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enter image description here

How we can find the answer in the simplest way between A and G without any advanced techniques.

My Approach

First, I noticed that there are twelve resistances arranged in the form of a cube, and I need to find the equivalent resistance between corners A and G. This problem is quite complex due to the symmetrical nature of the cube and the different resistances.

To start, I labelled the vertices of the cube for clarity. I see that vertices A, B, C, D, E, F, G, and H form the cube, and each edge has a specific resistance value. The cube's symmetrical properties might allow me to simplify the problem by considering the symmetry and equivalent paths for the current between points A and G.

My first step was to identify all possible paths from A to G:

  1. A → B → C → G
  2. A → D → C → G
  3. A → E → H → G
  4. A → E → F → G
  5. A → D → F → G

Next, I calculated the resistances along these paths and then combine them appropriately using series and parallel combinations.

For instance, in path 1 (A → B → C → G), I had:

  • A to B: 3Ω
  • B to C: 6Ω
  • C to G: 2Ω

These resistances are in series, so the total resistance for this path would be: R_1 = 3Ω + 6Ω + 2Ω = 11Ω

I repeated this for each path and then tried to find the equivalent resistance by combining these paths in parallel. However, the calculations got quite intricate.

Can anyone please help

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  • 1
    \$\begingroup\$ there's a strong 3-fold symmetry going on \$\endgroup\$
    – Neil_UK
    Commented Jun 19 at 13:27
  • 1
    \$\begingroup\$ You missed one: 6. A → B → F → G. Yes, symmetry! )) \$\endgroup\$
    – Astrogator
    Commented Jun 20 at 15:06

4 Answers 4

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  • Redraw it from a 3D shape to a 2D shape: -

enter image description here

  • Recognize the symmetry of the design

  • That symmetry means all the currents through the 3 resistors connected to (A) are equal

  • That means you have equal voltages at (E), (D) and (B)

  • Because they are equal you can apply the two shorts (see red text on image).

  • It's the same argument for currents through the 3 resistors connected to point (G)

  • Apply two more shorts between (F) and (H) and, (H) and (C).

  • Drill down until the number of resistors are reduced to the bare minimum (paralleling)

Can you take it from here now?

My answer is 3 Ω.

Or just simulate it: -

enter image description here

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  • \$\begingroup\$ Thank you so much for answering it is really helpful. \$\endgroup\$ Commented Jun 19 at 14:13
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As Andy aka suggested (in the now deleted comment), the easiest way is a simulator: connect your resistors, apply a 1 A current source, and the voltage you measure is U=RI = R * 1 A.

If you want to do it by hand, it all depends on your strengths:

  • if you are good at matrix manipulations: write all your equations and solve the matrix
  • you can probably use Millman's theorem to simplify calculations
  • or you can look in symmetries to simplify the problem: from node A, you always have a 3 Ω resistor, leading to a node with a 6 Ω and a 12 Ω resistor. The nodes at the ends of the 6 or 12 Ω resistors receive the other one (12 or 6 Ω) and lead to G through a 2 Ω resistor. So you have only 4 currents and 4 voltages to consider.
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Network analysis with three different methods. It seems to me that the equivalent resistance sought is 3_Ohm:

enter image description here

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  • \$\begingroup\$ This is an impressively large amount of math to get to the anticlimactically simple answer of.. 3. Thank you, @Franc, I learned something today. \$\endgroup\$ Commented Jun 20 at 18:50
  • \$\begingroup\$ @Stephan Samuel: This is what electrical engineering offers to tackle problems from the simplest to the most complicated. It is not simply academia. Inspection analysis is good for small networks. \$\endgroup\$
    – Franc
    Commented Jun 21 at 5:37
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In the general case, you can use the Star-Mesh transform (or its three-resistor simplification, the Y-Δ transform) to remove any one vertex that doesn't have an external connection. This leads to resistors in parallel, which can be merged (which leads to resistors in series). You can then repeat for another vertex until there's only one resistor between each external connection.

The Star-Mesh transform increases the number of resistors in the system, but the Y-Δ doesn't. They both lead to simplifications.

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