Thinking Recursively: How to Crack the Infinite Resistor Ladder Puzzle!
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- Опубликовано: 4 ноя 2021
- How do you find the total resistance of an infinite resistor ladder? The key is to think recursively! Get the notes for free here: courses.physicswithelliot.com...
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About the classic physics problem series:
In these intro-to-intermediate-level physics videos, I'll discuss classic physics "challenge" problems that you might meet in your introductory mechanics and electromagnetism classes. They might be based on simple concepts, but these problems can still get pretty tough!
About me:
I’m Dr. Elliot Schneider. I love physics, and I want to help others learn (and learn to love) physics, too. Whether you’re a beginner just starting out with your physics studies, a more advanced student, or a lifelong learner, I hope you’ll find resources here that enable you to deepen your understanding of the laws of nature. For more cool physics stuff, visit me at www.physicswithelliot.com. 
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So cool to see that turn into such a simple number at the end.
In a week, this has become my favorite RUclips channel.
Thanks Rob!!
A great set of physics lectures on your channel! would love to see some Electromagnetics lectures
Thanks. My new favorite channel.
thanks for the simple explanation!!
Amazing, I like it!!
Beautiful explanation this problem haunted me as an undergraduate!
Hi Elliot! Great walkthrogh of the puzzle, I'm trying to go through the last step of your notes, how do did you get the recursion relation for finite blocks? Cheers!
Brilliant. Simply great. How about QM,Electrodynamics & GTR lectures ?
Thanks Shishir!
Thank you ❤
When I saw the first two values were 3 and 11/4 I was hoping it would converge to e but hey, 1+sqrt(3) is cool too :D
I was thinking the same thing because of it being similar to compounding interest, but it's even wilder that it's NOT e yet only off by like 0.5%
Great problem! The key insight you used to rewrite the recurrence relation in terms of its infinite limit, which allowed you to solve it algebraically, is super cool. Is there a general procedure to solve this kind of non-linear recurrence relation without taking the infinite limit?
I think there are a number of strategies that can be applied to solving recurrence relations like this, but I don't think there is an algorithm that will always work
Hello! Thanks for posting such an incredible video! Your explanation is great and I have followed every part except the last one when you establish the quadratic equation. Could you explain how did you get from the inductive formula to the quadratic equation? Thank you so much!
Here is a possibily interesting question can you access (for any arbitrary number n of resistors) the current passing by each upper resistor, I said intersting because i personally like the method of solving this problem
The final resukt can ne used in any value of resitance? Be it a 5R or something
How did you get the simultaneous eqn
Question: If each parallel pathway had only one resistor (an infinite number of resistors in parallel with none in series), would the infinite ladder lead to a total effective resistance of 0.73R? Or would the total effective resistance go to zero? TIA
Meaning you want to erase al the horizontal resistors? Then the recursion relation is R_N = (1/R + 1/R_{N-1})^{-1}, and the solution goes to zero as N goes to infinity
What the f how do u have such small subscribers you deserve more then a million
He's just started 3 months ago, though mark my words he'll blow up in a year.
If R=1.. that infinite ladder of 1 ohm ans=?
Sir finally that is (1+squroot(0.38... ))R then only we get if R=1. As Requ=1.618 sir.... whether this is rgt r wrg?
You've shown that it's bounded but you didn't prove convergance. The solution is only valid if the series converges.
The convergence doesn't need to be proven due to the physical sense. Think a bit and you will see it. Also, you can convert this speculation based on physical sense into mathematical proof. To do that, you can get away from formulas for effective resistance and get to the lower level, considering voltages and currents. Besides, look at my solution (somewhere at the top level of comments). It uses only the standard formulas plus the concept of self-similarity of infinite structures. So, the video solution (or explanation) could be better.
I did it quite differently, I assumed that the I is the current coming out from the battery and then
We can see that the current will be decided into to different current
Say nI and (n_1)I .then we can use kircchoff's law to put two equations and solving for n
I don't really remember nor understand what i was saying 😅.
I guess I was trying to use some kind of recursion.
Here:
Let Rt be the total resistance of the set so if we added another set of these three resistors the whole this is really the same cause it's still infinite
Now we have Rf ( of the set after addition of the extra set )
= Rt
Rf it is r+r+(1/Rt +1/r)^-1 = Rt
Now rearrange
We get R^2-2rR-r^2=0
R=(1+√3)r
Discard the -ve
And yay🎉
Oh! my old me used Kirchhoff's
@@tryingtobeintelligentYOU ARE WELCOME 🤗
It’s the golden ratio! Really Mind-blowing! Thank you very much I enjoyed this.
It isn't... The result here is (1+sqrt(3)) The Golden Ratio is (1+sqrt(5))/2.
@@milobem4458 So typical! Many people called almost everything like that "Golden Ratio" and later they called almost everything a "fractal". The thinking of many people is purely mythological and is driven by trends.
I was participating physics olympiad exams in my country and I faced the exact same question but unfortunately it didn't come to my mind after equating it to infinity
👍👍👍👍
This is just a very standard and basic question for jee students
This was a neet pyq as well
use thevenin theorem repeatedly.
I was so close but I dint thought that we add the two Rs and I got either zero or infinity 😂
No, it wasn't even close. 🙂Everything is much simpler. You will find it if you read about the solution I've demonstrated in my top-level comment. It comes from the first glance.
No, this solution (or explanation) is not good enough. I figured out the solution at the first glance. Look, take the entire thing for x. Using the fact we have self-similarity with the infinite circuit, detach one infinite part of the circuit. It goes to the right of the first vertical resistor. It is perfectly identical to the entire circuit, so, this is also x. Then we combine the first two horizontal Rs, the first vertical R, and attach a single resistor in parallel, with the effective resistance of x to it. It gives us exactly the same thing as your green equation. The problem is solved.
I never solved exactly this problem, but it is a typical problem for Olympic games in physics for middle/high schools, for pretty low levels.