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(C) Let the equivalent resistance of the infinite network be $R$. Since the network is infinite,adding or removing one repeating unit does not change

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$\frac{1 + \sqrt{5}}{2}\,\Omega$

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(C) Let the equivalent resistance of the infinite network be $R$. Since the network is infinite,adding or removing one repeating unit does not change the total resistance $R$.The circuit can be viewed as a $1\,\Omega$ resistor in series with the parallel combination of a $1\,\Omega$ resistor and the equivalent resistance $R$ of the rest of the infinite network.Thus,the equivalent resistance $R$ is given by:$R = 1 + \frac{1 	imes R}{1 + R}$Multiplying both sides by $(1 + R)$:$R(1 + R) = 1 + R + R$$R + R^2 = 1 + 2R$$R^2 - R - 1 = 0$Using the quadratic formula $R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$R = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}$$R = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$Since resistance cannot be negative,we take the positive root:$R = \frac{1 + \sqrt{5}}{2}\,\Omega$.

The equivalent resistance between points $A$ and $B$ of an infinite network of resistances each of $1\,\Omega$ connected as shown,is

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