Calculating Resistance - Electrical Circuits & Networks

AI Thread Summary
The discussion revolves around a problem in an Electrical Circuits and Networks course, where a participant believes their professor made an error in calculating equivalent resistances, particularly noting that Rcd should be zero due to a short circuit. The conversation shifts to the concept of a Wheatstone bridge, with participants discussing how to visualize the circuit by rearranging it. They explore the topology of the circuit, comparing it to a tetrahedron and discussing resistance calculations between nodes. Additionally, there is curiosity about creating an Admittance matrix by inspection, indicating a deeper interest in the mathematical aspects of circuit analysis. The thread highlights the collaborative effort to clarify circuit concepts and calculations.
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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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?
 
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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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