Two infinite circuits (shown below) are calle | vedclass

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(D) For Circuit $A$: The infinite ladder can be simplified by replacing the repeating part with $x$. The equivalent resistance is $x = \frac{R(2R + x)

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All of the above

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(D) For Circuit $A$: The infinite ladder can be simplified by replacing the repeating part with $x$. The equivalent resistance is $x = \frac{R(2R + x)}{R + (2R + x)} = \frac{R(2R + x)}{3R + x}$.This simplifies to $x(3R + x) = 2R^2 + Rx$,which gives $x^2 + 2Rx - 2R^2 = 0$.Using the quadratic formula,$x = \frac{-2R + \sqrt{4R^2 - 4(1)(-2R^2)}}{2} = \frac{-2R + \sqrt{12R^2}}{2} = (\sqrt{3} - 1)R$.For Circuit $B$: The infinite ladder can be simplified by replacing the repeating part with $y$. The equivalent resistance is $y = 2R + \frac{Ry}{R + y} = \frac{2R^2 + 2Ry + Ry}{R + y} = \frac{2R^2 + 3Ry}{R + y}$.This simplifies to $y(R + y) = 2R^2 + 3Ry$,which gives $y^2 - 2Ry - 2R^2 = 0$.Using the quadratic formula,$y = \frac{2R + \sqrt{4R^2 - 4(1)(-2R^2)}}{2} = \frac{2R + \sqrt{12R^2}}{2} = (\sqrt{3} + 1)R$.Comparing the results: $x = (\sqrt{3} - 1)R \approx 0.732R$ and $y = (\sqrt{3} + 1)R \approx 2.732R$.Thus,$y > x$ is true,$y = (\sqrt{3} + 1)R$ is true,and $xy = (\sqrt{3} - 1)R \cdot (\sqrt{3} + 1)R = (3 - 1)R^2 = 2R^2$ is also true.Therefore,all of the above are correct.

Two infinite circuits (shown below) are called 'Circuit $A$' and 'Circuit $B$'. The equivalent resistance of 'Circuit $A$' is $x$ and that of 'Circuit $B$' is $y$ between terminals $1$ and $2$.

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