Nodal Analysis Question

Discussion in 'Homework Help' started by RoKr93, Jun 17, 2013.

  1. RoKr93

    Thread Starter New Member

    Jun 17, 2013
    Hello all,

    I've hit quite the rut on this nodal analysis problem and would really appreciate some help:

    In the circuit below,
    Vs1 = 10 V
    Vs2 = 4 V
    Is3 = 1 mA
    G1 = 0.4 mS
    G2 = 2 mS
    G3 = 3 mS
    G4 = 5 mS


    Use nodal analysis to find V1 and V2. Then compute the power delivered by the independent sources and the power absorbed by G2.

    Isn't V1 just equal to Vs1? Or do I have the wrong idea about that? As for V2, I don't feel like I have enough information to do nodal analysis for Vc or the unlabeled node to the right of Vc...I don't know the resistance/conductance over Vs2 and I'm not sure how to figure out the current/voltage over G4. I tried KVL for the bottom right loop but that got me nowhere, and I'm not sure how to proceed...would really appreciate some assistance on this one.

  2. WBahn


    Mar 31, 2012
    Only if you can justify the assertion that there is no current flowing in G1.

    What node to the right of Vc? The node labeled Vc would appear to be on the right-most edge of the circuit.

    Why do you need that?

    Think about this for a moment. You wanted to assert that V1 was simply going to be Vs1. Why don't you think that Vc isn't going to simply be Vs2?

    Well, if you tried something, then show your work for what you tried! We are not mind readers. How can we possibly tell you where to proceed if you don't show us where you are proceeding from? We need to see the work you have done up to this point.

    Your title says that you are doing Nodal analysis. So why are you writing loop equations?

    Q1) What are the nodes in your circuit? Identify them (labeling them, if needed).

    HINT: There are five total nodes.

    Q2) Which node is your reference (i.e., 0V) node?

    HINT: This one is already picked for you.

    Q3) Which nodes have trivial solutions?

    HINT: There are two of them.

    Q4) What are the node equations (think KCL) for the remaining two non-trivial nodes?
  3. RoKr93

    Thread Starter New Member

    Jun 17, 2013
    Thank you for the response.

    Sorry- I meant the left.

    In order to perform KCL for Vc.

    I forgot to mention that- I'm aware that Vc = Vs2.

    My work was straightforward and fruitless, and I definitely didn't get anything from doing KVL, so I didn't think it was worth showing. I'm sorry if I was unclear, but as you implied yourself later on, there's really no reason to do KVL, which I realized, so I didn't think providing that work would help at all.

    I just thought I'd try, since I wasn't sure how to continue with nodal analysis.

    Vc, the one to the left, and the one to the left of that (between G1 and G2). And then there's the bottom node. That makes four- I don't see a fifth one...?

    The bottom node (ground).

    Do you mean voltage? Because we have Vc = Vs2 and the node to the left of that = V2. But in terms of current/KCL, it's less straightforward.

    I don't see how I can do KCL for any of these given that I only have an expression for current for a couple of the branches in the entire circuit. I have Is3, V2*G3, and (V2 - Vs2)*G4 as far as current values go but that's it as far as I can tell. And that's not enough to write a node equation for any of the nodes.
  4. WBahn


    Mar 31, 2012
    Okay, we are making progress (even if it might not seem that way just yet). I am getting a better feel for how you are thinking about various aspects of the problem. The good news is that most of your thoughts are in the right direction and it is just a matter of bringing them together.

    Consider the following mark-up:


    Do you now see the five nodes?

    Since each of the four non-reference nodes already has a symbolic voltage that has that node as the positive side and GND as the negative side, we can just call the voltages from left to right:

    Vs1, V1, V2, Vs2

    Vs2 already has a name, namely V_C. Similarly, I added a name, V_A, for the left-most node (Vs1). You can use which ever ones you like. These are the trivial nodes:

    V_A = Vs1
    V_C = Vs2

    V1 and V2 are the non-trivial nodes.

    There is no need to do KCL for node V_C because the whole point is to find the voltage on the nodes; but since we already KNOW the voltage on node V_C, we don't need to use KCL on it at all! The same goes for V_A.

    So let's focus on the two remaining nodes.

    Apply KCL to each of these nodes and write down the node equations (The node equation is nothing more than KCL applied to a node and then rewritten to be a sum of terms on the left side where each term involves one of the unknown node voltages and everything else on the right side).

    So, for instance (just making one up), you might have:

    G7*V1 - G3*V2 = Is7 - 6mA + G7*Vs1

    But start with KCL applied to one of the nodes and then rearrange it as necessary.
  5. RoKr93

    Thread Starter New Member

    Jun 17, 2013
    Okay. I'm not sure if I'm interpreting where the voltage V1 is correctly, but for KCL on the V2 node, I have:

    V2*G3 + (V2 - Vs2)*G4 + (V2 - V1)*G2 = 0

    which, when I group it nicely to be ready to be put in matrix form, is

    -G2*V1 + (G2 + G3 + G4)*V2 = G4*Vs2 (which is a known value).

    Then for the V1 node, we have

    Is3 + (V1 - V2)*G2 + (V1 - Vs1)*G1 = 0

    grouped nicely:

    (G1 + G2)*V1 - G2*V2 = Vs1*G1 - Is3 (also a known value).

    I'm still a little bothered by V1...I'm not sure if I used it right, but it's really the only way I could find that made KCL make sense. When I throw that into a matrix equation and solve, I get V1 = 0.5 V and V2 = -2.1 V, which seems...sketchy.
  6. WBahn


    Mar 31, 2012

    Looks great!

    It's good that you are bothered by your answer, because it means that you are doing something that far too few people do -- ask if the answer makes sense!

    So -- check it to see if it is correct. That is one of the beautiful things about circuit analysis and engineering in general -- you can usually verify the correctness of an answer from the answer itself.

    There are a number of checks you can make here. Let's do KCL at V2.

    Q5) What is the current INTO the node from V1 through G2?

    Q6) What is the current INTO the node from Vs2 through G4?

    Q7) What is the current OUT of the node through G3?

    Do these three currents satisfy KCL?

    As for where you went wrong, there is no way for me to even guess since you didn't show your work. Your node equations are fine. I can tell you that because you showed your work up through that point. But then you just jump to the answers and so I can only sit there and repeat the mantra, "Show your work!"

    Often times, just organizing your work so that you can post it will result in you seeing where the mistake is.
  7. RoKr93

    Thread Starter New Member

    Jun 17, 2013
    Well...I jumped to the answers because MATLAB is a beautiful thing. I just finished up linear algebra in the spring and have dealt with more matrices than I ever cared to, so I figured I'd take the lazy way out this time around. ;)

    But after some checking, I realized for whatever reason I got an incorrect answer out of MATLAB the first time around, and it's actually V1 = 3.5 and V2 = 2.7. And when I run KCL on V2, everything checks out, so that's right!

    Thank you so very much for your patience and assistance. It really helped a lot. If I could pick your brain one last time- I found the notation in the diagram for V1 to be very confusing. I thought it meant the voltage drop over the independent source on the left, but apparently that's not the case. What exactly does that signify visually? I think if I had understood that better I could have avoided a lot of confusion with this problem.
  8. WBahn


    Mar 31, 2012
    Yes, MATLAB is a wonderful tool. But don't fall into the tarpit of letting the whizbang tools do your thinking for you. Keep yourself in practice with the basic techniques by using them whenever doing so it reasonably feasible. In this case, you had two equations in two unknowns that could be solved by hand without even touching a calculator.

    When I started college I was very good at mental arithmetic. Then I got into the habit of using a calculator for everything because it was now always with me and it was just easier to take the lazy way out. But one day I realized that I was actually reaching for my calculator to add a one digit number to a two digit number -- and that I had to struggle just a wee bit to actually answer it without the calculator! That was a watershed moment for me and ever since I try to do algebra and arithmetic manually as much as I reasonably can. I'm not a Luddite about it -- these tools are valuable and have their place, I just choose to keep them in their place as valuable tools and not as substitute brains.

    By doing this, I can keep my estimating skills pretty sharp so that I can estimate answers to most problems and then compare that to what I end up with (regardless of how I end up with it). That has saved my bacon several times.

    Ah, and now you have brought out my inner Units Nazi! V1 is NOT equal to 3.5. That's like saying that the distance from London to Paris is 214. Someone else might say it is 344. Somewhat else might say it is 1.13 million and yet someone else might say it is about 0.05. Which of these are right? Which are wrong? Any? All?

    In fact, ALL of them are wrong because distance is a "dimensioned quantity", meaning that it has a magnitude coefficient multiplied by a unit of measure. Without the units, it is simply wrong.

    The voltage on V1 is 3.5V and V2=2.7V.

    This isn't just nitpicking or harping on semantics. As an engineer or technician, you are going to be working a lot of problems, both big and small. You are going to make lots of mistakes along the way. If you religiously track your units throughout your work, them most of your mistakes will get caught right away because most mistakes will mess up the units. Any time the units don't work out, you KNOW the answer is wrong. Being able to detect that early allows you to avoid wasting time continuing on with a problem that has no hope of yielding a correct answer and lets you find and fix the error much more promptly.

    Thank you and it was my pleasure. It's part of how I give back to a field that has been very good to me and which I have enjoyed being a part of for quite some time.

    The key is to look at what node the '+' and the '-' symbols are next to. Those are the nodes that the voltage is measured between.