Hi. Anyone knows how to calculate the Req of this circuit? n tends to infinite.
Kinda.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?
Okay, I've moved your thread to the schoolwork forums where all schoolwork-type questions belong.Kinda.
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 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?
Well, once n tends to infinite, the resistance value is so high that can be considered as an open circuit.And then say how you would extend it to the infinite case in the problem?
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 likeHint 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?
Hmmm.. Seems to be incorrect.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 + ...
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...?*Shrugs*
I tried to solve it in another way, but I realized it was also incorrect, I don't know..
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...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.
Ohhhhhhh, yea.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)
2R2-9R-9=0.I'll bite. Could you give the quadratic even if not in LaTex.?
Yea same. :)While I knew this trick for the resistor ladder with equal resistances throughout, I couldn't extend that logic to this problem.
Well, by "simple" I meant it doesn't have capacitor and/or inductor.Looks like the circuit is not as "simple" as the title says .