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How to calculate the equivalent resistance in this Circuit?

  • Thread starter VitaminK
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  • #26
TSny
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Well, this is a correct drawing. But it is not what I had in mind. Make use of the fact that (a) and (b) have the same resistance R.
 
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  • #28
TSny
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  • #29
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I've now replaced circuit b) with R.
 
  • #30
scottdave
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View attachment 258304I've now replaced circuit b) with R.
So find an expression for the resistance of this circuit (leaving in the variable R). Then, as you figured out, this circuit has the same resistance as the one you drew in post #25.
 
  • #31
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is this the expression?
 
  • #32
TSny
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View attachment 258309is this the expression?
Yes. Good. But keep going. Can you continue to simplify this circuit by combining the two resistors in your diagram?
 
  • #33
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Think I got it now. Thanks everyone!
 
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  • #34
scottdave
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Only one of those values is possible in a real circuit since one is positive and the other is negative.
Now that you solved it, here's a check for your answer. Let's start several nodes out to the right.
We have the 10Ω in parallel with the unknown equivalent resistance to the right of that. Let's call that R0.

If we move 1 node to the left, then R1 = 10Ω + (10Ω)(R0) / ( 10Ω + R0) . This is the Product / Sum for parallel resistances. You can keep going: R2 = 10Ω + (10Ω)(R1) / ( 10Ω + R1) and so on.

So why am I doing all of this? You can iterate till it converges on a number. Even though we don't know what R0 is, we can take an initial guess. We know it should be larger than 10Ω, so pick 11 as an initial guess.

I used Excel. Put my guess for R0 in the first cell {A1}, then in A2 I put the following formula: =10+10*A1/(10+A1)

I copied this down several rows. You can see that it converges fairly fast to a number (numerically close to your answer).

But even if I put wild guesses like 0 or 1 million, it still converges pretty close after only 5 or 6 iterations.
 

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