A wire of resistance 9 Omega is bent to form | vedclass

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(C) The total resistance of the wire is $9 \ \Omega$. Since it is bent into an equilateral triangle,the wire is divided into three equal parts,each ha

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$2$

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(C) The total resistance of the wire is $9 \ \Omega$. Since it is bent into an equilateral triangle,the wire is divided into three equal parts,each having a resistance of $R' = 9 \ \Omega / 3 = 3 \ \Omega$.When we consider the equivalent resistance across any two vertices (say $B$ and $C$),the resistor between $B$ and $C$ is in parallel with the series combination of the other two resistors (between $A-B$ and $A-C$).The resistance of the branch $AB$ and $AC$ in series is $R_{series} = 3 \ \Omega + 3 \ \Omega = 6 \ \Omega$.Now,this $6 \ \Omega$ resistance is in parallel with the $3 \ \Omega$ resistor connected directly between $B$ and $C$.The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2}$.Therefore,$R_{eq} = 2 \ \Omega$.

$A$ wire of resistance $9 \ \Omega$ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be . . . . . . ohm.

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