Node voltage method (step 5)

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About this video

This video solves a resistive circuit with the node voltage method. The reference node and node voltages V1 and V2 are set up, the known source voltage is substituted, and a single KCL equation gives V2 = 1 V. The video then shows why KVL is automatically satisfied when writing node voltages, which is why this method is efficient and used by circuit simulators.

You'll be able to

  • substitute known source values into a node voltage equation and solve for the unknown node.
  • explain why node voltage equations satisfy KVL automatically.
  • recognize that only KCL equations are written in the node voltage method.

Key words

Node voltage
The voltage at a node measured relative to a chosen reference node.
Reference node
The node chosen as zero volts, against which all other node voltages are measured.
Element voltage
The voltage across a single component, which can be written as a difference of node voltages.
KCL
Kirchhoff's Current Law, which says the currents entering a node sum to zero.
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.

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