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(C) Let the equivalent resistance between points $A$ and $B$ be $R$. Since the circuit is infinitely long,the equivalent resistance of the remaining c

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$(\sqrt{3} + 1) \ \Omega$

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(C) Let the equivalent resistance between points $A$ and $B$ be $R$. Since the circuit is infinitely long,the equivalent resistance of the remaining circuit to the right of the first section (between points $C$ and $D$) will also be $R$.The circuit now consists of two $1 \ \Omega$ resistors in series with the parallel combination of a $1 \ \Omega$ resistor and the equivalent resistance $R$.The equivalent resistance $R$ is given by:$R = 1 + 1 + \frac{1 	imes R}{1 + R}$$R = 2 + \frac{R}{1 + R}$$R(1 + R) = 2(1 + R) + R$$R + R^2 = 2 + 2R + R$$R^2 - 2R - 2 = 0$Using the quadratic formula $R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$R = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)}$$R = \frac{2 \pm \sqrt{4 + 8}}{2}$$R = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}$Since resistance cannot be negative,we take the positive root:$R = (1 + \sqrt{3}) \ \Omega$.

The resistance between the terminal points $A$ and $B$ of the given infinitely long circuit will be

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