Introductory Physics: Connection of Resistors ...

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The discussion revolves around solving a resistor network problem using Kirchhoff's rules, where participants initially arrive at different answers for the equivalent resistance (R_eq). The proposed correct answer is 4/5 * R, which several users confirm through symmetry arguments and simplifications. There is a consensus that using symmetry can significantly ease the problem-solving process, often avoiding complex equations. Participants also discuss the importance of clear presentation of solutions, suggesting that images of handwritten work can be difficult to interpret. The conversation concludes with encouragement to focus on fundamental physics concepts, particularly electromagnetism, over circuit analysis.
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Homework Statement
Calculate R_eq in terms of R for the shown figure ...
Relevant Equations
Kirchhoff's rules.
IMG-20250522-WA0000.webp

This problem was sent to me by a good friend of mine. He said that he'd attempted it, but that he couldn't find an answer—at least, not an answer that matched the proposed answer (5/6 * R). So, I attempted it (armed with Kirchhoff's rules) and after a bit of running in circles, I got: 4/5 * R (picture of solution is attached beneath this wall of pre-text. I would have written it in LaTeX, but I neither have the time—revising for finals—or the wherewithal to. Apologies in advance).

What's interesting is I'd attempted it prior and got 1/2 * R as an answer (undoubtly, I must have done something wrong in either—or both—attempt). I come here seeking, what's hopefully, a correct solution I could share with my friend. Much appreciated, let me know if we need to have it written in LaTeX, I will find some time to.

(note that: my 3's and J's look very similar, but I did try to make it look as neat as possible. Also, in double-digit-subscripted currents, the first digit is the parent current and the second is the "order").

*Edit (1): Unfortunately, looking over my solution again, I have found that it isn't as orderly and neat as I would like it to be. The way my solutions go, they're quite fast and all over the place, not intended to be seen, so I do hope you appreciate the fact that I did all I could. Again, apologies.

*Edit (2): I just noticed the image was also incredibly down-scaled. It's truly horrible to try and make out the details. Do let me know if I need to post it page by page (I still have them here) or if I need to rewrite them better, or write them on LaTeX.

20250522_202949.webp
 
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Lunar Manatee said:
Do let me know if I need to post it page by page (I still have them here) or if I need to rewrite them better, or write them on LaTeX.
Separate pages would be better as much of your attachment is unreadable due to the small print and poor lighting. We generally discourage photos of work done for exactly these reasons.
 
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Mark44 said:
Separate pages would be better as much of your attachment is unreadable due to the small print and poor lighting. We generally discourage photos of work done for exactly these reasons.
Will get to doing that once I have the time—should have some tonight. Will rewrite them aswell since that'd probably be better.
 
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While you can do this with KVL/KCL, it's much easier if you use symmetry arguments to simplify the problem. You should be able to do it with almost no equations. Try it and ask us if you want more hints.
 
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Lunar Manatee said:
Homework Statement: Calculate R_eq in terms of R for the shown figure ...
Relevant Equations: Kirchhoff's rules.

I just noticed the image was also incredibly down-scaled. It's truly horrible to try and make out the details.
I'm not able to make out the equivalent circuit that you started to write down and scratched out, but were you trying to use symmetry arguments in that line of thought? I'm not great at these kinds of symmetric resistor problems, but can you say anything about what simplifications you might be able to do by cutting the circuit with a vertical line down the middle? Do you see any voltage arguments that may help you to redraw the circuit to make it easier to solve?
 
Lunar Manatee said:
I got: 4/5 * R
BTW, using symmetry and then simple parallel-series simplifications, I got the same answer as you. I'll need to re-check my work to be sure, though.
 
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berkeman said:
BTW, using symmetry and then simple parallel-series simplifications, I got the same answer as you. I'll need to re-check my work to be sure, though.
I think I realised by the end of the KVL/KCL that it was symmetric all around. The currents going into the center junction are the same ones going out, the horizontal currents up top and down bottom are also equal.

Rewriting the problem in LaTeX for future convenience (for other people) so people can be sure I did do the work.

*Edit: I forgot to thank you! Apologies! And many many thanks!
 
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berkeman said:
I'm not able to make out the equivalent circuit that you started to write down and scratched out, but were you trying to use symmetry arguments in that line of thought? I'm not great at these kinds of symmetric resistor problems, but can you say anything about what simplifications you might be able to do by cutting the circuit with a vertical line down the middle? Do you see any voltage arguments that may help you to redraw the circuit to make it easier to solve?
Yes, I was.
 
BTW, you are correct the answer is (4/5)R.
 
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  • #10
DaveE said:
BTW, you are correct the answer is (4/5)R.
Many thanks! And thank you for the tip, will be sure to look out for symmetry next time around; physics seems to like symmetry, hah!
 
  • #11
As a long time analog EE, I always look for symmetry and other graphical network transformations to simplify linear networks before I resort to KVL/KCL. You may not have seen it yet, and they aren't necessary here, but the Norton/Thevenin source transformations are really powerful.

If your KVL/KCL equations end up with 4th (or greater) order matrices, you're probably doing it the hard way.

PS: Also Miller's theorem, which is a bit more advanced, but great when you need it.
 
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  • #12
DaveE said:
As a long time analog EE, I always look for symmetry and other graphical network transformations to simplify linear networks before I resort to KVL/KCL. You may not have seen it yet, and they aren't necessary here, but the Norton/Thevenin source transformations are really powerful.

If your KVL/KCL equations end up with 4th (or greater) order matrices, you're probably doing it the hard way.
I am hopeful sometime in the future I will study EE; it's my favourite. And I am also hopeful that sometime soon, I will read up on that. I'd jumped headfirst into electromagnetism (Electromagnetic Fields by Wangsness, is it a good book? I find it's pretty great) back before the school year started, but ran out of time just after I'd read the Magnetic Multipoles chapter.

Once I find the time, I will reinvigorate my interest in EE and electromagnetism. For the meantime, thanks for the help, the sources, the tips and the encouragement.
 
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Lunar Manatee said:
I am hopeful sometime in the future I will study EE; it's my favourite. And I am also hopeful that sometime soon, I will read up on that. I'd jumped headfirst into electromagnetism (Electromagnetic Fields by Wangsness, is it a good book? I find it's pretty great) back before the school year started, but ran out of time just after I'd read the Magnetic Multipoles chapter.

Once I find the time, I will reinvigorate my interest in EE and electromagnetism. For the meantime, thanks for the help, the sources, the tips and the encouragement.
Sorry, I don't know that book. If you think it's great then I would stick with it. Within reason, the text you like and will read is the best one.

Ask in a new thread in the classical physics forum about EM texts, you'll get better answers.

I would encourage you to focus more on the physics, like EM, than circuits. You can study circuit analysis more easily on your own than the fundamental physics of EM. Analog EE circuit design is a bit of a dying or limited career IMO. Electricity & Magnetism will never go away.
 
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  • #14
Lunar Manatee said:
Many thanks! And thank you for the tip, will be sure to look out for symmetry next time around;
The basic way of using symmetry in a circuit is to carve the circuit up along symmetry axes.
In this case, one axis is ‘vertically' through the middle of the diagram and one horizontally.
Where the cut runs longitudinally through a resistor R, it produces two parallel resistors of 2R; where transverse through the middle of a resistor R, two of resistance R/2.
The top left quarter of the circuit now looks like (sorry, not good at online diagrams):
1747948065599.webp

At the cut through the vertical axis, being everywhere half way along the flow, all points must have the same voltage. This allows us to connect up C and D, making the R/2 (BC) and the R (BD) parallel, so they collapse into R/3.
This is in series with AB, so we collapse those into 4R/3.
Finally, that’s parallel with AD, giving 4R/5.
Lunar Manatee said:
physics seems to like symmetry, hah!
As do problem setters. The real world is often less kind.
 
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  • #15
haruspex said:
As do problem setters. The real world is often less kind.
Yes! Not my favorite kind of circuit HW because of that. In 30 years of analog design I never had this sort of problem IRL, at least in a discrete circuit form.
 
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  • #16
haruspex said:
not good at online diagrams
Me either. They are much more work than a photo of what you probably already sketched out..
1747953202618.webp
 
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