a
The circuit can be drawn as follows
Equivalent resistance $R = \frac{{3 \times (3 + 3)}}{{3 + (3 + 3)}} = 2\,\Omega $
Current $i = \frac{2}{2} = 1\,A.$ So,${i_1} = 1 \times \left( {\frac{3}{{3 + 6}}} \right) = \frac{1}{3}\,A$.
Potential difference between $A$ and $B =\frac{1}{3} \times 3 = 1\,volt.$
