# NVM Solving for variables?

#### thatsmessedup

Joined Feb 15, 2018
28
When we were doing KVL and KCL, it was trivial to perform reduced row echelon form with my calculator to solve for the variables. But for NVM I am unsure on how to do the actual solving? I have written all of the equations to account for all the variables but how do I plug them into my calculator?

#### WBahn

Joined Mar 31, 2012
24,699
You have an error in one of your equations. After setting up your equations, a very good habit to get into is checking each equation term-by-term to make sure that it is correct. When verifying an equation that you've already written down, you are looking at it with a different level of attention to detail than you are when you are figuring out what to write down to begin with.

You are not really applying NVM, but rather something that is moderately related to it.

Remember that NVM is nothing more than a formalized way of applying KCL at every essential node in terms of the node voltages. In your circuit you have three places in which a supernode should be used (at the level of detail that they seem to want you working at right now) whereas you only have one.

#### thatsmessedup

Joined Feb 15, 2018
28
Thanks! I see the mistake I made on 3 (C+B). So my main question is how to plug it into the calculator, easy peasy like I did with KVM and KCL type of questions?

#### thatsmessedup

Joined Feb 15, 2018
28
hold up wait a minite! think I got this.

#### thatsmessedup

Joined Feb 15, 2018
28
AAAha... I have figured it out. I had also messed up equation 7. Thanks again for the push in the right direction.

#### The Electrician

Joined Oct 9, 2007
2,737
Your result is not correct. Look at the circuit; the node you have colored red is connected to ground. Assuming the usual convention, the voltage at the reference node (ground) is taken to be zero volts. This means that Vy is zero, and your equation 6 is not correct. Since the dependent voltage source Vs1 has value 25*Vy, its output must also be zero; therefore Vb is also zero.

You may have other errors but I haven't tried to find any.

#### WBahn

Joined Mar 31, 2012
24,699
Your result is not correct. Look at the circuit; the node you have colored red is connected to ground. Assuming the usual convention, the voltage at the reference node (ground) is taken to be zero volts. This means that Vy is zero, and your equation 6 is not correct. Since the dependent voltage source Vs1 has value 25*Vy, its output must also be zero; therefore Vb is also zero.

You may have other errors but I haven't tried to find any.
How do you figure that Vy = 0 just because the red node is 0 V relative to ground?

In order for Vy to be 0 V, the red node would have to be at the same potential as the purple node. While I agree that the red node is at 0 V relative to ground, there is no basis (that I can see) for asserting that the purple node is also at 0 V relative to ground. What is your reasoning for concluding that it is? Even if you could conclude that ix = 0 A so that i_s2 = 0 A, that would not imply that the voltage across i_s2 was necessarily 0 V. If anything, it would suggest that it is not zero since it has to be whatever voltage is needed to prevent current flowing through it.

#### The Electrician

Joined Oct 9, 2007
2,737
How do you figure that Vy = 0 just because the red node is 0 V relative to ground?
It looks to me like Vy IS the red node; it's grounded, is it not?. I don't think that the purple node is at 0 V relative to ground.

#### WBahn

Joined Mar 31, 2012
24,699
It looks to me like Vy IS the red node; it's grounded, is it not?. I don't think that the purple node is at 0 V relative to ground.

Vy is defined at the voltage at the red node relative to the purple node.

#### The Electrician

Joined Oct 9, 2007
2,737
View attachment 149405

Vy is defined at the voltage at the red node relative to the purple node.
This is a bit of a trick question to my mind. We already have a definition for the voltage across the is2 source, Ve.

Since the red node is ground, "the voltage at the red node relative to the purple node" is -Ve. Saying this is to use the purple node as a reference for this one definition, but Ve is already defined relative to ground, isn't it?

It seems clear that the voltages Va, Vb, Vc, Vd, and Ve are defined relative to ground.

How is "the voltage at the red node" defined? Why would it be defined in any other way than relative to ground, as all the other voltages defined ? Vy is adjacent to the red node so like all the other designators which are taken to refer to the node they are closest to, it's reasonable to take Vy to be a designator for the red node, which is ground, so Vy is at zero volts if it is taken to be defined relative to ground as every other voltage is.

#### WBahn

Joined Mar 31, 2012
24,699
This is a bit of a trick question to my mind. We already have a definition for the voltage across the is2 source, Ve.
Ve is NOT defined as the voltage across the is2 source, it is defined as the voltage, relative to the common reference node, at node e. That it is numerically equal (in magnitude) to the voltage across the is2 source is a coincidence on two counts -- first it is a coincidence of the topology in that the other side of the source happens to be connected to the red node and, second, it is a coincidence of the arbitrary assignment of the red node as being the common reference node.

Surely you would agree that if someone happened to decide that the green node would make a better common reference node that the circuit's behavior will not change. But if we use Ve (or -Ve) as the controlling voltage for the vs1 dependent voltage source, it would change since Ve changes with a change in reference node. But vs1 is not controlled by the voltage at a node, it is controlled by the voltage difference between two nodes (which is virtually always the case with a voltage-controlled source), thus the circuit needs to define the controlling voltage as being the voltage difference between the correct two nodes and that definition is completely independent of which node we happen to call "ground".

#### The Electrician

Joined Oct 9, 2007
2,737
Ve is NOT defined as the voltage across the is2 source, it is defined as the voltage, relative to the common reference node, at node e. That it is numerically equal (in magnitude) to the voltage across the is2 source is a coincidence on two counts -- first it is a coincidence of the topology in that the other side of the source happens to be connected to the red node and, second, it is a coincidence of the arbitrary assignment of the red node as being the common reference node.
The behavior of any particular circuit is dependent on many coincidences of the topology. Because one end of is2 is connected to ground, defining the voltage across it to be equal to the voltage at Ve relative to ground (for the purpose of analysis of this circuit) is to use an equivalent definition, for this particular circuit, and it's this particular circuit with all the coincidences that accompany it that I'm concerned with. To say otherwise for this circuit, as the lawyers would say, is a distinction without a difference. For some other choice of reference node there would be a difference, but not for this circuit as presented, coincidences and all.

Why is it that the all of the other sources have their textual descriptions immediately adjacent to them? The Vy symbol is closer to the red colored parts than to the is2 source itself; why wasn't it placed immediately adjacent to is2 if it was intended to represent the voltage across is2? I don't think it's unreasonable to believe that Vy is a designator for the red colored node. That would make for a nice trick to ensnare the student.

Maybe the TS redrew the circu ciit, and chose the ground node. Look just below the green node. There's a small vertical line, and something just below that line that I can't make out. There are also 5 little horizontal lines just above the existing ground symbol. Do those things indicate that the TS modified the drawing he was given? If some node other than the red node (let's say the green node) were the reference node, then the proximity of the Vy symbol to the red color would even more strongly convince me that Vy would be the designator for the red node, and not the voltage across is2. Then Vy would represent the voltage of the red node relative to the green node, and 25 Vy as the controlling voltage of vs1 would workable even though this would give a different result than the given problem, because then Vy would not be equal to -Ve.

Since you are not the author of this problem, you are speculating as much as I am.

Surely you would agree that if someone happened to decide that the green node would make a better common reference node that the circuit's behavior will not change. But if we use Ve (or -Ve) as the controlling voltage for the vs1 dependent voltage source, it would change since Ve changes with a change in reference node.
If the green node were the reference node, and if it were wanted that the controlling voltage be the voltage across is2, then that controlling voltage would have to be specified in such a way that made that clear. An example of where an unambiguously specified controlling voltage is across a component without one end grounded is the voltage Vx here: https://forum.allaboutcircuits.com/threads/dc-circuit-help-with-dependent-current-source.36737/

If a controlling voltage were given as a difference betwen two node voltages, for example (Va - Vb) it would be clear. If a controlling voltage is given as a single node designator such as V2 here: https://forum.allaboutcircuits.com/attachments/q-png.55430/ then it's understood that this is a difference of two node voltages, namely V2 and the reference node.

You know all this, and I know all this, but I still think that the meaning of the Vy symbol in the circuit presented in this thread is ambiguous for the reasons I've given.

But vs1 is not controlled by the voltage at a node, it is controlled by the voltage difference between two nodes (which is virtually always the case with a voltage-controlled source), thus the circuit needs to define the controlling voltage as being the voltage difference between the correct two nodes and that definition is completely independent of which node we happen to call "ground".
I fully understand that you believe the Vy symbol would represent the voltage across is2 regardless of which node is chosen as the reference node. I've explained why I don't interpret Vy the same way you do. If the Vy symbol had been placed much closer to is2 I would agree, but it is closer to the red colored parts of the circuit than to is2.

However, it's easy to solve the circuit either way. Equation 6 of the TS's analysis now says Vy + Ve = 0. That is the same as saying Vy = -Ve, and I think that's the result of your interpretation (and the TS's interpretation apparently) for the choice of reference node this problem has given us. If the TS wanted to get the solution for my interpretation all he has to do is let equation 6 be Vy = 0.

Either way, the TS's solution is not correct because of other errors. If he ever returns we can help him eliminate them.

#### WBahn

Joined Mar 31, 2012
24,699
The behavior of any particular circuit is dependent on many coincidences of the topology. Because one end of is2 is connected to ground, defining the voltage across it to be equal to the voltage at Ve relative to ground (for the purpose of analysis of this circuit) is to use an equivalent definition, for this particular circuit, and it's this particular circuit with all the coincidences that accompany it that I'm concerned with. To say otherwise for this circuit, as the lawyers would say, is a distinction without a difference. For some other choice of reference node there would be a difference, but not for this circuit as presented, coincidences and all.
Then let's consider this circuit as presented. You said that we already had a defined voltage, namely Ve. That's not the case. The circuit, as presented, does not define ANY node voltage -- I'm not even sure that it defines the common reference node as the ground symbol appears hand drawn -- note particularly that it does not connect to the node at the connection dot, that the line connecting it actually crosses the node line, and there is no connection dot where it does cross. But even if it is defined, none of the other node voltages are defined by the problem as presented -- those are all names assigned by the TS.

Why is it that the all of the other sources have their textual descriptions immediately adjacent to them?
Simple. As you say, those other textual descriptions are "their" descriptions -- they define and describe the source. Vy is NOT the textual description for a source -- it is merely the definition of the controlling signal for one of the sources. Since the controlling signals are almost never immediately adjacent to the source they control, it is hardly reasonable to draw a conclusion based on this one not being close to the source it is associated with (which, by the way, is vs1). This would be the same as drawing some conclusion based on ix, the controlling signal for is1, not being drawn immediately adjacent to some source.

The Vy symbol is closer to the red colored parts than to the is2 source itself; why wasn't it placed immediately adjacent to is2 if it was intended to represent the voltage across is2? I don't think it's unreasonable to believe that Vy is a designator for the red colored node. That would make for a nice trick to ensnare the student.
The location of the label, as far as whether it is in the middle or closer to one node or the other, is irrelevant. The far more important information that trumps that is that it is located in line between the two polarity marks so as to associate them as a set. This is the preferred layout. Because of the limited horizontal extent of both nodes, there is no way to move them outward far enough to enable the label to be centered vertically while still being inline with the marks. So it either had to be above or below the outline of the source symbol and the usual convention when this is the case is to place it on the positive side of the definition.

Maybe the TS redrew the circu ciit, and chose the ground node. Look just below the green node. There's a small vertical line, and something just below that line that I can't make out.
That vertical line connects to the supernode definition. I can't make out what it says either. Looks like Σn, which is perhaps a shorthand used by the TS's professor for "sum of nodes" to indicate a supernode. But that IS pure speculation on my part. Since that label is never used anywhere in the analysis, I can't see that it matters.

There are also 5 little horizontal lines just above the existing ground symbol. Do those things indicate that the TS modified the drawing he was given? If some node other than the red node (let's say the green node) were the reference node, then the proximity of the Vy symbol to the red color would even more strongly convince me that Vy would be the designator for the red node, and not the voltage across is2. Then Vy would represent the voltage of the red node relative to the green node, and 25 Vy as the controlling voltage of vs1 would workable even though this would give a different result than the given problem, because then Vy would not be equal to -Ve.

Since you are not the author of this problem, you are speculating as much as I am.
I don't believe I am, precisely because I am using information in the circuit, as presented, that you are steadfastly refusing to consider.

You are choosing to ignore two very important pieces of information. Look at the information I circled in green. In addition to the name designator, there are the two polarity designators, a + and a - symbol. In order to maintain your interpretation, you need to provide reasonable explanations for what each of those mean, why they are where they are, and why they do NOT indicate that Vy is defined as the voltage of the red node relative to the purple node. The simplest (and as near as I can see only) reasonable way to interpret them is that they form the classic and usual definition of a named voltage differential between those two nodes.

#### thatsmessedup

Joined Feb 15, 2018
28
I can't begin to understand what you guys are talking about. I've solved the problem and have gotten the same answer as my teacher (ix = -0.379[A]). If there is a mistake in one of my labelings or maths please refer specifically to them in my work and I will be the wiser for it. Otherwise thank you for all your help in the matter.

#### WBahn

Joined Mar 31, 2012
24,699
I can't begin to understand what you guys are talking about. I've solved the problem and have gotten the same answer as my teacher (ix = -0.379[A]). If there is a mistake in one of my labelings or maths please refer specifically to them in my work and I will be the wiser for it. Otherwise thank you for all your help in the matter.
It's fine if you aren't following all the details at this point. Let them simmer in the back of your mind and, as you work more problems, particularly ones in which you have to come up with the schematics because you are involved in designing the circuit, the more subtle points being discussed will invariable raise their ugly heads again and you will be better able to reflect, even if just subconsciously, on discussions like this and make better sense of what you are doing.

I haven't looked in detail through your subsequent equations and so I don't know if you have any errors in them or not.

But now that you have an answer, you can (and should) confirm that it is the correct answer from the proposed answer itself. Don't just rely on "I got what the teacher got, so it must be right" approach. Teachers and and do make mistakes and they can and do make the same simple and easy mistakes that some of their students will make.Write the node voltages that you determined next to each node and, from that, start determining the currents in each branch. Work your way over to ix and see if they agree.

That is one of the most beautiful things about most fields of engineering -- even in the "real world" the correctness of the answer can usually be confirmed from the answer itself irrespective of how you got it.

#### Jony130

Joined Feb 17, 2009
4,982
I upload orginal question and circuit.

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#### WBahn

Joined Mar 31, 2012
24,699
I upload orginal question and circuit.
Thanks, Jony. I think that makes it pretty clear what Vy has to mean.

#### The Electrician

Joined Oct 9, 2007
2,737
I can't begin to understand what you guys are talking about. I've solved the problem and have gotten the same answer as my teacher (ix = -0.379[A]). If there is a mistake in one of my labelings or maths please refer specifically to them in my work and I will be the wiser for it. Otherwise thank you for all your help in the matter.
Here is the matrix you derived and its solution showing all of your variables, not just Ix. I checked the numbers in this matrix several times and I believe I'm using the same values shown on your work sheet.

To 8 digits, you got Ix = +0.37930552 A. I'm showing more digits than you did for a reason that will be explained. Notice that the sign which the matrix arithmetic delivers is positive even though you showed a result with a minus sign. Did your calculator actually return a result with a minus sign?

Each row in the matrix corresponds to one of your equations, i.e., row 1 is Eq. 1, row 2 is Eq. 2, etc.

In row 1 (equation 1) you left out the -12/20. This should have been added to the far right column.

In row 5, column 4 you have 1/22; it should be -1/22

In row 7, column 1 you have 1/20; it should be -1/20

In row 7, column 8 you have 12/20; it should be -12/20

If these changes are made this is the result:

Now the result for Ix is -0.37869988. Notice that if the magnitude of your result and of this result are rounded to 3 places, the result is .379

Apparently the errors you made sufficiently compensated for one another that your final result magnitude, rounded to 3 digits is the same as the rounded magnitude with your errors corrected. This could lead one to overlook the errors and think you had done everything correctly. The sign difference is another matter.

#### The Electrician

Joined Oct 9, 2007
2,737
Here is the result if Vy is taken to be zero:

The value of Ix rounded to 3 is -0.377; only a little bit different than the result with Vy = -Ve

#### WBahn

Joined Mar 31, 2012
24,699
Here's my annotated drawing that would be my starting point (though I would probably choose G as my reference node since that directly solves for node D as well, but I chose to assign the node with the most branches connected to it as the reference node as this is more often than not the best choice, all else being equal).

Notice that I have colored the essential nodes and the nodes involved in supernodes. I have chosen to use the same color for each of the nodes that are associated with a given supernode. Thus we will need four node equations and since we have three supernodes we will need three auxiliary (constraint) equations that stitch the three supernodes together. We also have the controlling equations for the two dependent sources, so that's a total of nine equations. As usual we get one for free (due to the arbitrary common reference node), so we need to come up with eight equations.

First, let's get the auxiliary equations out of the way:

Blue:
$$V_c \; = \; V_e \; + \; V_{s2}$$

Red:
$$V_b \; = \; V_{s1}$$

Green:
$$V_d \; = \; V_g \; + \; V_{s3}$$

We only need to write node equations for the blue, green, and purple nodes because all parts of the red node are taken care of by the auxiliary equations since the reference node is part of that supernode.

Blue:
$$V_c \(\frac{1}{R_3}$$ \; + \; V_e $$\frac{1}{R_8}$$ \; = \; -i_{s1}\)

Purple:
$$V_f \( \frac{1}{R_5} \; + \; \frac{1}{R_6}$$ \; - \; V_g $$\frac{1}{R_6}$$ \; = \; -i_{s2}\)

Green:
$$-V_b \( \frac{1}{R_2} \; + \; \frac{1}{R_7}$$ \; + \; Vd $$\frac{1}{R_2} \; + \; \frac{1}{R_4}$$ \; - \; V_f $$\frac{1}{R_6}$$ \; + \; V_g $$\frac{1}{R_6} \; + \; \frac{1}{R_7}$$ \; = \; 0\)

Next we have our two control equations:

i_s2:
$$i_{s2} \; = \; 0.2 \cdot i_x$$

v_s1
$$v_{s1} \; = \; 25 \cdot V_y$$

Finally, we need to relate our two control variables to the node voltages (since we are using the node voltage method):

$$V_y \; = \; -V_f$$

$$i_x \; = \; V_e \(\frac{1}{R_8}$$ \; + \; V_f $$\frac{1}{R_5}$$\)

Now that we have all of the setup equations written, the next (and most frequently overlooked step) is to go back and verify their correctness one at a time and term by term. ALL of the EE stuff related to this problem is captured by these equations -- everything from this point on is pure math. If we have made a mistake in any of these equations, the math can't catch it since we are simply solving the math associated with a different problem.

At this point we might notice that R1 appears in none of our setup equations. Does this make sense? If not, then NOW is the time to resolve this issue because, if we have made a mistake by not including it, then everything we do from this point on is pure wasted effort.

In this case, it DOES make sense, because it is connected in series to a current source and so the voltage at node a will become whatever is needed to make 3 A flow through R1 regardless of what (finite) value R1 happens to have.

Once we have verified the set-up equations, it is time to turn your favorite crank to solve them. After that, the second most frequently overlooked step is to check the correctness of the resulting answer.

EDIT: Added second controlling equation that got left out. Thanks to The Electrician for pointing it out.

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