Calculate the resistance Rab between A and B for an half-infinite ladder (i.e. a ladder that has a beginning but no end).

I'm trying to solve this problem. The solutions I've found seem to be based on the notion that if you add one more resistor in parallell, then that single resistor will have the same resistance as the whole resistance Rab. But why is that? Why does another resistor automatically have the same resistance as the whole chain?

 

The fundamental unit of the chain is a 3 resistor unit. The chain consists of an infinite number of those units in cascade.

It's not true that another single resistor will have the same resistance as Rab.

What is true is that adding another unit of 3 resistors to the left end won't change the resistance Rab.

Whatever the value of Rab, adding another resistor R in parallel with Rab, and then adding two more resistors in series (this is what you are doing when you add another fundamental unit at the left) just gives Rab again.

Set up an equation representing this fact and solve for Rab in terms of R.

 

What is true is that adding another unit of 3 resistors to the left end won't change the resistance Rab.

Whatever the value of Rab, adding another resistor R in parallel with Rab, and then adding two more resistors in series (this is what you are doing when you add another fundamental unit at the left) just gives Rab again.

Why is that?

 

I get an answer of 2,732016501 R as the total resistance between A and B

 

Why is that?

Because when you add another unit to the left, the whole semi-infinite chain is unchanged. Adding another section doesn't increase the number of sections in the chain since the chain is infinitely long.

 

I get an answer of 2,732016501 R as the total resistance between A and B

This is the homework help forum. You're not supposed to solve the problem for the student; you're supposed to give them help to solve it themselves.

 

I didn't solve the problem ... I just gave the answer . Actually I am asking if it's correct because I myself do not know how to solve this . I used a cheat method and now i'm asking if the answer is right.

Actually , Tezsla , if you want you can try it yourself

http://www.falstad.com/circuit/e-ohms.html

It's a java realtime circuit simulator.
Just erase the circuit (it's editable right there on the page) that is there and right click to add components . Build it yourself (put a voltage source between A and B)and see how the circuit behaves, it might help you in formulating a mathematical approach to solving it.

 

@Shagas: I second the request not to give answers. Keep in mind that, increasingly, homework is graded automatically and so the person may only have to input their final answer into a webpage to get credit. The way to approach something like this is to say something like, "I came up with an answer that is between 2.5R and 5R. Does that sound reasonable?"

@Tezla: One of the first things I would recommend is to look at the bounding cases to put outside limits on the answer. What simplications can you make that will always result in the effective resistance being larger that it should be but allow you to come up with an answer by inspection? That gives you a result that you know the actual answer must be smaller than. Now do the opposite and come up with a result that you know the actual answer must be larger than.

But, for the record, your answer is wrong. Then again, when you report an answer to ten sig figs, you are claiming that it is correct to ten sig figs.

The key observation that you need to make, which The Electrician pointed out, is that that you can replace a semi-infinite chain of resistors with its equivalanet resistance, even if that is presently not known (just do it symbolically). If you cut the original chain vertically at any point, you have a finite set of resistors to the left of the cut. But what do you have to the right?

 

But, for the record, your answer is wrong. Then again, when you report an answer to ten sig figs, you are claiming that it is correct to ten sig figs.

One of the hazards of using a "cheat" method I guess.

 

Actually, one of the hazards of claiming more precision than is justified.

 

Actually, one of the hazards of claiming more precision than is justified.

Indeed - I guess one can always blame the calculator or ones spectacles, and so forth ......

 

Ah, yes, the old, "I couldn't see clearly and so I had to just keep writing more digits down," excuse.

 

Well I just throught that since we were looking at theoretically perfect values where R= R = R = R etc then the precision would be valid

 

Again, if you report a value to ten sig figs, you are claiming that the value you are reporting is correct to ten sig figs.

Your simulation is not valid to an arbitrary number of digits because your simulation is not the actual circuit, but rather an approximation of it.

Why don't you do this and report back (and I'm serious, I think it would be useful to you and others):

Take the circuit you used and short the final resistor. How much does that change your result by? That gives you a good indication of how many sig figs you are justified.

What would be particularly intersting would be to chart the lower/upper bound resistances as a function of the number of resistor triples that are included in the circuit.

 

Oddly, your value is low when I would expect it to be high. I'm assuming that your simulation used an integer number of resistor triples. I was curious to see if I could figure out how many triples you used. I am guessing six (a total of eighteen resistors).

My guess is that you are hitting the round off error of the simulator you are using. Given the amount of the error you have, I would say that the value you are using from the simulator has 5 sig figs. If you are applying a 1V source, then it is probably reporting a current of 366.03μA after six stages, which would closely match the value you gave.

 

Well yes, I vastly overdid it with the digits
I used a 1v source with 1ohm resistors and got a simulation current of xxx.xx mA so 0.xxxxx Amps so i'm guessing that my result can not be more accurate than 2,73201 assuming that the simulation is perfect (edit: Which I don't see why it wouldn't be since we are working with abstract values)

Shorting the last resistor had no effect . Only when shorting the 3rd external chain parallel resistor had a (by the software) visible change of 0.04 mA through the supply .
I see your point ... sortof

So what is your thought on this ?
(I hope i'm derailing this thread into oblivion)

 

it is probably reporting a current of 366.03μA after six stages, which would closely match the value you gave.

377.22uA at the end of the 5th stage

 

Oddly, your value is low when I would expect it to be high. I'm assuming that your simulation used an integer number of resistor triples. I was curious to see if I could figure out how many triples you used. I am guessing six (a total of eighteen resistors).

I did the calculations in a spreadsheet and it had converged by the ninth step. I believe it had converged by the sixth step, but I need to recreate the spreadsheet since I didn't save it. BTW, it converged to an integer + irrational number.

 

When you want to measure resistance in a simulation, apply a 1 Amp current source. Then the resistance will be numerically identical to the voltage across it. No calculation (i.e., reciprocal) is required.

 

377.22uA at the end of the 5th stage

Is this the current in the last resistor with 1 volt applied to the left end of the chain?

For comparison with what I get, would you list the currents in the last resistor for chains of 1, 2, 3, 4 and 5 sections?