Finding Resultant Resistance of Electrical Circuit - Gamma

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Homework Help Overview

The discussion revolves around finding the resultant resistance of a complex electrical circuit. Participants explore various methods and reasoning related to circuit analysis, particularly focusing on series and parallel resistor combinations.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the challenges of applying Kirchhoff's law and consider alternative methods for simplifying the circuit. Some express uncertainty about the correct approach to combining resistors, particularly in complex configurations.

Discussion Status

There is an ongoing exploration of different strategies to simplify the circuit. Some participants suggest redrawing the circuit to clarify the arrangement of resistors, which has proven helpful for one contributor. However, there is no explicit consensus on a single method to find the resultant resistance.

Contextual Notes

Participants mention specific resistor values and configurations, indicating that the circuit's complexity may hinder straightforward analysis. There are references to assumptions about current flow and the behavior of resistors at junctions that are under discussion.

Gamma
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I had a complicated electrical circuit which I was able to reduce to the following (attached). How do I go about finding the resultant resistance of this circuit? Is there a short way of doing this?

Thanks,

Gamma
 

Attachments

  • Circuit.JPG
    Circuit.JPG
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Using Kerchoff's law would be complecated. I tried that. Did not proceed with the calculation. I am sure there is an easy way. If some body can show me how I would proceed that would be great.

Thanks.
 
I reduced it by allowing a known current (1 amp) to pass through the complecated circuit (not the R and 2R which is in series with the battory).

Finally I am getting a number which is 3R + 16/50 R.
 
For this problem, you should be able to replace a pair of resistors in parallel or series by its equivalent resistance. Continue this process until you end up with one equivalent resistance.
 
The circuit in the lower right hand coner seems that it can not be reduced using parallel/series method. If you look at a junction, For example the junction of 4R/3 and R/2 ( in the far right), the current does not simply divide between these two resistors. Rather the current at this juction sees 4R/3 in one branch and some other effective resistance in the other branch. So can I take 4R/3 and R/2 as parallel?
 
yes, but there's other resistors that should also be included.
 
A lot of times it helps to redraw this in a way that makes it easier to see each branch before trying to calculate anything.

From the positive terminal, you pass through a 2R resistor. Then you come to a junction that branches off. One of those branches is a straight line that goes to another junction. If you redraw this circuit, the straight line does you no good at all. What you really have is a junction with three different branches. Of the three branches, one has a resistor in series with two parallel resistors. The other two branches have one resistor each.

Once you reduce the more complicated branch down to one equivalent resistance, your problem gets a lot simpler. You'll have three resistors in parallel with each other.

And, of course, you finally go through one more resistor in series.
 
Thanks every one. Redrawing helped simplify the problem a lot.

Finally I am getting a number which is 3R + 4/25 R which is same as what I got before, but it took less time.

regards,

gamma
 

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