Calculating Resistance - Electrical Circuits & Networks

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Discussion Overview

The discussion revolves around calculating equivalent resistances in electrical circuits, specifically addressing a problem presented by a professor. Participants explore concepts related to circuit topology, the Wheatstone bridge, and resistance calculations in a network of resistors.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that the equivalent resistances provided in the problem are incorrect, particularly noting that Rcd should be zero due to a short circuit.
  • Another participant emphasizes the absence of a connection dot at the junction where wires cross, which may affect the circuit's interpretation.
  • Discussion includes a reference to the Wheatstone bridge, with a participant expressing confusion about its relevance to the current problem.
  • A suggestion is made to rearrange the circuit diagram to better visualize the connections and identify the bridge.
  • One participant describes visualizing the circuit as a tetrahedron with resistances on its edges, proposing a method to project it flat for clarity.
  • Extra credit questions are posed regarding the calculation of resistance seen by a source and the creation of an Admittance matrix.
  • Participants express curiosity about the mathematical aspects of resistance calculations and matrix representations in circuit analysis.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the equivalent resistances and the identification of the Wheatstone bridge. The discussion includes both agreement on the need for clarification and ongoing uncertainty regarding specific calculations and interpretations.

Contextual Notes

Some assumptions about circuit connections and configurations remain unresolved, and participants rely on visual representations that may not be fully detailed in the discussion.

Hidd
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Hello!

I've took a course about Electrical circuits and Networks at the university, and i found out a problem which is uploaded by my professor, and i think he made a mistake!

i3ii4k.jpg


this is the solution

10gwl04.jpg


i think that the equivalent resistances are wrong (beside Rab), for example if you look at Rcd it must be 0, because it's short circuited.
And i coudn't find "the bridge" that he mentioned.

Thanks for help!
All the best,
 

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Hidd said:
i think that the equivalent resistances are wrong (beside Rab), for example if you look at Rcd it must be 0, because it's short circuited.
Note that there is no connection "dot" where those wires cross :wink: Every other junction has a dot.
 
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gneill, Thank you for your reply and i think you're right! but what about the "bridge"?
we worked on "Wheatstone bridge" which looks like that:
https://www.electronics-tutorials.ws/wp-content/uploads/2013/10/wheatstone1.gif

but i couldn't find similarities with the one that i mentioned earlier!
 
You may need to re-draw your circuit (rearrange the diagram) in order to be able to visualize the bridges. For example, if you remove the resistor that connects node A to node B, then you have:
upload_2018-2-15_17-38-37.png


Which can be rearranged as:
upload_2018-2-15_17-39-15.png


Note that the same resistors still connect to nodes with the same labels; the topology of the circuit is the same as before. Now can you "see the bridge"?
 

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Yep! now i see it, Thank you very much gneill.
 
You can think of ABCD as a tetrahedron, with resistance R on each of the edges.
I was having a hard time drawing one, so I found an image of one.
z1bpW8g.jpg

Then you can look at it from above A, and project it flat, so A is in the center of the triangle, and it connects outward to B C and D.
 

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For extra credit. what is the resistance seen by the source Uo in Fig. 3.1 if ALL (including Rd) of the resistors are 50 ohms?
 
@The Electrician. Can I get extra credit for clarifying the diagram and setting the source to one amp ?
Network_R.png
 

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For doubleplusgood credit, calculate the resistance from every node to every other node. First number all the nodes in the original network.

Network.jpg


Then do the math:

All Resistances.png

In the final result matrix the resistance from node j to node k is given by the (j,k) element of the matrix. The resistance from node j to k is, of course, the same as the resistance from node k to node j, hence the matrix is symmetric. And we see all zeroes on the main diagonal because the resistance from a node to itself is zero.
 

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Last edited:
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  • #10
The Electrician said:
For doubleplusgood credit, calculate resistance from every node to every other node...
Then do the math:

View attachment 221247
I am now curious, since I just went through a Linear Algebra course: what is the key to creating the Admittance matrix by inspection?
 
  • #11
scottdave said:
I am now curious, since I just went through a Linear Algebra course: what is the key to creating the Admittance matrix by inspection?

 
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  • #12
The Electrician said:

Thanks.
 

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