Why Is Nodal Analysis Confusing in This Circuit Example?

Click For Summary
SUMMARY

The discussion centers on the application of nodal analysis to a specific circuit example, focusing on the preparation of the conductance matrix. The participants clarify that the equivalent resistance of three resistors of value R and two resistors of value 2R connected in parallel from node 1 to ground results in R/4. They emphasize that while simplifying the circuit is recommended, it is not mandatory for solving using Kirchhoff's Current Law (KCL). The conversation concludes with a participant successfully drawing the circuit and arriving at the solution as per the textbook.

PREREQUISITES
  • Nodal analysis and conductance matrix setup
  • Understanding of Kirchhoff's Current Law (KCL)
  • Basic circuit theory involving resistors in series and parallel
  • Familiarity with circuit simulation tools like LTspice
NEXT STEPS
  • Study the principles of nodal analysis in electrical circuits
  • Learn how to derive equivalent resistance for resistors in parallel
  • Explore the use of LTspice for circuit simulation and analysis
  • Review examples of KCL applications in complex circuits
USEFUL FOR

Electrical engineering students, circuit designers, and anyone looking to deepen their understanding of nodal analysis and circuit simplification techniques.

arhzz
Messages
284
Reaction score
58
Homework Statement
Use nodal voltage analysis to solve the circuit
Relevant Equations
Circuit Analysis
Hello! Consider this circuit

Snimka zaslona 2022-06-05 214729.png


Now I am susposed to solve this using nodal analysis. The first step is to "prepare the network" so that I can set up the conductance matrix.

Now I've tried it for a bit and was not really able to do it. So I went ahead and looked at the solutions and this is how it looks like.

weqweqwe.png


Now I think I've figured out how the got the right part of the circuit (the K2). Now the reference node is clear all of the wires are connected to K0 to it. Now looking at the voltage source that is to the right we can transform that into a current source and add the conductances of the resistors (R 2R and 4R).

## \frac{1}{4R} + \frac{1}{R} + \frac{1}{2R} ## this gives us the conductance G ## G =\frac{7}{4R} ## and if we want the resistance we need to take 1/G and we get the solution from the book. Now the current that the source is "emitting" is simply voltage divided through resistance,so and the nodal voltage is from K2 to K0, so its ##U_{20} ## The right part is pretty much clear but the left one is confusing me.

Now we can transform Uq1 just like we did for Uq3 but,I need to add the two resistors that are infront of it and behind it(since they are in series) .If we do that the current will be calculated like Uq1/2R. Now what I do not understand is how they get to the value of R/4? I've tried redrawing the circuit many times and I have not been able to do it. I can see that one of the resistors is short circuted but I can't seem to get to the R/4. This is how I have the circuit drawn
kita.png


Take note that the current source at R1 and R4 are missing since I can't find them in LT spice. Now if I calculate the total resitance here I don't get the R/4 like in the solution.I've tried it a bunch of times and it feels like guessing at this point. What am I doing wrong ?Thanks and excuse the long post.
 
  • Like
Likes   Reactions: Delta2
Physics news on Phys.org
arhzz said:
Now I am susposed to solve this using nodal analysis. The first step is to "prepare the network" so that I can set up the conductance matrix.
Since you haven't received a reply yet...

Are you required to try to simplify the circuit before solving it? For me, the beauty of using KCL to solve networks is that you can write a number of equations that equals the number of unknowns, so you should be able to solve for the circuit voltages and currents without needing to simplify it. Have you tried solving it yet without simplifying? At the very least, it will be a good cross-check of your answers if you do end up simplifying the circuit before writing KCL equations.
 
  • Like
Likes   Reactions: hutchphd
Just looking at this and wondering which are the knowns and which are the unknowns ...
:rolleyes:

##\ ##
 
berkeman said:
Since you haven't received a reply yet...

Are you required to try to simplify the circuit before solving it? For me, the beauty of using KCL to solve networks is that you can write a number of equations that equals the number of unknowns, so you should be able to solve for the circuit voltages and currents without needing to simplify it. Have you tried solving it yet without simplifying? At the very least, it will be a good cross-check of your answers if you do end up simplifying the circuit before writing KCL equations.
Well it is recommened that you simplify the circuit, but it is not mandatory
 
BvU said:
Just looking at this and wondering which are the knowns and which are the unknowns ...
:rolleyes:

##\ ##
To ne honest that has not been stated in the question. They want me to set up the conductance matrix, the voltages are than from node 1 to node 0, node 2 to node 0.
 
  • Like
Likes   Reactions: BvU
Resistors in series with current sources and resistors in parallel with voltage sources have no effect on the larger circuit. So the resistor in series with current source Iq2 can be eliminated.

I see 3 resistors of value R connected from node 1 to ground, and 2 resistors of value 2R connected from node 1 to ground. The equivalent resistance of those 5 resistors in parallel is R/4
 
  • Like
Likes   Reactions: arhzz
The Electrician said:
Resistors in series with current sources and resistors in parallel with voltage sources have no effect on the larger circuit. So the resistor in series with current source Iq2 can be eliminated.

I see 3 resistors of value R connected from node 1 to ground, and 2 resistors of value 2R connected from node 1 to ground. The equivalent resistance of those 5 resistors in parallel is R/4
I think I see it as well now,I'll try drawing it now,to see if I understood it well
 
arhzz said:
I think I see it as well now,I'll try drawing it now,to see if I understood it well

Okay took me a litle long while but this is how I've drawn the circuit.And from this point on I have been able to get to the solution like in the book.

Thanks for the help everybody!

Simplified circuit dark 02.png
 
Last edited by a moderator:
  • Like
Likes   Reactions: dlgoff

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K