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Bardia Sahami
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Hi. Anyone knows how to calculate the Req of this circuit? n tends to infinite.
Kinda.The numbers are based on ohm, so if number 1 in the first resistor means a 1 ohm-resistor.berkeman said: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.Bardia Sahami said: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.berkeman said: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.berkeman said:And then say how you would extend it to the infinite case in the problem?
berkeman said: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?
Hmmm.. Seems to be incorrect.Bardia Sahami said: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 + ...
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?Bardia Sahami said:we have 2R parallel with the first block (R),
*Shrugs*berkeman said:What is the multiplier then?
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...?Bardia Sahami said:*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...Bardia Sahami said: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.
Well, I see a n*2.4 (n*12/5) in my calculations..berkeman said: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...?
I don't know. Something like ~5.4 (or less)??berkeman said:Can you say what you got for the approximate final resistance with that method?
Any helps?berkeman said:
Would you explain this part please? Why the other part is 2x the first block network?hutchphd said:but I converted the network into itself (2x the network attached to the three leftmost resisters is the network again)
Ohhhhhhh, yea.hutchphd said: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)
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...Bardia Sahami said:Ohhhhhhh, yea.
Thanks. I finally realized.
I'll bite. Could you give the quadratic even if not in LaTex.?hutchphd said:R=1ohm+2ohm + (3ohm in parallel with 2R)
Gives me a quadratic for R (pardon my Latex-lazy presentation).
2R2-9R-9=0.Tom.G said:I'll bite. Could you give the quadratic even if not in LaTex.?
Yea same. :)cnh1995 said: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.cnh1995 said:Looks like the circuit is not as "simple" as the title says .
An equivalent resistor in a simple circuit is a single resistor that has the same resistance as the combination of multiple resistors in the circuit. This allows for easier analysis and calculation of the circuit's behavior.
The equivalent resistor in a simple circuit can be calculated using the formula Req = R1 + R2 + R3 + ..., where R1, R2, R3, etc. are the individual resistances in the circuit. This formula is valid for resistors in series.
The equivalent resistance for resistors in parallel can be calculated using the formula Req = 1/(1/R1 + 1/R2 + 1/R3 + ...). This formula takes into account the inverse relationship between resistance and current in parallel circuits.
When resistors are added in series, the equivalent resistance increases. This is because the total resistance is the sum of individual resistances, and the current has to pass through each resistor in the series.
When resistors are added in parallel, the equivalent resistance decreases. This is because the total current is divided among the parallel resistors, resulting in a lower overall resistance.