Finding Resultant Resistance of Electrical Circuit - Gamma

AI Thread Summary
To find the resultant resistance of a complicated electrical circuit, it is recommended to simplify the circuit by identifying and reducing resistors in series and parallel. Initially, using Kirchhoff's law was deemed too complex, but redrawing the circuit helped clarify the branches and junctions. The process involves combining resistors step-by-step until reaching a single equivalent resistance. The final calculation yielded a resultant resistance of 3R + 4/25 R, demonstrating that a clearer visualization can significantly expedite the problem-solving process. Simplifying the circuit layout is crucial for efficient analysis.
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
 

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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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