# Homework Help: Finding resistance of (semi-)infinite resistor chain between two points.

1. May 13, 2010

### physicsforus

1. The problem statement, all variables and given/known data
What is the resistance of the (semi-)infinite resistor chain below, between points A and B, if R = 25 ohms?

3. The attempt at a solution
I am not sure where to begin exactly, but I am thinking of this formula:
VAB=VB-VA=∑ε-∑i.R
or
[PLAIN]https://www.physicsforums.com/latex_images/12/1284517-1.png [Broken]

Just not quite for sure.

Last edited by a moderator: May 4, 2017
2. May 13, 2010

### collinsmark

Hello physicsforus,

Welcome to Physics Forums!

Nah, the way to solve this is to start by actually finding some equivalent resistances.

I just solved this myself, and I admit its a bit tricky. There are a number of possible semi-infinite problems such as this, and I'll at least give you the strategy on how to solve this one, and then you can use the same strategy for other similar problems in the future.

Start with the right half of the figure, and draw a new resistor network with the 3 rightmost resistors, except, replace the 2R resistor with the variable 'x'. So in other words, you have 3 resistors, R, R, and x, all connected in series, and you're trying to find Req, as measured across one of the R resistors.

Why do we use the variable x? because at the moment, we don't know what left-looking equivalent resistance is at that node. Hypothetically, we could measure x by removing the two rightmost R resistors from the circuit, and place an Ohmmeter across x. That of course is assuming that we actually had such a circuit hooked up; but of course we don't, so we'll just call it 'x' for now. This concept is important though, and we'll come back to it later.

Solve for Req (of the three resistor circuit) in terms of R and x.

Now suppose you were going to take an Ohmmeter and measure x. Rip away the two R resistors on the right, and take a look at the new big circuit. Now look at the circuit closely. Recall that both circuits go out to infinity and end up with infinite valued resistors at the far left end (both the new and old big circuits are the same in that respect). The new big circuit (with the equivalent resistance of 'x') is the same as the old big circuit, except with all the resistor values doubled! That means,

x = 2Req

Well, you already have a function for Req in terms of R and x, so multiply it by 2 and set the result equal to x, and solve for x.

After that you can get an expression for Req as a function of R alone (without being a function of x). (Hint: Req = x/2 )

Last edited by a moderator: May 4, 2017
3. May 13, 2010

### physicsforus

I think I understand. (I have never done this sort of thing before, so please don't get too irritated)

Req= R/n

So given that R=25 ohms

I take 25/3=8.33 ohms

x=2(8.33)=16.67 ohms

Am I close at all? or am I still not understanding?

4. May 13, 2010

### collinsmark

Sorry, but that's not what I meant.

Starting with original circuit, consider that 2R resistor and everything to the left of it, has a combined equivalent resistance of x. See the figure below.

Next find the equivalent resistance, which we call Req, across the nodes AB.
[You have one resistance R in parallel with another resistance (R + x). Find the parallel resistance of R and (R + x). Express this equivalent resistance, Req, in terms of R and x.]

Note that in general, two resistances in series take the form,

$$R_s = R_1 + R_2$$

And two resistances in parallel take the form,

$$R_{||} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$

Does that make more sense?

#### Attached Files:

• ###### Infinite R-network.gif
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Last edited: May 13, 2010
5. May 13, 2010

### physicsforus

Thanks for your help, but I am still completely confused. This was for a game I am playing with a bunch of people and we are allowed to use any resources. I was hoping to learn something new, but this is way over my head. Hopefully one of the other people will understand all of this.

Thanks again.

6. May 14, 2010

### collinsmark

Don't give up yet. :tongue:

There's a little bit more math involved, but let me get you started.

$$R_{eq} = \frac{1}{\frac{1}{R} + \frac{1}{R+x}}$$

Do a little algebra, then go back to my previous post.

[Edit: Or perhaps better yet, go back to my previous post first and then do a little algebra.]

Last edited: May 14, 2010
7. May 14, 2010

### physicsforus

$$R_{eq} = \frac{1}{\frac{1}{25} + \frac{1}{25+x}}$$

Req=1/.08x

Req=12.5x

x=2(12.5x)

x=25x

x + -25x= 25x + -25x
x+-25x =0
-24x=0
x=0

so between A and B there is no resistance?

(my algebra may be a little rusty especially at 5:20 in the morning)

Last edited: May 14, 2010
8. May 14, 2010

### physicsforus

I know 0 ohms is not right.

so new calculations:

1
________
1 + 1 + 1
_ __ __ = 3 25x
25 25 X ___ = ____ = 8.3333X

25x 3

Now I have just confused myself....grrrr. so frustrating.

9. May 14, 2010

### collinsmark

It might just be easier to keep things in terms of R and x for now (although not really necessary -- you could substitute in the 25 Ohm if you wish).

$$R_{eq} = \frac{1}{\frac{1}{R} + \frac{1}{R+x}}$$

$$= \frac{1}{\frac{R+x}{R(R+x)} + \frac{R}{R(R+x)}}$$

$$= \frac{1}{\frac{R+x+R}{R(R+x)}}$$

$$= \frac{1}{\frac{x+2R}{R(R+x)}}$$

$$= \frac{1}{\frac{x+2R}{R^2 + Rx}}$$

which becomes,

$$R_{eq} = \frac{R^2 + Rx}{x+2R}$$

Now look at your original circuit again (the one you posted in the original post). What we called x is the same thing as if the two rightmost R resistors were removed. It should be obvious to you that the result is that x = 2Req. If this isn't obvious immediately, keep looking at the circuit, and think about it for a few minutes. It's because what we are calling x is the same thing as Req, except with all the resistor values doubled!

In other words,

$$x = 2 \left( \frac{R^2 + Rx}{x+2R} \right)$$

Solve for x.

As a final step, get back to your Req, noting that Req = x/2.
(Or alternately, noting that x = 2Req, make that substitution beforehand and solve for Req directly.)

By the way, in case you haven't seen this:
http://xkcd.com/356/

:rofl:

Last edited: May 14, 2010
10. May 14, 2010

### physicsforus

By golly, Jeeves I think I have it.

I did the Math and got

-25$$\sqrt{}2$$

which broke down to
-35.3553/2

which came to
-17.67767ohms

(or that might be positive not quite sure)

That is actually the second time I got some kind of nerd snipe thing because of this question. Besides those two times, I never seen it before. (No more physics for me. I am sticking to chemistry and biology)

11. May 14, 2010

### collinsmark

Yes, it's positive. Req = 17.6776... Ohms.

And good job! http://www.websmileys.com/sm/cool/653.gif

'Nothing wrong with chemistry and biology. But physics is phun! http://www.websmileys.com/sm/cool/717.gif