The equivalent resistance of the infinite net | vedclass

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(C) Let the equivalent resistance of the infinite network be $x$.Since the network is infinite,adding one more section to the front does not change th

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

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(C) Let the equivalent resistance of the infinite network be $x$.Since the network is infinite,adding one more section to the front does not change the total resistance. Thus,the network can be represented as two $1\,\Omega$ resistors in series with the parallel combination of a $1\,\Omega$ resistor and the equivalent resistance $x$.The equivalent resistance $x$ is given by:$x = 1 + \left( \frac{1 	imes x}{1 + x} ight) + 1$$x = 2 + \frac{x}{1 + x}$$x - 2 = \frac{x}{1 + x}$$(x - 2)(x + 1) = x$$x^2 + x - 2x - 2 = x$$x^2 - 2x - 2 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)}$$x = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}$Since resistance cannot be negative,we take the positive value:$x = (1 + \sqrt{3})\,\Omega$

The equivalent resistance of the infinite network given below is:

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