- #31
VitaminK
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- 4
How to calculate the equivalent resistance in this Circuit?
- Thread starter VitaminK
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Homework Help Overview
The discussion revolves around calculating the equivalent resistance of an infinite circuit with multiple resistors arranged in both series and parallel configurations. Participants are exploring the implications of the circuit's infinite nature on the ability to determine an exact resistance value.
Discussion Character
- Exploratory, Assumption checking, Conceptual clarification
Approaches and Questions Raised
- Participants discuss starting points for calculating equivalent resistance, with some suggesting beginning from the left side of the circuit. Others question the treatment of resistors in series and parallel, particularly in relation to the infinite aspect of the circuit.
Discussion Status
There is an ongoing exploration of different approaches to the problem, with participants sharing insights about patterns in the circuit and the potential for calculating resistance. Some guidance has been offered regarding the relationship between parts of the circuit, but no consensus has been reached on a specific method or solution.
Contextual Notes
Participants note that the circuit extends infinitely, which complicates the calculation of equivalent resistance. There are also references to specific resistor values and configurations that may influence the overall approach to the problem.
- #32
TSny
Science Advisor
Homework Helper
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Yes. Good. But keep going. Can you continue to simplify this circuit by combining the two resistors in your diagram?VitaminK said:
- #33
VitaminK
- 46
- 4
Think I got it now. Thanks everyone!
- #34
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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.
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.