# Using Nodal Analysis to Solve Independent Voltage and Current Sources Circuit

Thread Starter

#### project_science

Joined Sep 14, 2018
21
Hi,

Here's the goal of my attached HW: I have to use nodal analysis to solve the circuit for Vo. Vo is the voltage across the open ports.

Quick notes
1. Next to the voltage source, Vs, there's an 'Ao' written. Please ignore it.
2. The two attached pages start numbering on p.3 and go to p.4

My question
I solved the circuit as best I could. I want to know if my approach and answer to solving for Vo looks correct. Vs is unknown at the start, and thus the answer must be implicit and contain Vs.

Thanks!

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#### Zeeus

Joined Apr 17, 2019
615
Hi,

Here's the goal of my attached HW: I have to use nodal analysis to solve the circuit for Vo. Vo is the voltage across the open ports.

Quick notes
1. Next to the voltage source, Vs, there's an 'Ao' written. Please ignore it.
2. The two attached pages start numbering on p.3 and go to p.4

My question
I solved the circuit as best I could. I want to know if my approach and answer to solving for Vo looks correct. Vs is unknown at the start, and thus the answer must be implicit and contain Vs.

Thanks!
1st equation : V1 - v2 is not Vs...
Solved it
Looking your solution. what was your answer with superposition?

2nd and 3rd equation correct

Think only problem is equation 1

#### RBR1317

Joined Nov 13, 2010
610
One needs to develop a consistent approach to writing node equations. My recommendations are:

1. Identify the reference node with a ground symbol.
2. Name any nodes that don't already have an assigned name.
3. Colorize the nodes with different colors. I use felt tip markers on the schematic & only mark the nodes that need node equations. (If the reference node is complicated, I like to color that blue even though the reference node won't have a node equation.)
4. Write each node equation as the sum of all currents equals zero. Begin each term of the node equation with the voltage of that node, in the form: (Vnode - Vadjacent node)/(resistance to adjacent node). Specific currents leaving the node are positive, entering the node are negative.

So in this schematic the label V1 is not necessary since the node is already identified as AoVs, and V3 is not necessary since it is already labeled Vo. Nor is it necessary to redraw the schematic so that all lines converge at the point V2. Also, do not write a node equation for a node whose voltage is equal to a voltage source. Whatever Ao & Vs may be, just use the voltage AoVs as the adjacent node voltage in the other node equation(s). So it is likely that AoVs will end up being part of the final answer.

Here is an example of identifying the nodal currents. Add all the currents at a node and set the result equal to zero. Last edited:

#### WBahn

Joined Mar 31, 2012
26,398
Hi,

Here's the goal of my attached HW: I have to use nodal analysis to solve the circuit for Vo. Vo is the voltage across the open ports.

Quick notes
1. Next to the voltage source, Vs, there's an 'Ao' written. Please ignore it.
2. The two attached pages start numbering on p.3 and go to p.4

My question
I solved the circuit as best I could. I want to know if my approach and answer to solving for Vo looks correct. Vs is unknown at the start, and thus the answer must be implicit and contain Vs.

Thanks!
One of the things that you want to get in the habit of always doing is asking if the answer makes sense. This will let you catch a lot of the mistakes you make -- or at least identify that a mistake was made.

Here you claim that

Vo = (16 V - 2·Vs) / 7

Does this make sense?

What if Vs = 0? That would mean that the Vo is a little over 2 V. Look at the original circuit and short out the independent voltage source. Isn't the 8 A source going to send current UP through the 4 Ω resistor such that the output voltage would be negative?

Now do the analysis for the special case of the original circuit with Vs = 0. Do it any way you want -- this one you can actually do in your head if you think about it right. Now you have a data point that your full solution has to match.

Another sanity-check case you can look at is what happens as Vs gets very, very large. Do you expect Vo to go up or go down? Your answer has it going down and then going negative as Vs gets larger and larger. Does that make sense?

Let's walk through this limiting case just to see how it works. As Vs get very large, we expect more and more current to flow out of it and into the 1 Ω resistor. It then splits three way with 8 A going through the source and the rest of it going equally down the two 6 Ω paths. If Vs is high enough so that 8 A is a tiny fraction of the total current, then it has very little effect. So what we see is approximately 1 Ω in series with two 6 Ω paths in parallel giving a total of about 4 Ω. So the total current is going to be about Vs/4Ω and half of that will flow though the branch with Vo, so the current in the 4 Ω resistor will be 0.5Ω·Vs/4Ω yielding a Vo of Vs/2. So now you know that, as Vs gets very large, the equation should Vo should get closer and closer to being Vs/2.

Another thing that will help is to keep your work symbolic as long as possible. So don't use 8 A, instead use Is. Also, use R1 and so on instead of the actual resistor values. That will let you sanity check your results by being able to ask simply questions like whether the solution does what it should if the current source were turned off or if a particular resistor were removed or shorted. That will not only help you confirm that your answer is likely correct, but if it isn't it will really help you figure out where your work went awry. And you don't have to wait to the end to do these checks -- you should be asking these kinds of questions at nearly every step. Remember, once you make a mistake in your work, then every thing you do after that point is wasted effort. Far better to catch a mistake early on than after two or three pages of work.

Thread Starter

#### project_science

Joined Sep 14, 2018
21
Well I certainly appreciate the responses! I did my HW a 2nd (Try 2 PDF) and 3rd (Try 3 PDF) time to incorporate many of your remarks.

In Try 2, that my v1 eq. was wrong, where v1 should be, v1 = Vs. So my v1 eq. should be:

Vs = Iv*(1 ohm) + v2

Resolving the problem, and keeping v2 and v3 the same as before gave me:
Vo = 8V / (Iv * 1ohm)

However, that can't be because it would mean Vo is dimensionless. But I'm not sure where I went wrong in my approach to solve it.

In Try 3 (PDF), I tried the approach of keeping variables in my equations until the end, before using numbers. However, my answer for Vo is long, and it looks problematic, as some of the terms are dimensionless.

I'm more concerned to be setting things up correctly with the equations, as I assume much of the hand calculations will be resolved once my class starts using a circuit simulator.

Thoughts?

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#### WBahn

Joined Mar 31, 2012
26,398
That you are looking at the dimensions is an extremely good thing. But don't just look at the final result. Walk through you work line by line and ensure that the dimensions work out at each and every step. At some point you made a mistake that messed up the units. Fix that and move on from that point. In the future, get in the habit of doing that check at every line or two so that when you make such mistakes, you catch them almost immediately. It will save you a ton of time in the long run.

#### The Electrician

Joined Oct 9, 2007
2,824
These remarks are about Try3.

For the first equation you have unnecessarily introduced a variable Iv. Since there is a voltage source Vs connected directly to that node, you don't use a KCL equation; just use a constraint equation V1 = Vs.

For the second equation everything looks ok.

Your work for the 3rd equation has an algebra error, which messes up your units.

At a certain point you have: $$\frac {V2-V3}{R3}=\frac {V3}{R4}$$

But next you have: $$\frac {V2}{R3}-\frac {V3}{R3}=\frac {V3 R3}{R4}$$

Why did you multiply the numerator on the right hand side by R3? That's a mistake.

If you will make equation 1 just a constraint equation, and fix the error in your work for equation 3, I think you have it all correct.

#### WBahn

Joined Mar 31, 2012
26,398
These remarks are about Try3.

For the first equation you have unnecessarily introduced a variable Iv. Since there is a voltage source Vs connected directly to that node, you don't use a KCL equation; just use a constraint equation V1 = Vs.

For the second equation everything looks ok.

Your work for the 3rd equation has an algebra error, which messes up your units.

At a certain point you have: $$\frac {V2-V3}{R3}=\frac {V3}{R4}$$

But next you have: $$\frac {V2}{R3}-\frac {V3}{R3}=\frac {V3 R3}{R4}$$

Why did you multiply the numerator on the right hand side by R3? That's a mistake.

If you will make equation 1 just a constraint equation, and fix the error in your work for equation 3, I think you have it all correct.
This underscores the point about continuous units checking. Notice how the units work out in the first equation (a current equals a current) but a simple algebra goof (the kind we all make on an all-too-frequent basis) results in the next line being a current minus a current equals a voltage. If we check the units we catch the mistake right there instead of wasting a bunch of effort on work that is guaranteed to be wrong from that point on.

You'll be amazed at how quickly checking the units will become so second nature that you won't even realize that you are doing it -- you'll just somehow know that things aren't working out almost immediately after making such a mistake. It also gets to the point where you will have a hard time working a problem in which the units are not being properly set up or properly tracked and will have to go in and make them correct before you can work it.

Thread Starter

#### project_science

Joined Sep 14, 2018
21
This attempt on Try 4 (attached) makes a lot more sense. I've worked the nodal equations (v1 and v3), and the result was a lot simpler, where Vo = (Vs - 8V / 2).

It makes sense that as Vs increase, Vo would too, especially as Vs gets very large.

I then took the case where Vs = 0, which results with Vo = -4V. Since R4 = 4 ohms, 1A of the 8A current supply must go "up" it. After simplifying this circuit, and using a current divider for the current across R4, I got 1A. When taken to mean it actually goes "up" R4 instead, it gives Vo = -4V.

Thanks a lot everyone for your help!

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#### WBahn

Joined Mar 31, 2012
26,398
This attempt on Try 4 (attached) makes a lot more sense. I've worked the nodal equations (v1 and v3), and the result was a lot simpler, where Vo = (Vs - 8V / 2).

It makes sense that as Vs increase, Vo would too, especially as Vs gets very large.

I then took the case where Vs = 0, which results with Vo = -4V. Since R4 = 4 ohms, 1A of the 8A current supply must go "up" it. After simplifying this circuit, and using a current divider for the current across R4, I got 1A. When taken to mean it actually goes "up" R4 instead, it gives Vo = -4V.

Thanks a lot everyone for your help!
Nice job tracking your units.

You made a fair amount of extra work for yourself by not identifying the essential nodes. An essential node is one that is connected to more than two branches. If it is only connected to two branches, such as the case with V3, then you know that the current leaving the node into one of the branches is equal and opposite to the current leaving the node into the other.

That, combined with the observation that the node equation for V1 is trivial, namely V1 = Vs, you actually only have a single node equation to set up and solve.

$$\frac{\(V_2 \; - \; V_s$$}{R_1} \; + \; \frac{$$V_2$$}{R_2} \; + \; I_s \; + \; \frac{$$V_2$$}{$$R_3 \; + \; R_4$$} \)

Once you know V2 solving for V3 is just a simple matter of treating it as a voltage divider.

One thing that I would recommend is to split your work into two distinct phases (with a line drawn across the page between them to emphasize this). The first is setting up your system of equations. Just set them up in whatever form makes the most sense to you and that allows you to readily verify that they are correct. Don't do any manipulation on them -- just set them up. This is where all of the electrical engineering knowledge and principles come into play. Everything after this is just math. You should be able to give these equations to someone that knows algebra but has never heard of KCL or anything else remotely EE related and they should be able to solve them. By splitting the effort this way you get to focus on the EE stuff while you are developing your equations without getting distracted by any of the algebra concerns. When you then turn to the equations, you can apply all of your focus to getting the math right without being distracted by also having to apply EE principles here and there.

Another point is that this other person that is doing the math (including some form of equation solving software) cannot get the right results if the equations were not set up correctly. So check them carefully once you have set them up to be positive that they are, indeed, correct. Force yourself to look at what you actually wrote and not get blinded by what you thought you wrote or what you intended to write -- any mistake that doesn't get caught here means that all of the subsequent effort to solve the equations (which is usually the bulk of the work) is pure wasted effort.

#### RBR1317

Joined Nov 13, 2010
610
project science,
You might find it interesting to see how your circuit can be solved by putting the nodal equations in matrix form and letting a modern mathematical software do the algebra drudgery.
I don't quite understand why anyone would put node equations into any form other than node equations. Modern mathematical software, such as Maple, handle node equations rather well. (See attached screenshot from Maple) #### MrAl

Joined Jun 17, 2014
8,066
Hi,

Here's the goal of my attached HW: I have to use nodal analysis to solve the circuit for Vo. Vo is the voltage across the open ports.

Quick notes
1. Next to the voltage source, Vs, there's an 'Ao' written. Please ignore it.
2. The two attached pages start numbering on p.3 and go to p.4

My question
I solved the circuit as best I could. I want to know if my approach and answer to solving for Vo looks correct. Vs is unknown at the start, and thus the answer must be implicit and contain Vs.

Thanks!

Hi,

The basic idea is that you can write the equations the same way you do with any other type of source you just keep it more symbolic.

A common example is a voltage controlled voltage source. These sources have a 'gain' let's call it 'A'. That means the output is related to the input by the gain A and so we have that in the equation too.
A regular voltage source might be just:
Vout=E

a VCCS would just be:
Vout=A*E

where A is the gain. So it's not that much different.
If for example there is a resistor on the output, then the current is:
I=Vout/R=A*E/R

and you use that A*E/R in your equations in the same manner you would use E/R with a constant voltage source of voltage E.

When you solve the equations, you either keep that A as a symbol or replace it with it's value. If A=10 then you simply replace it:
Vout=10*E
I=10*E/R

you can then go and solve the set of equations.

#### RBR1317

Joined Nov 13, 2010
610
You should be able to give these equations to someone that knows algebra but has never heard of KCL or anything else remotely EE related and they should be able to solve them.
I'll go one better than that. Since writing node equations is just a standard procedure (Σ{Vnode - Vadjacent}/{resistance to adjacent node}), you should be able to give a color-coded schematic to someone that knows algebra and they should be able to write the node equations and solve them. The EE principles are important in identifying the nodes and supernodes for color-coding the schematic, and for those circuits with dependent sources, being able to express the control variables in terms of the node voltages. But once the schematic has been fully annotated, that is the end of the EE stuff.

#### WBahn

Joined Mar 31, 2012
26,398
I'll go one better than that. Since writing node equations is just a standard procedure (Σ{Vnode - Vadjacent}/{resistance to adjacent node}), you should be able to give a color-coded schematic to someone that knows algebra and they should be able to write the node equations and solve them. The EE principles are important in identifying the nodes and supernodes for color-coding the schematic, and for those circuits with dependent sources, being able to express the control variables in terms of the node voltages. But once the schematic has been fully annotated, that is the end of the EE stuff.
I don't think that quite captures it. Take this circuit, for instance. How would they use just (Σ{Vnode - Vadjacent}/{resistance to adjacent node}) to write the node equation for V2? They would only be able to write it in terms of (V2 - V1). Okay, fine. But now they need a node equation for V1, What would they write using just this standard procedure?

#### RBR1317

Joined Nov 13, 2010
610
But now they need a node equation for V1,
That is where the EE stuff comes in. V1 is just a supernode with the reference node so you would never write a node equation for V1. All that one needs to do with supernodes is to identify the offsets.

#### WBahn

Joined Mar 31, 2012
26,398
That is where the EE stuff comes in. V1 is just a supernode with the reference node so you would never write a node equation for V1. All that one needs to do with supernodes is to identify the offsets.
But the whole point is that the person you are giving this color-coded schematic to doesn't know any EE stuff! That's the whole point. Above that line is the EE stuff, below that line is the math stuff. Anything that requires EE stuff belongs above the line with as little math stuff as possible. Below that line there should be NO knowledge of EE stuff required so that the person can focus on the getting the math right.

#### RBR1317

Joined Nov 13, 2010
610
But the whole point is that the person you are giving this color-coded schematic to doesn't know any EE stuff!
I agree that's the whole point. When one has a fully color-coded & annotated schematic, all the EE stuff has been done. The next step is actually writing the node equations. Anyone can be taught to write node equations (even type them directly into a symbolic algebra engine) once the EE stuff has been finished.

#### WBahn

Joined Mar 31, 2012
26,398
I agree that's the whole point. When one has a fully color-coded & annotated schematic, all the EE stuff has been done. The next step is actually writing the node equations. Anyone can be taught to write node equations (even type them directly into a symbolic algebra engine) once the EE stuff has been finished.
But didn't we just agree that if all they are given is this color coded and annotated schematic that all of the EE stuff has NOT been done? They don't know what to do with V1; you said it yourself, "That is where the EE stuff comes in." But once we hand the schematic off then we are past the point of being able to apply any EE stuff.

#### RBR1317

Joined Nov 13, 2010
610
They don't know what to do with V1;
But of course they would know what to do with V1, because there would not be any node label at that location. It was an error in the first place to give that point a V1 node label. On a properly annotated schematic that point would only be identified as a supernode offset voltage from the ground reference.