Equivalent Resistor of a simple circuit

[Bardia_Sahami]

Hi everyone,
Is there anyone knows how to to calculate the Req of this circuit? n tends to infinite.

Thanks in advance.

PS - the numbers you see are based on ohm. like the resistor with "3" means a 3 ohm-resistor. (Numbers are the resistance value based on ohm)

 

[Bardia_Sahami]

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.

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 solve.

 

[KlausST]

Hi,

I assume it is school homework, so it´s not very helpful if we just give the solution.

Why don`t you simplify it?

What about an Excel chart? where n is on the X-Axis?
Or a little software program?

Show us what you have done so far.


Klaus

- - - Updated - - -

Hi,

again: Show what you have done so far:

Hint to simplfy:
* divide it into blocks. (you almost have done already)
* Take one block and simplify it.

Klaus

 

[albbg]

It's a differential ladder network. First of all transform it to a single ended one. This is very simple since the serie resistors of each cell simple adds-up.
For instance you'll have:


---1------2------ ........................ ----3-----6-------
.........3.......6 .......transforms to...............3.......6
---2------4------..........................------------------

after this manipulation there are tricks to calculate recusively the input resistance Rn.
I don't know if the above sketch can be understood, I have had to use dots instead of spaces.

 

[KlausST]

Hi,

I have had to use dots instead of spaces.

ASCII art can be improved by using code tags.

Klaus

 

[Ratch]

Hi everyone,
Is there anyone knows how to to calculate the Req of this circuit? n tends to infinite.
View attachment 158900
Thanks in advance.

PS - the numbers you see are based on ohm. like the resistor with "3" means a 3 ohm-resistor. (Numbers are the resistance value based on ohm)

I don't know if there is a closed form solution to your problem, because each resistor segment you add to the right of your network is a different ohmage value. However, with a computer, one can easily iteratively add a a large finite number of increasing resistor value segments and observe where the total resistive value of the network is heading. The computer program below does exactly that. It adds 100 resistor segments to the 3 ohm resistor on the left and computes the final value to 50 decimal places. It doesn't take many brain cells to see that the convergence is at 2.0 . Add the leftmost 1 ohm and 2 ohm resistors and one gets the final network resistance of 5 ohms for an infinite sequence. I will leave it to someone else to figure out a closed form solution. Ratch

 

[std_match]

I don't know if there is a closed form solution to your problem, because each resistor segment you add to the right of your network is a different ohmage value. However, with a computer, one can easily iteratively add a a large finite number of increasing resistor value segments and observe where the total resistive value of the network is heading. The computer program below does exactly that. It adds 100 resistor segments to the 3 ohm resistor on the left and computes the final value to 50 decimal places. It doesn't take many brain cells to see that the convergence is at 2.0 . Add the leftmost 1 ohm and 2 ohm resistors and one gets the final network resistance of 5 ohms for an infinite sequence. I will leave it to someone else to figure out a closed form solution. Ratch

View attachment 159006

I don't understand the syntax, but I think something is wrong with the program.

I get the total resistance to 2.25 + sqrt(2.25^2 + 2*2.25) = 5.3423292....

Since this probably is a school assignment, I will not reveal the method now.

 

[wwfeldman]

i used a spreadsheet over 5 or six stages
i got a resistance of about 6 ohms.

it is fairly clear that each stage added is a larger resistor in parallel with one resistor of the earlier stages, which only reduces the net resistance
since the first stage is 6 ohms, i think that puts an upper limit on the actual resistance

as demonstration, the next stage (12 ohms) is in parallel with the 3 ohm part of the first stage
thus the 3 ohm looks like a little less than 3 ohms, so the first stage is a little less than 6 ohms

 

[albbg]

Is is possible to find a closed form solution if the network goes to infinite. Remember that if I divide all the resistor of the network for a quantity "m", then the totale resistance will be divided by "m":

 

[Ratch]

I don't understand the syntax, but I think something is wrong with the program.

I get the total resistance to 2.25 + sqrt(2.25^2 + 2*2.25) = 5.3423292....

Since this probably is a school assignment, I will not reveal the method now.

Yes, I think you are right. My program is faulty. I was adding the resistor segments across the total network instead of the rightmost resistor. Back to the programming board.
Ratch

 

[c_mitra]

Is there anyone knows how to to calculate the Req of this circuit? n tends to infinite.

First steps:

1. Consider the first step of the ladder: ignore the rest: it is 1+3+2=6

2. Consider the next step of the ladder: ignore the previous and the following steps: you get 2+6+4=12=6x2

3. The n-th step of the ladder will be, by induction: 6x2^(n-1)

4. All the steps are in parallel connection: so you need to add the reciprocals:

((1/6)*(1/1 + 1/2 + 1/4 + ... to infinity)

5. The sum to infinity is not tough to calculate. It will be 2.

The equivalent resistance will be 1/((1/6)*2)=1/(1/3)=3.

 

[std_match]

First steps:

1. Consider the first step of the ladder: ignore the rest: it is 1+3+2=6

2. Consider the next step of the ladder: ignore the previous and the following steps: you get 2+6+4=12=6x2

3. The n-th step of the ladder will be, by induction: 6x2^(n-1)

4. All the steps are in parallel connection: so you need to add the reciprocals:

((1/6)*(1/1 + 1/2 + 1/4 + ... to infinity)

5. The sum to infinity is not tough to calculate. It will be 2.

The equivalent resistance will be 1/((1/6)*2)=1/(1/3)=3.

This is not correct. All "steps" are not in parallel. The 12 Ohms of the second step is not in parallel with the 6 Ohms of the first step. The 12 Ohms are in parallel with the 3 Ohm resistor in the first step. The connections are similar in the following steps. The resistance will be higher than your calculation.

My answer in post #7 and the solution by albbg in post #9 are identical (we used the same general method), and they should be correct. The total resistance is 5.3423292.... Ohms.

 

[c_mitra]

This is not correct. All "steps" are not in parallel.

Sorry, I goofed! Thanks for pointing out.


1. calculate partial sums: S1 is the first step of the ladder, S2 is upto the 2nd step and so on. Ignore the steps beyond.

2. S1= 1+3+2 -> 1+2+3 -> 3+3

3. S2=3+3||(2+6+4) ->3+3||(6+6)

4. S3=3+3||(6+6||(4+12+8)) ->3+3||(6+6||(12+12))

This is a continued fraction.

There is a simpler way:

Take the sum as S

Note that if you remove the first step of the ladder, you are left with a ladder whose value is 2S

So we get S=1+3||(2*S) +2=3+3*2*S/(3+2*S)

Solve this equation to get S; one root will be negative and ignore that.

 

[std_match]

There is a simpler way:

Take the sum as S

Note that if you remove the first step of the ladder, you are left with a ladder whose value is 2S

So we get S=1+3||(2*S) +2=3+3*2*S/(3+2*S)

Solve this equation to get S; one root will be negative and ignore that.

Yes, this is basically the same method that I used for post #7 and that albbg used for post #9.

S = 2.25 + sqrt(2.25^2 + 2*2.25) = (9 + sqrt(153)) / 4

 

[c_mitra]

this is basically the same method that I used for post #7

Only minor problem with post #7 is that it is incomprehensible.