# Homework Help: Resistive circuit - ladder problem

1. Sep 1, 2014

### Rectifier

Hey!
1. The problem
Figure shows a very long ladder where every part has a resistanse of R. What is the resistance between A and B if the ladder can be approximated to be semi-infinite (the ladder has a beginning but no end).

2.The figure

3. The attempt
The first three resistances have a resistance of 3R the next step add 3R in paralell to R. The next step after that provides 3 more resistances in paralell with on of the resistances in step two.

Step 1: $3R$ [$2R+\frac {3R}{3}$]
Step 2: $\frac{3R \cdot R}{R+3R}+2R=2R+\frac{3R}{4}$
Step 3: $2R+\frac{3R}{5}$
Step 4: $2R+\frac{3R}{6}$
Step 5: $2R+\frac{3R}{7}$

There is a pattern here. Namely
$Rab=2R+\frac {3R}{2+n}$ for n> 0 and n being integer.

The answere is 2.73 but i get 2.000... something for high values on n.

Last edited: Sep 1, 2014
2. Sep 1, 2014

### phinds

You might want to recalculate step 3

3. Sep 1, 2014

### Staff: Mentor

There was one of these questions just recently. The technique for solving it involves drawing a vertical line and saying the resistance to the right of that line equals the resistance seen from terminals A and B. Let's call it M. You can redraw the circuit so it looks much simpler now, and solve for M.

4. Sep 2, 2014

### Rectifier

Is this right for step 3?
$\frac{41R}{15}$

5. Sep 2, 2014

### Rectifier

I am not that I have heared of this technique before. I will try to use it anyway. At what "step" should I draw the line?

Btw, is there a name for it so I can read more about it?

6. Sep 2, 2014

### Staff: Mentor

Draw it where the network remaining to its right will look exactly like the network you started with.

I can't think of a good name for it right now.

7. Sep 2, 2014

### ehild

We think the way that the resultant resistance X does not change if one unit of the ladder is added or removed from the infinite number of units. See picture of the ladder and the equivalent circuit. Solve for x.

ehild

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8. Sep 2, 2014

### Rectifier

Okay. Lets say I draw a vertical line at step 2 "above" the resistor that is connected in parallell with the rest of the crcuit. Now I am left with 2 identical circuits.

How do I continue now?

9. Sep 2, 2014

### BvU

The technique is called: realizing you have the same situation if you cut off the leftmost three resistors. So apparently the whole thing has the same resistance as 2R + R//(the whole thing).

// meaning: parallel to in my sloppy notation

(If you are as lazy as I am, you want to draw the line at the first step where it makes sense, but not so far as to cause more work than absolutely necessary).

I don't have a good word for the approach: assume you know the answer and work backwards isn't good enough. Someone else ?

10. Sep 2, 2014

### Rectifier

Thank you for your replies I will read and comment back in some minutes.

EDIT:
I will have a lecture now and I will ask the professor if he has time. Sorry that I am being dumb on this one.

Last edited: Sep 2, 2014
11. Sep 2, 2014

### Staff: Mentor

Let's say the total resistance of the circuit on the right is equivalent to a resistance M.

You can also say that when you add 3 resistors to that resistor M you make a circuit which again has an equivalent resistance M.

Solve for M.

Perhaps this approach could be called a recursive technique!

12. Sep 2, 2014

### BvU

Re technique: at some point I learned to deal with geometric series, e.g.$$S =1+{1\over2}+{1\over 4} + {1\over 8} + \dots$$and the way to tackle them is to distinguish that $${1\over 2}S = {1\over2}+{1\over 4} + {1\over 8} + \dots$$so that $$S = 1+ {S\over 2}$$see the analogy ?

13. Sep 2, 2014

### phinds

No, it's not what I get. How about you show your work (in detail, step by step, and DRAW the exact circuit you are analyzing in that step)

14. Sep 2, 2014

### ehild

Yes, it is right.
What is the next?

Evaluating the results, you have a sequence 3R, 2.75R, 2.733R ... They appear to tend somewhere. The differences get smaller and smaller. At the end, the difference between the last two elements of the sequence is so small that they can be considered the same.
If you have the nth term, the n+1th is Rn+1=2R+RRn/(R+Rn), and at the end you can consider Rn+1=Rn. What do you get for Rn?

ehild

15. Sep 2, 2014

### phinds

Damn. I guess *I* must have made a mistake. Perhaps you can show me where I went wrong.

OOPS. NEVER MIND. I just didn't expand it the way you did. We agree.

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16. Sep 2, 2014

### ehild

ehild

17. Sep 2, 2014

### Rectifier

This is what my professor told me too. But I dont understand why you get the same resistance when you add stuff :,(.

18. Sep 2, 2014

### Rectifier

Thanks for the comment.

Why not?

19. Sep 2, 2014

### Rectifier

Here it is again

20. Sep 2, 2014

### Rectifier

"At the end, the difference between the last two elements of the sequence is so small that they can be considered the same."

$R_{ab}=R+R//R_{ab}+R \\ R_{ab}=2R+\frac{R \cdot R_{ab}}{R+R_{ab}} \\ R_{ab}(R+R_{ab})=2R(R+R_{ab})+R \cdot R_{ab} \\ R_{ab} \cdot R+R_{ab}^2=2R^2+2R_{ab}\cdot R+R \cdot R_{ab} \\ R_{ab}^2=2R^2+2R_{ab}\cdot R \\ 2R^2+2R_{ab}\cdot R-R_{ab}^2=0 \\$

Solve for $R_{ab}$ where $R_{ab}>0$

$R_{ab}^2-2 R \cdot R_{ab}-2R^2=0 \\ R_{ab}= R \pm \sqrt{R^2+2R^2} \\ R_{ab}= R \pm R\sqrt{3}$

$R_{ab}= (1 + \sqrt{3})R$ since $R_{ab}= (1 - \sqrt{3})R$ is undefiened.

Yey! :D It is the right answer.

Last edited: Sep 2, 2014
21. Sep 2, 2014

### ehild

Only the + sign gives valid solution. Rab can not be negative. But you have solved it! Congratulation!

The sequence of resistances, as you calculated them, converges very fast to (1+√3)R = 2.7320R. The fourth is 2.7321R already.

ehild

22. Sep 2, 2014

### Rectifier

No, sorry. I am not so farsighted :,(. I guess it something similar with my series of resistances. I am afraid I am not that good at series.

$$S = 1+ {S\over 2}$$ is true indeed. I couldn't have seen that from the equation.

23. Sep 2, 2014

### Rectifier

Thank you to all of you who took their time and helped me. I really appreciate that.

24. Sep 2, 2014

### Staff: Mentor

Well, if you added some resistance in series with the input you would increase the resistance, and if you added some resistance in parallel you would decrease the resistance, so if you add some of both---correctly chosen---you can reproduce the same resistance!