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enter image description here

To determine the equivalent resistance across \$AB\$,I noticed that the three resistors in the lower face of the cube are in parallel combination. So I replaced all three with a single resistor \$\frac{R}{3}\$ and the resulting circuit looks like the following(there will be another resistor \$R\$ along the diagonal of the upper face): enter image description here

Now no two resistor are seemingly in parallel or series combination and neither is there any kind of symmetry persisting in the circuit. Also I doubt if the way I placed the resistor \$\frac{R}{3}\$ in the correct way by removing the previous three wires. Adding a voltage source and using KVL also seems cumbersome. Any idea on how to simplify the circuit will be really helpful.

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  • \$\begingroup\$ @greybeard yes,could you please share where you will be applying the transformation? \$\endgroup\$
    – a_i_r
    Commented Dec 27, 2023 at 8:56
  • \$\begingroup\$ @a_i_r look at the vertical faces of your cube \$\endgroup\$ Commented Dec 27, 2023 at 8:58
  • \$\begingroup\$ @ConstantineA.B. I did but I am not sure of how to convert those star networks to delta since the branches are non planer which makes it difficult for me to visualize the delta network. A drawn circuit will be much appreciable. \$\endgroup\$
    – a_i_r
    Commented Dec 27, 2023 at 9:06
  • \$\begingroup\$ @a_i_r KCL says it's just \$\frac{139}{280}R\$. You can rest easy, now. \$\endgroup\$ Commented Dec 27, 2023 at 9:12
  • \$\begingroup\$ Some ideas instead of just an answer will be really appreciable @periblepsis. I just want to know how to approach such circuits. \$\endgroup\$
    – a_i_r
    Commented Dec 27, 2023 at 9:17

2 Answers 2

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One thing you should be able to see right away is that the resistor going directly from A to B may be removed before analysis, if you do these things by hand. It's directly across the terminals, so it will be in parallel with whatever you work out after first removing it. So I'm just tossing that out there in case it helps in some later practical cases.

KCL doesn't care, as it's just another graph edge. So I'll keep it for those purposes.

Here's your graph with the resistor you could remove (and later apply) circled in blue. The graph is otherwise labeled for KCL purposes below:

enter image description here

Node B has been grounded and I will inject a current of \$1\:\text{A}\$ into node A (also called \$v_1\$ above.) That will cause the voltage at \$v_1\$ to directly reflect the resistance magnitude.

I used SymPy/Python/Sagemath (all free to download and use) here.

e1 = Eq( 3*v1/R, v2/R + v5/R + 1 )               # nodal for v1 with 1 A injected
e2 = Eq( 4*v2/R, v1/R + v4/R + v5/R )            # nodal for v2
e3 = Eq( 5*v3/R, v4/R + v5/R )                   # nodal for v3
e4 = Eq( 3*v4/R, v2/R + v3/R + v5/R )            # nodal for v4
e5 = Eq( 4*v5/R, v1/R + v2/R + v3/R + v4/R )     # nodal for v5
solve( [ e1, e2, e3, e4, e5 ], [ v1, v2, v3, v4, v5 ] )[v1]
139*R/280

That's it. Feel free to ask for clarifications.

I discuss a related application of graph theory here. (KCL uses directed graph theory.) You can use graph theory to provide you directly with the meshes, if you wanted to apply mesh analysis, instead. I discuss that option there in the link (look for discussion of null spaces.)

Here's a Spice run using \$R=1\:\Omega\$:

enter image description here

The green circled value shows the ratio and is consistent with \$\frac{139}{280}R\approx 0.496428571\,R\$ when \$R=1\:\Omega\$.

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I have different results, I wonder where the elusive error could be:

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schematic

simulate this circuit – Schematic created using CircuitLab

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