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Sorry for the bad image quality

For those who can't see the image: SO there is an outer delta with two 1 ohms and a 2 ohm. Then an inner Y with three 2 ohms. Within the Y, there is a delta formed on it's two lower legs with three 1 ohms. The question is to find the resistance between the upper arm above the inner delta.

Now, my question is, which branches should I convert and also, what would it look like if I convert those branches?

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  • \$\begingroup\$ why are you asking? .... just pick one .... your question is like what will it look like if i pile up sand at a beach? \$\endgroup\$ – jsotola Mar 19 at 15:58
  • \$\begingroup\$ @jsotola since I am trying to find the resistances b/w AB, I should not disturb that branch right?(I heard that in class, no idea what that means) How do I do that? \$\endgroup\$ – Febin .K.Dominic Mar 19 at 16:32
  • \$\begingroup\$ " I should not disturb that branch right?" Well.. no. You can convert that branch as well. Just take care that if you do a conversion that you keep A and B in the same place. e.g. if you do a D-Y conversion including the AB 2 Ohm resistor you end up with two resistors between A and B. \$\endgroup\$ – Oldfart Mar 19 at 16:44
  • \$\begingroup\$ A and B will always exist, no matter how many conversions you do, so, disturb away to your heart's content \$\endgroup\$ – jsotola Mar 20 at 0:03
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Homework, only guidance:

Start by removing the 2 Ohm from AB. Later you will put it back when you have calculated the rest of the circuit. Let Rab be =X when that 2 Ohm is removed. The wanted Rab=X in parallel with 2 Ohm.

You must do two resistor removals and putting it back to get the result and I showed the first one. Series and parallel resistance formulas are as essential as delta-star formula.

After taking off that 2 Ohm there's only one delta. Convert it to star. Then you hopefully see another possibility to advance towards series-parallel circuit.

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  • \$\begingroup\$ hm.. couldn't we transform the inner 3x 1Ohm delta into a Y? the arms of the Y would align and go in series with AB resistor (and analogous bottom-left and bottom-right resistors) and then we should end up with a big triangle of 1ohm:1ohm:2ohm and an inner Y of (2+x):(2+x):(2+x). Then transform the inner Y to a triangle again, and have a big triangle with 2-parallel resistors on each edge? \$\endgroup\$ – quetzalcoatl Mar 19 at 20:34
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    \$\begingroup\$ ...and where's B after this all? If you can track it then draw it and write an answer. Check the result with a circuit simulator at first. \$\endgroup\$ – user287001 Mar 21 at 18:12
  • \$\begingroup\$ Hahha, you're totally right! My bad, I focused on the net diagram, forgetting what I should calculate :) thanks. I think I'll do some drawing actually! \$\endgroup\$ – quetzalcoatl Mar 21 at 20:23
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Here's my take on the problem. I'll do it graphically, just to show the transformations I'd do, no math. I'll use IEC+ANSI instead of NEMA symbols, because I'm more used to them, and it's easier to draw a rectangle in MS Paint than a sawtooth line.

enter image description here

  • We keep A-B intact (red), along with resistor between them
  • Focus on B-C-D delta (green), transform into star BCD+E
  • Notice R1a+R1b are in series, just like R2a+R2b
  • Focus on F-E-G delta, transform into star FEG+H
  • Notice R4a+R4b, and R5a+R5b, and R6a+R6b are respectively in series

We end up with image [5(first)] and with a bit of different layout [5(second)] we see that R4 and R6 are in parallel. Then in series with R5. Then in parallel with Red resistor. Then we'got just one resistor left = the answer.

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