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(A) When three resistors of $2\, \Omega$ each are connected in a triangle,let us find the equivalent resistance between any two vertices,say $P$ and $

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$4/3\, \Omega$

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(A) When three resistors of $2\, \Omega$ each are connected in a triangle,let us find the equivalent resistance between any two vertices,say $P$ and $Q$.One resistor is directly connected between $P$ and $Q$.The other two resistors are connected in series with each other,and this combination is in parallel with the first resistor.Resistance of the series branch $= 2\, \Omega + 2\, \Omega = 4\, \Omega$.Now,this $4\, \Omega$ resistor is in parallel with the $2\, \Omega$ resistor connected directly between $P$ and $Q$.The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{4} = \frac{2+1}{4} = \frac{3}{4}$.Therefore,$R_{eq} = \frac{4}{3}\, \Omega$.

Three resistors each of $2\, \Omega$ are connected together in a triangular shape. The resistance between any two vertices will be

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