Solving equivalent resistance in a triangular resistor network.

#12

Joined Nov 30, 2010
18,167
This isn't even a delta-wye transform. It's just a great heap of series-parallel resistances.
Start reducing at the top of the triangle.
 

#12

Joined Nov 30, 2010
18,167
After you reduce the top triangle, hold it separate and solve the second row. Rinse and repeat. Then stack the results back together and solve that. The method depends on how you break the segments apart to the point that no two people would do it exactly the same way. It's just tedious if you overlook the fact that this has no practical application. If you don't overlook its uselessness, it's just mental masturbation.
 

WBahn

Joined Mar 31, 2012
24,964
After you reduce the top triangle, hold it separate and solve the second row. Rinse and repeat. Then stack the results back together and solve that. The method depends on how you break the segments apart to the point that no two people would do it exactly the same way. It's just tedious if you overlook the fact that this has no practical application. If you don't overlook its uselessness, it's just mental masturbation.
I'm not picturing the approach you are talking about. How can you stack things back together after reducing the lower row? Interested to find out more.
 

dannyf

Joined Sep 13, 2015
2,197
Very easy.

Collapse the top 1 and 1ohn resistors and then parallel it with the base resistor.

Because the circuit is symmetrical, split that resistor into two and parralel each of them to their respective resistors on the side.

And carry it on and on.
 

WBahn

Joined Mar 31, 2012
24,964
Hello,
I was solving some questions related equivalent resistance in circuits.
This (see attatched file) question has been troubling me a lot.
I found something very similar to this circuit in the forums ( http://forum.allaboutcircuits.com/threads/triangular-resistor-network.43005/ )but even after using the info there, I couldn't solve this question.
Does anyone have any idea on solving this question?
If you use symmetry, then you can argue that the voltage everyone along a vertical line drawn right up the middle must be the same. If that line crosses a resistor, simply replace that resistor with two series resistors half the size, one on each side of the line. Now imagine cutting the triangle in half along that line and connecting all the nodes that are cut together (since they are at the same potential this changes nothing). Now you can work with a much simpler circuit and find the resistance of that and then multiply it by two to get the total resistance.

At the final step I simply bounded the results and ended up with pretty tight bounds. Only one of the answers falls between the bounds (assuming I didn't mess up the math).
 

#12

Joined Nov 30, 2010
18,167
I just don't have time to go down this Alice-in-Wonderland rabbit hole right now. I just finished chasing the lawn mower around for half an hour and I'm on a, "sweat break" until I go to the motorcycle wiring job.
In 40 minutes I have only reduced this from 18 resistors to 14 resistors.:(
I have a drawer full of 10 ohm, 1% resistors. I could build this and measure it faster than I can do the math.:mad:

(Runs out the door with a fresh cup of coffee...)
 

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WBahn

Joined Mar 31, 2012
24,964
I definitely see how a series of delta-wye transforms can be used to reduce it to a single resistor. I've sketched it topologically (without values) and it can be done. No big surprise there. The first several are symmetric which makes them easy, but the symmetry seems to break down pretty quickly so the math might get un-fun before it ends.
 

The Electrician

Joined Oct 9, 2007
2,751
Let that vertical line up the middle be a conducting wire, connecting everything it touches together. Let that wire be ground. Now you have 4 nodes (including A). Connect a 1 amp current source to A and solve the 4 nodal equations. Double the voltage at A and you have the desired equivalent resistance.
 

WBahn

Joined Mar 31, 2012
24,964
Let that vertical line up the middle be a conducting wire, connecting everything it touches together. Let that wire be ground. Now you have 4 nodes (including A). Connect a 1 amp current source to A and solve the 4 nodal equations. Double the voltage at A and you have the desired equivalent resistance.
That's essentially what I'm suggesting (or, rather, one way to exploit what I'm suggesting). Note that you can eliminate one of the nodes pretty trivially before applying a source. Once you've removed that node you also have a symmetric delta staring you in the face and if you transform that to a wye you are left with a very straight-forward series-parallel resistor network.

Stopping at the earlier bounds I mentioned previously, which can be arrived at within a couple of minutes, puts the minimum and maximum possible values less than 0.02 Ω apart, which is less than 1% on either side of the actual value. To get the actual value takes a bit more time (I think it took me about five minutes once I decided to do it).
 

The Electrician

Joined Oct 9, 2007
2,751
If you use symmetry, then you can argue that the voltage everyone along a vertical line drawn right up the middle must be the same. If that line crosses a resistor, simply replace that resistor with two series resistors half the size, one on each side of the line. Now imagine cutting the triangle in half along that line and connecting all the nodes that are cut together (since they are at the same potential this changes nothing). Now you can work with a much simpler circuit and find the resistance of that and then multiply it by two to get the total resistance.

At the final step I simply bounded the results and ended up with pretty tight bounds. Only one of the answers falls between the bounds (assuming I didn't mess up the math).
Doing this, I also end up with a couple of rather tight bounds which bracket the correct answer. I got to this point with a series of series/parallel reductions. What I ended up with is an unbalanced Wheatstone bridge. The bounds are obtained by changing the resistor across the middle of the Wheatstone bridge to either a short or an open--hence the two bounds. But, a single delta-wye can solve an unbalanced bridge. This would be probably the simplest route to an exact calculated value.
 

WBahn

Joined Mar 31, 2012
24,964
Doing this, I also end up with a couple of rather tight bounds which bracket the correct answer. I got to this point with a series of series/parallel reductions. What I ended up with is an unbalanced Wheatstone bridge. The bounds are obtained by changing the resistor across the middle of the Wheatstone bridge to either a short or an open--hence the two bounds. But, a single delta-wye can solve an unbalanced bridge. This would be probably the simplest route to an exact calculated value.
That's exactly the approach I took. And of the two delta choices you have, one of them is balanced. You can solve it for the exact solution without using a calculator in just a few minutes.
 
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