Why Is Nodal Analysis Confusing in This Circuit Example?

[arhzz]

Homework Statement

Use nodal voltage analysis to solve the circuit

Relevant Equations

Circuit Analysis

Hello! Consider this circuit

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.

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

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.

 

[berkeman]

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.

 

[BvU]

Just looking at this and wondering which are the knowns and which are the unknowns ...


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[arhzz]

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

 

[arhzz]

Just looking at this and wondering which are the knowns and which are the unknowns ...


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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.

 

[The Electrician]

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

 

[arhzz]

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]

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!