frac{1}{3}\right| = \frac{4}{3} \text{ V} \] Step 6: State the answer. So the potential difference between A and B is $\dfrac{4}{3}$ V, which is option 3. \[ \boxed{V_{AB} = \frac{4}{3} \text{ V}} \]
