Can you assist with solving an infinite resistor network problem?

AI Thread Summary
The discussion focuses on solving an infinite resistor network problem involving resistances R_0 and R_1. The key point is to demonstrate that the equivalent resistance, R_eq, can be expressed as R_eq = R_1 + √(R_1² + 2R_1R_0). The method involves visualizing the infinite network as a single resistor and adding three resistors back to maintain the equivalent resistance. After calculations, it is concluded that the negative solution for R_eq is invalid, confirming the positive expression for equivalent resistance. The final result is established as R_eq = R_1 + √(R_1² + 2R_1R_0).
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Could someone be so kind as to assist me in solving the following problem? I have had some basic knowledge of how the effect of multiple resistors add up in parallel in and in series, but I'm pretty much lost on how to solve this problem.

Consider an infinite network of resistors of resistances R_0 and R_1 shown in the figure.

Show that the equivalent resistance of this network is

R_{eq} = R_1 + \sqrt{{R_1}^2 + 2R_1R_0}
 
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Since the network is infinite, if you extend the network by one more copy of the pattern, the equivalent resistance shouldn't change. So represent the network's resistance with a single resistor with Req resistance. Now, can you add 3 resistors to the Req resistor to make it back into the infinite network? The resistance of this new network is easily calculated and it should have resistance Req.

For an alternative view, replace all resistors to the right of the leftmost 3 (top, bottom, and middle) as a single resistor with resistance Req. Do you see why yuo can do that? Now, calculate the resistance of this new network in terms of Req, R0, and R1. Shouldn't the resistance of this network be equal to Req? Do you understand why this is also true? Use this to set up and equation and solve for Req.

--J
 
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Thank you. I do see it now.

Through the steps you outlined, I managed to derive the following:

R_{eq} = R_1 \pm \sqrt {{R_1}^2 + 2R_1R_0}

However, since \sqrt {R_1(R_1 + 2R_0)} \geq R_1, this means that R_{eq} = R_1 - \sqrt {{R_1}^2 + 2R_1R_0} is not valid and must be rejected.

Hence, R_{eq} = R_1 + \sqrt {{R_1}^2 + 2R_1R_0}
 
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