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Equivalent resistor of a simple circuit

  • #1
Hi. Anyone knows how to calculate the Req of this circuit? n tends to infinite.

q1.png
 

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  • #2
berkeman
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Is this question for schoolwork?

And are the numbers by the resistors the resistance value, or are they just the number of the resistor and all resistor values are equal?
 
  • #3
Is this question for schoolwork?

And are the numbers by the resistors the resistance value, or are they just the number of the resistor and all resistor values are equal?
Kinda.


The numbers are based on ohm, so if number 1 in the first resistor means a 1 ohm-resistor.
 
  • #4
berkeman
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Kinda.
Okay, I've moved your thread to the schoolwork forums where all schoolwork-type questions belong. :smile:

And part of the rules for schoolwork is that you show your attempt at a solution before we can offer tutorial help. Can you show us how you would solve this for some finite cases? And then say how you would extend it to the infinite case in the problem?
 
  • #5
And part of the rules for schoolwork is that you show your attempt at a solution before we can offer tutorial help. Can you show us how you would solve this for some finite cases?
Yea, I tried to find a pattern with n=1 (the first block), n=2 (first and second blocks), n=3 and n=4, but it didn't have any kinds of patterns/valid sequences.


And then say how you would extend it to the infinite case in the problem?
Well, once n tends to infinite, the resistance value is so high that can be considered as an open circuit.
The thing is, it's possible to calculate the Req by checking a few Ns and you can see that the value of Req would almost remain the same, but it's not a valid mathematical solution.
 
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  • #6
berkeman
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I think I see a pattern that may help to solve the problem.

Have you studied infinite series in your class work so far? It might make the final solution a little easier to calculate, but you also may not need it. I'm not sure yet.

Hint about the pattern -- When you look at one of the vertical resistors in the middle of that ladder, what do the 3 resistors just to the right of that resistor total up to compared to that vertical resistor? And can you use that fact to help express the total resistance as a series? :smile:
 
  • #7
Hint about the pattern -- When you look at one of the vertical resistors in the middle of that ladder, what do the 3 resistors just to the right of that resistor total up to compared to that vertical resistor?
I guess I got the idea of your hint, but I'm not sure if it's correct. So if we only consider the resistance value of the first block (n=0 only) as R, in n=1 circuit (first and second blocks only), we have 2R parallel with the first block (R), so we get something like
ans1.png

And Req = 1/R + 1/2R + 1/4R + ...
 
  • #8
I guess I got the idea of your hint, but I'm not sure if it's correct. So if we only consider the resistance value of the first block (n=0 only) as R, in n=1 circuit (first and second blocks only), we have 2R parallel with the first block (R), so we get something like
View attachment 261270
And Req = 1/R + 1/2R + 1/4R + ...
Hmmm.. Seems to be incorrect.
 
  • #9
berkeman
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we have 2R parallel with the first block (R),
No, I don't think that helps as much as referencing the next stage to the value of just the previous vertical resistor. What is the multiplier then? :smile:
 
  • #10
What is the multiplier then? :smile:
*Shrugs*
I tried to solve it in another way, but I realized it was also incorrect, I don't know..
 
  • #11
berkeman
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*Shrugs*
I tried to solve it in another way, but I realized it was also incorrect, I don't know..
Can you show what the series looks like if you use my hint? What is the resistance of each next stage that is in parallel with each vertical resistor...?
 
  • #12
berkeman
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The thing is, it's possible to calculate the Req by checking a few Ns and you can see that the value of Req would almost remain the same, but it's not a valid mathematical solution.
Can you say what you got for the approximate final resistance with that method? I'd like to check the answer that I got with an infinite series... :smile:
 
  • #13
Can you show what the series looks like if you use my hint? What is the resistance of each next stage that is in parallel with each vertical resistor...?
Well, I see a n*2.4 (n*12/5) in my calculations..
 
  • #14
Can you say what you got for the approximate final resistance with that method?
I don't know. Something like ~5.4 (or less)??
 
  • #16
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All correct answers are equivalent (!) but I converted the network into itself (2x the network attached to the three leftmost resisters is the network again) to get a quadratic equation for the network R value. No explicit series need be evaluated. Whatever works!!
 
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  • #17
but I converted the network into itself (2x the network attached to the three leftmost resisters is the network again)
Would you explain this part please? Why the other part is 2x the first block network?
 
  • #18
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It is my favorite trick (really my only trick for series so I use it whenever)
Let the unknown value of the network resistance be R.
II multiply every value in the network by 2 then the value of that network is obviously 2R.
But look at that network.....if connect the 1,3 and 2 ohm resistors to the front end I again reproduce the original network. So I know that R=1ohm+2ohm + (3ohm in parallel with 2R)
Gives me a quadratic for R (pardon my Latex-lazy presentation).
 
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  • #19
Let the unknown value of the network resistance be R.
II multiply every value in the network by 2 then the value of that network is obviously 2R.
But look at that network.....if connect the 1,3 and 2 ohm resistors to the front end I again reproduce the original network. So I know that R=1ohm+2ohm + (3ohm in parallel with 2R)
Ohhhhhhh, yea.
Thanks. I finally realized.
Also thanks Berkeman for helping me out.

This can be closed I guess.
 
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  • #20
berkeman
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But I, but I, I have three pages filled with work! I'm sooo close to the solution! Just give me another 12 hours. That's all I ask! Pleeezzzee! o0)

Excellent trick by @hutchphd -- I think there is a puzzle on my math calendar from last month much like this problem. When I finally get to go back to work, I should be able to solve it finally!

Thanks guys. Fun stuff. :smile:
 
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  • #21
berkeman
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Ohhhhhhh, yea.
Thanks. I finally realized.
So when you amaze your friends and instructor with your solution, be sure to give attribution to this PF thread and, well, you know, that PhD guy... :wink:
 
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  • #22
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And learn it well......it is like free money when it works.!!!! I had to show it off.....
 
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  • #23
Tom.G
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R=1ohm+2ohm + (3ohm in parallel with 2R)
Gives me a quadratic for R (pardon my Latex-lazy presentation).
I'll bite. Could you give the quadratic even if not in LaTex.?
(I seem to be slow tonite. :nb) )

Tnx,
Tom
 
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  • #24
cnh1995
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Brilliant trick @hutchphd!

While I knew this trick for the resistor ladder with equal resistances throughout, I couldn't extend that logic to this problem.

Looks like the circuit is not as "simple" as the title says :wink: .


I'll bite. Could you give the quadratic even if not in LaTex.?
2R2-9R-9=0.
 
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  • #25
While I knew this trick for the resistor ladder with equal resistances throughout, I couldn't extend that logic to this problem.
Yea same. :)

Looks like the circuit is not as "simple" as the title says :wink: .
Well, by "simple" I meant it doesn't have capacitor and/or inductor.
 
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