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(A) The wire of resistance $9 \, \Omega$ is cut into $3$ equal parts,so each part has a resistance of $3 \, \Omega$. These are connected to form an eq

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(A) The wire of resistance $9 \, \Omega$ is cut into $3$ equal parts,so each part has a resistance of $3 \, \Omega$. These are connected to form an equilateral triangle $ABC$,where $R_{AB} = 3 \, \Omega$,$R_{AC} = 3 \, \Omega$,and $R_{BC} = 3 \, \Omega$.When a $2 \, V$ cell is connected across $B$ and $C$,the branch $BC$ (resistance $3 \, \Omega$) is in parallel with the series combination of branches $AB$ and $AC$ (total resistance $3 + 3 = 6 \, \Omega$).The equivalent resistance $R_{eq}$ of the circuit is given by $\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$,so $R_{eq} = 2 \, \Omega$.The total current from the cell is $I = \frac{V}{R_{eq}} = \frac{2}{2} = 1 \, A$.Using the current divider rule,the current $i_1$ flowing through the branch $AB$ is $i_1 = I 	imes \left( \frac{R_{BC}}{R_{AB} + R_{AC} + R_{BC}} ight)$ is incorrect; rather,$i_1 = I 	imes \left( \frac{3}{3 + 6} ight) = 1 	imes \frac{3}{9} = \frac{1}{3} \, A$.The potential difference across $AB$ is $V_{AB} = i_1 	imes R_{AB} = \frac{1}{3} 	imes 3 = 1 \, V$.

$A$ uniform wire of resistance $9 \, \Omega$ is cut into $3$ equal parts. They are connected in the form of an equilateral triangle $ABC$. $A$ cell of $e.m.f.$ $2 \, V$ and negligible internal resistance is connected across $B$ and $C$. The potential difference across $AB$ is ............... $V$.

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