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Node voltage method (step 5)

Last step of the Node voltage method for analyzing circuits: Solve the system of simultaneous equations. Created by Willy McAllister.

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  • orange juice squid orange style avatar for user JoshH.Ong13
    Khan Academy,
    Would you please do a more complex example?
    (33 votes)
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  • blobby green style avatar for user bdincer
    I don't understand the fact that if V1 is 15 volt and V2 is 1 volt. How will this equation be true:
    Vs = V1 + V2? If Node 1 is 14. But instead: V1 = Vs - V2 which is 14. Then I understand this expression but I am just curious if this is right what I just said. Thanks in advance.
    (5 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      Good question. You mixed up node voltage and element voltage. In your KVL equation, (the sum of element voltages around a loop), be sure to insert element voltages at each spot (not node voltages). In particular, the element voltage across R1 is (V1 - V2), not just V1.

      The element voltage across R2 is (V2 - V3). Since we defined node 3 to be our reference node, that means V3 is zero by definition, and the element voltage across R2 is (V2 - 0) = V2. This is a little bit of a trap to recognize. Because R2 is connected directly to the reference node, the element voltage across R2 is the same value as the node voltage at the top of R2 (node 2), namely V2. Don't let this lull you into thinking that all element voltages are the same as all node voltages.

      The KVL equation with element voltages should read as Vs = (V1 - V2) + V2.
      We end up with Vs = V1.
      (12 votes)
  • duskpin tree style avatar for user Jeff Smith
    When you add a current source to a circuit, what is supplying that current?
    I understand that when using wire, the electrons are pushed from the unstable atoms by the force of the voltage applied, but when you draw a current source "separate" from the voltage and somehow create another current source, where is that added current coming from?
    (3 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      When talking about the abstract mathematical construct of a constant voltage source or constant current source, the current (and voltage) provided by either is generated inside the object.

      A constant voltage source creates a constant voltage at its terminals and provides whatever current is demanded by the things it is connected to.

      A constant voltage source creates a constant current in the wires attached to it, and generates whatever voltage is demanded by the things it is connected to. For example, if you connect a 1mA current source to a 1kohm resistor, there is a sense that the resistor is "demanding" a voltage of 1v from the current source.

      In real-life sources, the voltage and current from a voltage source comes from chemical energy if the source is a battery. If the source is a DC power supply the current and voltage come from the AC wall plug. The internal circuit of the power supply performs the AC to DC conversion. A real-life current source plugged into the wall is also a complicated circuit.

      We learn about the ideal current source because we use it to model a transistor (bipolar or MOSFET). This image comes from the Wikipedia MOSFET page. https://en.wikipedia.org/wiki/MOSFET#/media/File:IvsV_mosfet.svg

      It plots the current through the device (vertical axis) against the voltage across the device (horizontal axis). Each blue line is for a different voltage between the gate and source terminals. Notice each blue line has a horizontal section on the right side. That horizontal part is modeled by a constant current source. (The i-v graph of a constant current source is a horizontal line.)
      (3 votes)
  • mr pink orange style avatar for user Austin Michael
    why don't you ever write the unit ohm when you do circuit analysis? In physics 1 the teachers are very strict about units.
    (1 vote)
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    • spunky sam orange style avatar for user Willy McAllister
      Notice I use the Omega symbol when drawing the schematic. As I build the equations that need to be solved my habit is to omit the Omega when writing various forms of Ohm's Law. All of the terms are of the form v/R or iV. I suppose at :30 I could write 15V/4kOmega. I would do that if my teach told me it was important.

      My habit, once I've constructed the proper equations, is to leave out most units as I do the algebra. It's just math after all. At the end I apply the correct units to the answers.
      (3 votes)
  • duskpin ultimate style avatar for user Rax parker
    can we choose any one potential to be zero and then solve using any method? Since we only care about potential diff.?
    (1 vote)
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    • old spice man green style avatar for user Willy McAllister
      Yes. The choice of the reference node (ground) is arbitrary when doing analysis. Your voltage results will be with respect to that node, so you won't get the same numbers as someone else who selected a different reference. The currents in each circuit element will come out the same for all references. This will work with any analysis method.

      There are habits you will fall into for picking the reference. Often the negative terminal of the power source is picked as ground. But for the mathematics the reference doesn't matter.
      (2 votes)
  • primosaur seed style avatar for user old_painles
    i dont understand the 3,75mA when there is nowhere
    that amount of current in our circuit

    14 volts over R1 =
    14 / 4000 = 3,5mA

    1 volt over R2 =
    1 / 2000 = 0,5mA

    i1 - i2 - Is = 3,5mA - 0,5 mA - 3mA = 0

    nowhere in this circuit i see a 3,75mA flowing
    (1 vote)
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    • spunky sam orange style avatar for user Willy McAllister
      The voltage source is 15V, not 14.
      15V/4000ohms = 3,75 mA
      This is is an intermediate term that appears while solving the equation with real component values. The term happens to have units of milliamps, which makes it tempting to look for in the circuit, but it is not there. If you look back through the algebra steps you can see where it jumps out an individual term in the KCL equation. I wouldn't put too much effort into figuring out if this current is "real", it is just an artifact of solving the math.
      (2 votes)
  • aqualine ultimate style avatar for user Mateo Ceballos Querol
    im confused why there are 2 voltage drops across r1?
    (1 vote)
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  • orange juice squid orange style avatar for user ConnorBecz
    Hi Willy. I have trouble following the calculations in this example after the point where you've stated that current i-1 minus current i-2 will be equal to the 3mA of current Is. I see that you then subtract the current on the positive side of R1 from current Is and factor out V2 from both resistors, saying that the current on the negative side of R1 and the positive side of R2 added together are equivalent to the current Is minus the current on the positive side of R1.
    I.e., -V2(1/4k+1/2k)=3mA-15v/4k

    Firstly, how can you be taking current values for different sides of a component [e.g. 15v/4k or -V2(1/4k+1/2k)] as I thought current through a circuit was supposed to be constant unless it branches off into multiple paths?

    Secondly, can you elucidate in what sense -V2(1/4k+1/2k)=3mA-15v/4k? What does the 3/4k value which you then produce from 1/4k+1/2k correspond to in terms of the circuit? Similarly, why would multiplying it by negative V2 be equivalent to Is minus the putative current of 3.75mA on the positive side of R1? At the last step where you find V2 by multiplying .75mA by 4k/3, what is the significance of this reciprocal value of 3/4k and how does it help to find V2?
    (1 vote)
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    • spunky sam orange style avatar for user Willy McAllister
      This is a complex question. Let me try a high-level answer, and then we we work on the details.

      The video starts with the Kirchhoff's Current Law equation for the node at the top-center of the schematic (the equation with the check mark).

      By you have captured the connectivity of the circuit in a single KCL equation and it's time to go to work to solve it. It is time to take off your Circuit hat and put on your Algebra hat. Everything from here on is a bunch of algebra. We quickly discover there is only one real variable, v2. So it's a matter of cranking the math to get v2 isolated.

      During this algebra phase of the problem it is tempting to interpret the individual parts of the equation as electrical quantities. But it is usually fruitless to try to make sense of these terms as electrical quantities. Just use your Algebra brain to solve an equation.

      I can answer more questions if you have them. One thing that helps me find where you are talking about is to use time stamps. If you write 1 : 0 0 you get this symbol: and it will jump right to that point in the video.
      (1 vote)
  • blobby green style avatar for user Salvatore  Turano
    Hi i am wondering how you flipped the 3/4K to 4K/3 in the 3rd green step to the 4th green step before the final answer for V2? Could you please help
    (1 vote)
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  • orange juice squid orange style avatar for user raineeee
    This may be off topic, but how and where do I download SPICE?
    (1 vote)
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Video transcript

- [Voiceover] And now we're down to solving this circuit. What I want to do now is put in the con pointed values and solve this specific circuit. Let me move the screen up again. We'll leave the list of steps up there so we can see them. Let's go to work on this equation now. We have a little bit of algebra, and we can plug in values where we need to. We can plug in 15 volts for V1. And for R1, we can plug in 4000 ohms. We can put in, for V2. V2's still unknown. And that's divided by 4K. In this expression here we still have V2 unknown over 2K And let's put IS over on the other side. IS was three milliamps. Let's just keep working at this now. V2 times 1/4K plus 1/2K equals three milliamps. Oh, get my minus signs right. Minus sign over here. Let's bring the constant term over to this side, so this is 15 volts divided by 4K. And continuing over here, we have minus V2. Let's combine those two resistor terms. So it's going to be 1+2 over 4K, equals three milliamps. And 15 volts divided by 4K is 3.75 minus 3.75 milliamps. Moving on, minus V2 equals -0.75 milliamps, times 4K over three. Let's get rid of the two minus signs. We don't need those anymore. And V2 equals one volt. That's good; we solved it. We solved our prior two voltages. We have one here and we have one over here. And we can check off the last step. So that's our first application of the node voltage method. I want to show you one more thing that is a powerful part of this technique. Let me quickly sketch the schematic again. So this was our schematic, and we assigned node voltages. We assigned node voltages here. V1, and V2, and we made this our reference node. And one of the things we did not do as part of the node voltage method, we did not us KVL to write equations around these loops. And one of the features of the node voltage method is that the KVL equations, because we're using node voltages, the KVL equations are automatically satisfied. And I'll show you why. I want to put one more label on here, which is the element voltage. We'll call this VR1. That's the element voltage across here. The element voltage here is just V2. So in this case for R2, V2 is the element voltage and the node voltage at the same time. VR1 is an element voltage. And now we're going to write KVL, starting from this point, and going around the loop in this direction. And what we have is... let's get all our labels on here. The loop voltages, we start with a rise of +VS, then we take away VR1, and then we take away V2, and that equals zero. So that would be the KVL equation for this circuit. Now I'm going to plug in, using node voltages, I'm going to write VR1. So I get +VS minus, VR1 is node voltage V1 minus node voltage V2. V1 minus V2, minus V2 equals zero. And we'll just do one more substitution. But I forgot VS and V1 are the same voltage. So this is actually V1 minus V1, minus V2, minus V2 equals zero. And if we look at this equation, that goes, that goes, plus V2, minus V2. This equation is automatically true, if we write Kirchhoff's Current Law in terms of node voltages. That always turns out to be the case. That's why we don't bother to do it. We know it's going to be true. So that's a nice feature of the node voltage method. It's a really efficient way to write equations. You only write KCL equations. And this is such a good method in fact, that circuit simulators, like, you may come across a circuit simulator called Spice. Almost every circuit simulator uses this node voltage method to do its computations.