Resistor Network: Find the Equivalent Resistance Across AB

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Homework Help Overview

The problem involves determining the equivalent resistance across points A and B in an infinite resistor network. Participants are exploring various interpretations of the circuit and the implications of its infinite nature on resistance calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss initial confusion about where to start and question the relevance of certain resistors in the network. Some attempt calculations for finite segments of the network, while others suggest a recursive approach to define the total resistance. There is also mention of trends in resistance values as more resistors are added.

Discussion Status

The discussion is active, with participants sharing different perspectives and calculations. Some guidance has been offered regarding the recursive nature of the problem, but there is no consensus on the correct approach or final answer. Multiple interpretations of the resistance values are being explored.

Contextual Notes

Participants are grappling with the implications of an infinite resistor network and the assumptions that come with it. There is mention of a specific answer being suggested, but confusion remains about how to arrive at that conclusion.

lingling
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A resistor network is built up indefinitely as shown in the figure. The equivalent resistance across AB is
A. 1 ohm
B. 1.24 ohm
C. 2 ohm
D. unable to be determined.

-> I don't know where to start with. Can anyone give me some hints?
 

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Am I missing something? It looks to me like the picture says that the resistance across AB is 2 ohms. The rest of the circuit is irrelevant.
 
HallsofIvy said:
Am I missing something? It looks to me like the picture says that the resistance across AB is 2 ohms. The rest of the circuit is irrelevant.
If you put one end of a battery at A and the other end at B, current will flow not just across the 2ohm resistor you speak of but all the other resistors aswell.

A quick calculation indicates the following.

The resistance for a network of length 1 is 4/3 ohms since 1/(1/2+1/4)=1/(3/4)=4/3

The next value obtained by extending the network by an extra 3 resistors is can be found by again just using the rules for parallel and series resistors. I found this to be equal to 5/4 ohms.

Hence the trend appears to be

4/3
5/4

I would be extremely surprised if this trend did not continue and the next value would be 6/5. In that case if the network is infinitely long the resistance would tend to 1ohm
 
But the answer is 1.24 ohm.
I can't understand. Why it is not 2 ohm...?
 
Interesting question...

Assume that the total resistance is R. The resistor network right-of (and including) the second 2ohm resistor from the left is exactly the same as the entire resistor network. Thus, we have
R = 2 || (2+R)
Solving for R gives you the desired answer.
 
I still cannot understand.
Is there any simpler approach?
 
lingling said:
I still cannot understand.
Is there any simpler approach?
doodle has the answer. There is no easier way. There is no end to the chain of resistors, and if you think about how the current will flow you will realize that each 2-ohm resistor will have less current than the previous 2-ohm resistor. If you removed all the resistors touching points A and B, you would have exactly the same net resistance you have with them there.

This problem is a precursor to the important concept of characteristic impedence of transmission lines.

http://www.allaboutcircuits.com/vol_2/chpt_13/3.html
 

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