Resistor pattern - Equivalent resistance

In summary, the conversation discusses how to calculate the equivalent resistance of an infinite chain of resistors. Multiple methods are suggested, including using mesh analysis and simplifying the circuit by cutting it at a certain point. The correct answer is determined to be R( (1 + root(5)) /2 ).
  • #1
jegues
1,097
3

Homework Statement


See figure


Homework Equations



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

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?
 

Attachments

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  • #2
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
[tex](2R)i_1 + (-R)i_2 = 1[/tex]
[tex](3R)i_2 + (-R)i_3 = 0[/tex]
[tex](3R)i_3 + (-R)I_4 = 0[/tex]
... and so on toward infinity

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

But if you keep substituting in, due to the repeated division of 3, you end up with:
[tex]i_1 = \frac{i_\infty}{\infty} + \frac{1}{2R} = \frac{1}{2R}[/tex]
So the equivalent resistance then equals:
[tex]R_{eq} = \frac{v_{applied}}{i_1} = \frac{1}{\frac{1}{2R}} = 2R[/tex]
which is wrong!
 
Last edited:
  • #3
Does that mean I can simpy do this? (See figure)

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

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  • #4
jegues said:
Does that mean I can simpy do this? (See figure)

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

Yes, I believe so.
 
  • #5
So then my answer would simply be 2R, no?
 
  • #6
jegues said:
So then my answer would simply be 2R, no?

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.

[tex]R_{eq} = R + (\frac{1}{R} + \frac{1}{R_{eq}})^{-1}[/tex]
 
  • #7
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
jegues said:
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?

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:
  • #9
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
xcvxcvvc said:
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.

[tex]R_{eq} = R + (\frac{1}{R} + \frac{1}{R_{eq}})^{-1}[/tex]
I calculated it using the equation I posted, and got one of those answers. Solve for Req!
 
  • #11
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
xcvxcvvc said:
I calculated it using the equation I posted, and got one of those answers. Solve for Req!

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

[tex] Req = R + (\frac{RReq}{R+Req}) [/tex]

??

I think I got it, is it R( (1 + root(5)) /2 ) ?
 
  • #13
jegues said:
Is this the right algebra problem I'm looking at then:

[tex] Req = R + (\frac{RReq}{R+Req}) [/tex]

??

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

yes.
 

1. What is a "resistor pattern"?

A resistor pattern refers to a specific arrangement or configuration of resistors in an electrical circuit. This arrangement determines the overall resistance of the circuit and is important in determining the flow of current through the circuit.

2. What is "equivalent resistance"?

Equivalent resistance is the combined resistance of multiple resistors in a circuit. It is the single resistance value that would produce the same effect on the flow of current as the combined resistors in the circuit.

3. How is equivalent resistance calculated?

Equivalent resistance is calculated using Ohm's Law, which states that resistance is equal to the voltage divided by the current. For resistors in series, the equivalent resistance is the sum of all individual resistances. For resistors in parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of each individual resistance.

4. Why is it important to calculate equivalent resistance?

Calculating equivalent resistance is important in understanding the overall behavior of an electrical circuit. It helps determine the amount of current that will flow through the circuit, which in turn affects the voltage and power in the circuit. It is also useful in troubleshooting and designing circuits.

5. Can equivalent resistance ever be lower than the lowest individual resistance?

No, equivalent resistance can never be lower than the lowest individual resistance. This is because adding resistors in parallel increases the total number of pathways for current to flow, which decreases the overall resistance. However, adding resistors in series only increases the overall resistance, making it impossible for the equivalent resistance to be lower than the smallest individual resistance.

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