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  4. A uniform wire of resistance 9 Ω is cut into 3 equal parts. They are connected in the form of equilateral triangle A B C. A cell of e.mf 2 V and negligible internal resistance is connected across B and C. Potential difference across AB is

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Q. A uniform wire of resistance $9 \Omega$ is cut into $3$ equal parts. They are connected in the form of equilateral triangle $A B C$. A cell of e.mf $2 V$ and negligible internal resistance is connected across $B$ and $C$. Potential difference across $AB$ is

ManipalManipal 2018

Solution:

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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