[SOLVED] Node voltage analysis

[Teszla]

I want to do a nodal analysis in order to find out the voltage V0.

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Doing KCL on the node, I get:

\[\frac{V_{0}}{1\Omega }+\frac{V_{0}+3V}{3\Omega }-2A=0\\ \frac{V_{0}}{1\Omega }+\frac{V_{0}}{3\Omega }=2A-\frac{3V}{3\Omega }\\ \frac{4V_{0}}{3\Omega }=1A\\ V_{0}=\frac{3}{4}V\]

This is obviously wrong though. How would the correct expression be? Do I need to analyse several nodes?

 

[KerimF]

I see 3 nodes here, don't you see them? Let us call them a, b and c. Could you?
Actually the are 4 nodes (the ground is also a node) but since here it has two branches only, we can ignore it.

 

[Teszla]

Sure.

**broken link removed**

I assume that node a has the same voltage as v0.

I'm mostly wondering about the analysis between node a and b. I wrote before that it's

\[\frac{V_{0}+3V}{3\Omega }\]

But I suspect something is wrong about this. How would the correct expression be?

 

[KerimF]

Now we have Va, Vb and Vc.

From the circuit:
Va = Vo
Vc = 6V
Vb is unknown.

So now we have at node (a) :
( Vo + 3 - Vb ) / 3 + Vo / 1 - 2 = 0

Could you write the equation of node (b)?

 

[Teszla]

Thanks, I solved it.

 

[KerimF]

I knew you can do it. And someday you will surprise yourself and solve one or more problems that no one thought solving yet.
Studying (in any field) is like playing a game. One learns how to win as long he enjoys playing it

 

[Teszla]

Since you were so helpful maybe I could ask about NVA for another circuit in the same thread.

**broken link removed**

Here I wonder, when doing NVA for V0, how do you treat the dependant voltage source?

 

[KerimF]

Sorry, where are the labels of nodes on the new circuit?

 

[Teszla]

Let's say V0 is the node in the top right corner and V1 in the top left corner, and the bottom being GND.

 

[KerimF]

In general, it is better selecting node labels that are different from all other symbols on the circuit.
And in such analysis, it is better referring to each node with one letter (to simplify the writing of the equations).

a is for your GND node.
b is for your Vo node.
c is for the third node.

Now we have:
Vb - Va = Vo
Vb - Vc = 2Ix

First, let us write the nodal equation at "a".

 

[Teszla]

First, let us write the nodal equation at "a".

Do you usually write equations for the GND node? I would guess it's:

(Va-Vb)/2Ohm-2A+(Va-Vc)/2Ohm+4A=0

Is this correct? What about the equations for the other nodes? I especially wonder about how to treat the dependant voltage source.

 

[KerimF]

Please, for clarity, don't add units (like Ohm and A) while writing an equation.

So we have now:
Vb - Va = Vo
Vb - Vc = 2Ix
( Va - Vb ) / 2 - 2 + ( Va - Vc ) / 2 + 4 = 0

It is time to write the last equation of Ix. Could you?

 

[Teszla]

Ix = Vc/2 ?

Is there any special reason for including node Va? Since Va is GND, isn't Va by definition = 0 V?

 

[KerimF]

You can do it, I mean assuming Va = 0

So, we have now:
Vb = Vo
Vb - Vc = 2Ix
- Vb / 2 - 2 - Vc / 2 + 4 = 0

And the ( Vc - Va ) / 2 = Ix becomes as you wrote:
Vc / 2 = Ix
or
2Ix = Vc

Now we have an algebra exercise; 3 equations and 3 unknowns (Vb, Vc and Ix).
I hope you can solve them to get Vb because Vb = Vo here, (by eliminating Ix and Vc in a way or another).

 

[Teszla]

Aha.

Just out of curiousity, is it possible to do KCL on nodes a and c also? If so, how would they look? I.e. how would one treat the dependent voltage source?

 

[KerimF]

You can do it. But we have to add another unknown first as the current Ibc (from b to c for example) and write the equations at node b and c. Obviously, we will get more equations and if solved (by eliminating Ibc too) we will surely get the same answer for Vb (Vo).

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