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Equivalent Resistance

  • Thread starter rbwang1225
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  • #1
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Homework Statement


A circuit extends forever to the right, and all the resistors have the same value ##R##. Calculate the equivalent resistance measured across the two terminals at left.
circuit.jpg



Homework Equations


The series and parallel equivalent resistance equation.


The Attempt at a Solution


I have no idea to begin, could someone give some suggestions for me.
Sincerely.
 

Answers and Replies

  • #2
CWatters
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There is a paper on this on the web but if I post a link to it that will give you too much help.

If the chain is infinite then adding another "stage" to the input won't change the resistance because it's equivalent to shifting the whole chain physically one place to the left.

So call the existing resistance Z and write an equation for the resistance with an additional stage added...

Z = Z//R + 2R

Over to you..
 
Last edited:
  • #3
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Dear CWatters:

Could you post the link to me?
I want to understand it more clearly.

Regards
 
  • #4
CWatters
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The forum rules discourage us giving away the whole solution so before I do that which bit are you struggling with?

Here is a diagram explaning of why adding a stage doesn't change the resistance. It works because ∞ + 1 = ∞.

Although you are adding a stage to the left the result is indistinguisable from adding a stage to the right hand side which is at infinity.
 

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  • #5
CWatters
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Here is how you derive the equation for Z that you need to solve...
 

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  • #6
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I think I understand the argument, so the equations are as follows:
##Z=Z//R+2R=(\frac{1}{Z}+\frac{1}{R})^{-1}+2R=(\frac{R+Z}{RZ})^{-1}+2R=\frac{ZR}{Z+R}+2R##
##ZR+Z^2=ZR+2R^2+2ZR##
##Z^2-2RZ-2R^2=0##
##Z=\frac{2R±√(4R^2+8R^2)}{2}=(1±√3)R##
There is a problem for choosing the sign.
Is there any argument that one of which is right?
If my calculations are correct, could you please let me see the paper?
Thank you for kind reply!
 
  • #7
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I think that Z should be positive, therefore, ##Z=(1+\sqrt 3)R##.
 
  • #9
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Dear CWatters:

Thank you for kind help!

Regards.
 

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