A uniform wire of resistance $9$ $\Omega$ is cut into $3$ equal parts. They are connected in the form of equilateral triangle $ABC$. A cell of $e.m.f.$ $2\,V$ and negligible internal resistance is connected across $B$ and $C$. Potential difference across $AB$ is ............... $V$
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 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 $B =\frac{1}{3} \times 3 = 1\,volt.$

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