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anonymous
 4 years ago
Could someone explain, stepbystep, how to find the resistance of an infinite ladder circuit with resistors?
anonymous
 4 years ago
Could someone explain, stepbystep, how to find the resistance of an infinite ladder circuit with resistors?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Something similar to this:

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3This should help: http://www.crbond.com/papers/ent23.pdf

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks, I've already been directed to that page before, however. I was wondering if you could explain what they wrote there: Z = R + R  Z

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3That's just a notational short cut for what comes next. If we added just one more resistor, and not in parallel, then the resistance would be Z + R where Z is the original resistance and R is the resistance of one resistor. If we added just one resister in parallel, the new resistance would be \[ \frac{RZ}{R+Z} \] Now, having added one more step in the ladder, the new resistance is \[ R + \frac{RZ}{R+Z} \] This must be equal to the original resistance; hence \[ Z = R + \frac{RZ}{R+Z} \ \ \ \ \ \  (*)\] The notation \[ Z = R + R  Z \] is just a short way of writing the equation (*)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, alright. It took me a bit, but I think I get it now. Thank you very much!

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3Sure. I haven't looked at this problem for a while and it's nice to see it again. I'm particularly happy to see the quantity \[ \frac{1 + \sqrt{5}}{2} \] in the answer. This is the Golden Ratio of mathematics and I'd forgotten that it turns up here as well.
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