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(C) Let the equivalent resistance of the infinite network be $R_{eq}$. Since the network is infinite,adding one more section of $1 \,\Omega$ resistors

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the equivalent resistance of the infinite network be $R_{eq}$. Since the network is infinite,adding one more section of $1 \,\Omega$ resistors will not change the equivalent resistance. The circuit can be viewed as a series combination of a $1 \,\Omega$ resistor and a parallel combination of a $1 \,\Omega$ resistor and the equivalent resistance $R_{eq}$. Thus,$R_{eq} = 1 + \frac{1 \cdot R_{eq}}{1 + R_{eq}}$. Multiplying by $(1 + R_{eq})$,we get: $R_{eq}(1 + R_{eq}) = (1 + R_{eq}) + R_{eq}$ $R_{eq} + R_{eq}^2 = 1 + 2R_{eq}$ $R_{eq}^2 - R_{eq} - 1 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get: $R_{eq} = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}$. Since resistance cannot be negative,we take the positive root: $R_{eq} = \frac{1 + \sqrt{5}}{2} \,\Omega$.

What is the equivalent resistance between $A$ and $B$ in the given infinite network?

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