# Resistor pattern - Equivalent resistance

1. Jun 16, 2010

### jegues

1. The problem statement, all variables and given/known data
See figure

2. Relevant equations

Rs = R1 + R2
Rp = (R1*R2)/(R1+R2)

3. The attempt at a solution

I tried completing one repitition of the pattern to see if I could figure out what the Req would be so:

R + R in series on the end so 2R // R so 2/3R in series with R so 5/3R and the pattern repeats.

How do I figure this out?

#### Attached Files:

• ###### ResistorProblem.JPG
File size:
11 KB
Views:
163
2. Jun 16, 2010

### xcvxcvvc

since the chain is infinite, you can break the chain at any point that gives you some finite arrangement of resistors connected to an infinite chain of resistors as long as the new infinite chain begins exactly like the old infinite chain began.

Then, the equivalent resistance will equal some transformation of those finite resistors and the equivalent resistance, because the new infinite chain can be thought of as Req as well.

Below is wrong but I wonder about it
Let's say we were going to do some mesh analysis on this infinite chain with an applied 1V source. 1v/i1 = Req
$$(2R)i_1 + (-R)i_2 = 1$$
$$(3R)i_2 + (-R)i_3 = 0$$
$$(3R)i_3 + (-R)I_4 = 0$$
... and so on toward infinity

We can then solve for i_1:
$$i_1 = \frac{1}{2R} + \frac{i_2}{2}$$
but what is i_2?
$$i_2 = \frac{i_3}{3}$$
and i_3?
$$i_3 = \frac{i_4}{3}$$
and so on...

But if you keep substituting in, due to the repeated division of 3, you end up with:
$$i_1 = \frac{i_\infty}{\infty} + \frac{1}{2R} = \frac{1}{2R}$$
So the equivalent resistance then equals:
$$R_{eq} = \frac{v_{applied}}{i_1} = \frac{1}{\frac{1}{2R}} = 2R$$
which is wrong!

Last edited: Jun 16, 2010
3. Jun 16, 2010

### jegues

Does that mean I can simpy do this? (See figure)

If I cut it there it would repeat the same pattern after, no?

#### Attached Files:

• ###### ResistorProblem.JPG
File size:
11.9 KB
Views:
98
4. Jun 16, 2010

### xcvxcvvc

Yes, I believe so.

5. Jun 16, 2010

### jegues

So then my answer would simply be 2R, no?

6. Jun 16, 2010

### xcvxcvvc

NO! That mesh analysis I posted was wrong, and I even labeled as such. I was hoping someone could tell me the flaw in my reasoning that led the to wrong answer.

$$R_{eq} = R + (\frac{1}{R} + \frac{1}{R_{eq}})^{-1}$$

7. Jun 16, 2010

### jegues

I'm still confused how to get the answer then, you told me I could cut my circuit where I had labeled in the image above. Would it not reduce to 2R?

8. Jun 16, 2010

### xcvxcvvc

No, it reduces according to the equation I wrote. Let me draw a picture. Wait a second.

edit:
[PLAIN]http://img16.imageshack.us/img16/5953/42170211.jpg [Broken]

Last edited by a moderator: May 4, 2017
9. Jun 16, 2010

### jegues

How do I solve that though? The choices I have to choose from are listed as follows:

R/2, R( (1+root(3)) /2 ), R ( 1 + root(3) ), R( (1 + root(5)) /2 ), and 2R

I can't seem to reach any of those from your drawing.

Any more ideas?

10. Jun 16, 2010

### xcvxcvvc

I calculated it using the equation I posted, and got one of those answers. Solve for Req!

11. Jun 16, 2010

### vk6kro

The problem is that adding extra resistors reduces the output from the first two.

If you want to just get an answer, there is a fairly easy way to do it.

Assume all resistors are 1000 ohms and a power supply of 1000 volts on the input.

Work out the current in the first resistor when 4 pairs of resistors are included. Then try it with 5 pairs.
If the result is the same within 1 or 2 % then the effect of adding extra pairs will be negligible.

Knowing this current and the supply voltage, you can work out the effective resistance and the ratio of this to 1000 ohms.

12. Jun 16, 2010

### jegues

Is this the right algebra problem I'm looking at then:

$$Req = R + (\frac{RReq}{R+Req})$$

??

I think I got it, is it R( (1 + root(5)) /2 ) ?

13. Jun 17, 2010

yes.