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(A) Let the equivalent resistance between points $A$ and $B$ be $r$. Since the circuit is an infinite ladder,adding one more section to the input does

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

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(A) Let the equivalent resistance between points $A$ and $B$ be $r$. Since the circuit is an infinite ladder,adding one more section to the input does not change the total resistance. Thus,the circuit can be represented as two resistors of resistance $R$ in series with the parallel combination of a resistor $R$ and the equivalent resistance $r$.The equivalent resistance $r$ is given by:$r = R + R + \frac{R \cdot r}{R + r}$$r = 2R + \frac{Rr}{R + r}$Multiplying by $(R + r)$:$r(R + r) = 2R(R + r) + Rr$$Rr + r^2 = 2R^2 + 2Rr + Rr$$r^2 - 2Rr - 2R^2 = 0$Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$r = \frac{2R \pm \sqrt{(-2R)^2 - 4(1)(-2R^2)}}{2(1)}$$r = \frac{2R \pm \sqrt{4R^2 + 8R^2}}{2}$$r = \frac{2R \pm \sqrt{12R^2}}{2}$$r = \frac{2R \pm 2R\sqrt{3}}{2}$$r = R \pm R\sqrt{3}$Since resistance must be positive,we take the positive root:$r = (1 + \sqrt{3})R$

The resistance between points $A$ and $B$ is

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