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Nodal Analysis Circuits - Physics · Acedra
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Nodal Analysis Circuits - Physics
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0:00
By the end of this video, I hope to teach you how to use nodal analysis, including super nodes to solve pretty much any circuit. To start, let’s look at this simple circuit. The key to nodal analysis is understanding node voltages. Now you should know what a node is from the ohms law and kirchhoff’s laws video, but to quickly recap, a node is one end of a circuit element together with all of the conductors attached to it. So in this circuit,
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we have 5 nodes. Notice that this whole bottom is a single node. If it’s hard to see, remember that conductors can be bent like this so you can see it all comes together. Let’s talk about the main things you need to know about nodal analysis, and then we will do a bunch of examples to really understand it. The first step in nodal analysis is choosing a reference node, and you label it
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with the ground symbol. You should choose a node that has the most connections to it. So here, that would be this bottom node since it has all of the other conductors connected to it. So why do we do that? Well, now we are going to write the voltage of the other nodes with respect to our reference node. So let’s say we figured out that the voltage of this node is 5 V, what we are saying is that
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with respect to our reference node, the voltage here is 5 V. In other words, this 5v node is at a higher potential than the ground node, which makes sense, since the ground node is considered to be 0 V. Each of these nodes will have a voltage, let’s label them from 1 to 4. Again, every single one of these node voltages is with respect to the ground node. Next, we guess the directions of current. It doesn’t really matter which way you pick, because if you get a negative answer,
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then you know it’s opposite to your assumption but there are certain things you need to keep in mind to make your life easier. I will explain those things during the examples. Remember to label positives and negatives on each resistor properly based on which direction you assumed your current to flow. So current enters the positive side. Now there is also voltages in each of these resistors.
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Let’s label them like this. Here is the important part. What can we say about the current across this resistor? We know from ohm’s law, current is voltage divided by the resistance. So here, that would be Va divided by R1. But what can we say about this voltage VA? Well, VA is actually equal to the difference between the voltage of this node and this node. So we assumed our current is travelling left to right which means we are saying V1 is at a higher potential than V2. Remember,
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current travels from the higher potential to the lower potential. So then the voltage in this resistor is V1 minus V2 which gets us the voltage difference. That means we can write every single current in this circuit using nodal voltages just like we did for this one. You must keep this in mind because this is what nodal analysis is built off of. The rest is just solving KCL equations but it’s much easier to teach that part by going through examples so let’s get to it.
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Let’s take a look at this question where we need to figure out V1 using nodal analysis.
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This is a question where we have just resistors and independent current sources. We will pick the bottom node to be ground because it has the most connections. What that means is, we now consider this bottom node to be zero volts. Now we can label this as node 1 and this as node 2. I will assume the currents to be like this. Here is the first point you should keep in mind. When I assumed the current on this branch, why is it beneficial to pick the current to flow this way?
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Remember, current flows from a higher potential to a lower potential, and we said our bottom node is our ground. So relatively speaking, voltage up here is at a higher potential than down here, which means current must travel down towards ground. You don’t have to do it this way, but this makes the math much easier and you are prone to less mistakes. Also, in this problem, since we are given the positives and negatives on the 3kohm resistor, it’s better to assume the current in
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this conductor will flow this way, since current enters the positive side of the resistor. On this side, we aren’t given any positives or negatives for our resistor, but again, the current will flow down towards ground since we are saying the top is at a higher potential. The same is true for I3. But what about I2? This is just a guess, but what we are saying by showing a current flow from left
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to right is that the voltage of this node is at a higher potential than this node. We need to keep this in mind when we write our equations. Now we can label the rest of the resistor voltages along with the positives and negatives based on our assumed current directions. Next, we can label the node voltages. Since the numbers are taken, we will label them VA and VB. Now let’s consider node 1 and write a KCL equation. Any current coming in to the node
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will be considered positive while any current leaving the node will be considered negative. We have I1 leaving, I2 leaving and 6 ma leaving. All of that is equal to 0. I am going to multiply all the terms by negative one, so we just have positives to work with. Now that we have our KCL equation for our node, the next step is to define our currents in terms of voltages and resistance using ohms law. Remember, current is equal to the voltage divided by the resistance. So looking at
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I1, we see that it’s the current flowing through this 3k ohm resistor. So that would be voltage 1 divided by the 3k ohm resistor. Next, we have the I2 current, which is the current through this 2k ohm resistor. So we can write that as V2 divided by the 2k ohm resistor. The 6 mA we can take to the other side of the equation. It should be noted here that the k represents 1000, and the m represents 10^-3. This makes it easier to work with rather than writing down 3000ohms
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or 2000 ohms. Now we can multiply both sides by 6k, eliminating our fractions. Note that this also gets rid of the mili part. This part can be confusing to some, so let me quickly explain what’s happening here. This equation can be written like this. By multiply everything by 6k, which is 6000, we are effectively getting rid of the decimal places, and removing the fractions at the same time. Anyways, the next step is to express these voltages in the resistors using
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our nodal voltages. So V1 is the difference in voltage between node 1 and ground. We can write that like this. Remember, we showed our current to travel down, so we are saying VA is at a higher potential. So it’s always higher potential minus the lower potential. So in this case, it’s VA minus 0. Let’s look at V2. V2 is the difference in voltage between node 1 and node 2. Our current
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is leaving Node 1, which means V2 is equal to VA minus VB. So the difference in voltage between the nodes. Why is that? Again, that’s because current always travel from a higher voltage potential to a lower voltage potential and we assumed current to travel this way. That means we assumed VA to be at a higher potential than VB. So we subtract the lower voltage potential value from the higher one,
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giving us the voltage in the resistor. In the future, we will not be writing down VA-0, since writing down zeros is redundant, but this illustrates how we use our nodes relative to the ground node to get the equations. Let’s simplify this equation. This is our first equation. Now we look at node 2 and write a KCL. We have I2 coming in, I3 leaving, I4 leaving and 4 mA leaving. If you are comfortable, you can directly write your equations in terms of node voltages
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without considering voltages in the resistors but for this example, we will go step by step. So we need to write these currents in terms of the resistor voltages and resistance. I2 is V2 divided by the 2k ohm resistor, I3 is V3 divided by the 2k ohm resistor, and I4 is V4 divided by the 2k ohm resistor. Let’s multiply everything by 2k and we can bring the 8 to the other side. The next step is to write all of these voltages in terms of node voltages. So V2, is the same as before,
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VA minus VB. Nothing changes because the current is still travelling from VA to VB. V3 is just node voltage B minus 0, but we aren’t going to write minus zero anymore, and the same is true for V4, it’s also just the node voltage at B. Let’s simply this equation. This is our second equation. Now we have 2 equations with 2 unknowns. We can solve for them using substitution. So let’s isolate for VA, and then plug that into our first equation. Now we can solve for VB.
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Using this VB value, we can plug it back into our VA equation and solve for it. We get -11 volts, which is also the value of V1 so that’s our answer. The negative voltage simply means that node 1 and node 2 are actually at a lower potential relative to our ground node. There is something important that should be noted from this problem. We can see that while this circuit had 3 nodes, we only needed 2 equations to figure out the unknowns. This
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is the case for all circuits. If you have, for example 5 nodes, then you only need 4 equations to solve for all the unknowns. Let’s take a look at this question where we need to figure out I0. Looking at the circuit, we see that it has 3 nodes that means we will need 2 equations to solve it. This bottom node can be our ground since it has the most connections. Let’s label this point as node 1 and this as node 2. Now it might be a bit hard to see how this circuit
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only has 2 nodes excluding our ground node, but remember, there is nothing between these points, so it’s just extra-long conductors. Even though we pick singular points to label as our nodes, all of this is node 1, and all of this is node 2. Now we can assume the directions of the current. Here, once again, keep in mind that current should flow towards the ground, because we assume these nodes
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to be at a higher potential. So for example, here, our current arrow should face this way. Also, if we are given current sources, we don’t assume any direction for them, we just follow the directions of current indicated by the source itself. Let’s label our currents. Now we can label our node voltages so this will be V1 and this V2. The next step is to write our KCL equations. Let’s start with node 1. We have I1 leaving, I2 leaving, 4 mA coming in and 2 mA leaving. It might be a
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bit confusing seeing I2 be part of node 1, but remember, all of this is node 1. Now instead of writing the voltages in the resistors like we did before, we are going to go one step ahead and write everything in terms of node voltages. If you are not comfortable jumping to this step, that’s okay, simply write each current in terms of the voltages in each resistor and then go to the next step. Anyways, so I1 is V1 divided by the 2k ohm resistor. I2 is shared between
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node 1 and node 2. The current leaves node 1, which means we assumed V1 to be at a higher potential so it’s V1 minus V2. That’s divided by the 4k ohm resistor. Lastly, 4 minus 2 is 2, giving us 2 milliamps. Now we can multiply everything by 4k, getting rid of the fractions. Let’s simplify. This is our first equation. Now I am going to pause here show you what happens if we assumed our current in this branch to go this way instead. What we are saying now is that
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our ground node is at a higher potential than V1. So if we were to write our KCL equation for node 1, it would be like this. We have I1 coming in, I2 leaving, 4 ma coming in and 2 ma leaving. Then we would write these in terms of node voltages. But, remember, we said the ground is at a higher potential, so now we need to write our voltage as 0 minus V1. It’s always higher potential minus the lower potential, but by saying our current travels bottom to top, we are essentially saying
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V1 is at a lower potential. So, again, it will be 0 minus V1. You still end up with the same equation when you simplify, but this can make you more prone to errors. So to overcome this, point your current arrows towards ground. Now let’s go back to our problem. So next comes node 2. Let’s write our KCL equation. We have 6 mA coming in, 4 mA leaving, I0 leaving, I2 coming in, and I3 leaving. Again, if it’s hard to see, remember that all of this is node
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2. Now we can write our currents in terms of node voltages. I0 is V2 divided by 12k, I2 is shared between node 1 and 2 as before, so it’ll be V1 minus V2 divided by 4k, and I3 is V2 divided by the 4k ohm resistor. We can multiply everything by 12k. Let’s simplify this equation. This is our second equation. We now have 2 equations, and you can solve them using substitution like we did in the previous question. To solve for I0, we just need to divide node
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voltage 2 by the 12k ohm resistor. We get 0.44 mA and that’s our answer. Now we are going to look at a question with an independent voltage source. Here, we need to figure out V0. We see that there are 4 nodes in this circuit which means we will need 3 equations to solve it. We will make the bottom node our ground and then we can label our nodes. Now we will assume current directions and label them. Lastly, we can label our node voltages. Since node
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3 gives us V0, I will label it V0. Let’s start writing our KCL equations, starting with node 1. All of our currents, i1, i2, and i4 are leaving the node so let’s multiply our equation by -1 to get rid of our negatives. Now we can write our currents in terms of our nodal voltages and resistances. So I1 is V1 over the 1k ohm resistor. I2 is leaving node 1 and going towards node 2, so V1 is at a higher potential. That means our voltage would be V1 minus V2 over the
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2k ohm resistor. Lastly, we have I4, which is going from node 1 to node 3, with node 1 being at a higher potential, so we have V1 minus V0 over 1k. Remember, it’s always higher potential minus the lower potential. Now we can multiply everything by 2k to get rid of the fractions. Let’s simplify. That’s our first equation. Next, I am going to skip node 2 for now and go straight to node 3. Let’s write our KCL. We have I3 coming in, I4 coming in and I5 leaving.
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Let’s write these in terms of nodal voltages. I3 is leaving V2, so it’s at a higher potential. That means it’s V2 minus V0 over the 2k ohm resistor. Then we have I4, which is the same as before, V1 minus V0 over 1k. And lastly, I5 is just v0 over the 1k ohm resistor. Let’s multiply everything by 2k to get rid of the fractions. Now we can simplify and this is our second equation. Let’s go back to node 2. What can we say about the node voltage of node 2? Notice how it’s directly
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connected to the positive side of our 12 V source. That means the voltage difference between node 2 and our ground is exactly 12 volts. We can write that like this. So the voltage difference is V2 minus the ground voltage, which is 0 V, and that’s equal to our 12 V power source. In other words, V2 is equal to 12 volts. That’s our 3rd equation and now we can solve for V0 and V1 using substitution. We get 4 V and that’s our answer. Let’s take a look at this question where we need
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to find V0. Now we are heading into the super node territory. We see that there are 4 nodes in this circuit. That means we will need 3 equations to solve it. We will make the bottom node our ground. Let’s label our nodes. Now we will assume current directions. Lastly, we can label our node voltages. Let’s focus on node 1 and node 3. What can we say about the voltage between these nodes? Notice how there is a voltage source directly connected between
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them. Also notice how the positive side of the voltage source is connected to V0, so that side is the higher potential side. That means we can say V0 minus V1 is equal to 12 V. That’s going to be our first equation. The second equation can be written about node 2. Let’s write our KCL. We have I2 coming in, I3 leaving, and 4 mA coming in. Now we can write our currents in terms of nodal voltages and resistances. I2 is leaving V1 and coming towards V2, so we have V1 minus V2 over
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the 1k ohm resistor and I3 is V2 minus V0 over the 2k ohm resistor. Let’s multiply both sides by 2k. Now we can simplify. That’s our second equation. Now we just need one more equation but the problem is, we can’t write one for node 1 or 3. Why is that? Because we have no idea about the current flowing through the 12V power source. Here is where our KCL knowledge comes into play. Assume that we create an imaginary boundary like this. Notice how it covers node1, node 3,
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and our power source. Now remember, KCL must hold true even for a surface like this. This boundary we created is called a super node. So we are going to write a KCL equation for it. Let’s go through it step by step. We have I1 coming out, I2 coming out, I3 going in, and i5 coming out. It’s important to note that we don’t care about node 2, the reason i3 is accounted for is because it’s going to travel this way and enter our super node. This is also why we don’t care
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about the 4 mA because it doesn’t enter our super node. Now we can write our currents in terms of node voltages. This part is the same as all of the other questions we’ve done so far. Let’s multiply everything by 2k. Now we can simplify. This is our third equation. We can now solve the three equations using substitution. Just want to point out that if you are allowed a calculator in your class, most of them can solve 3 simultaneous equations so check your calculator manual.
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Let’s take a look at this question where we need to find V0. Now we tackle dependant current and voltage sources. Looking at the circuit, we have 4 nodes, which means we will need 3 equations to solve it. Ground was already picked for us, so we can start labeling the rest of the nodes. Since V0 was already assigned to the top node in the question, I will label it node 0. Keep in mind, while we selected this point as node 0, all of this is still node 0. Now we can assume
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current directions and label them. Lastly, let’s label the node voltages. Let’s start by looking at node 1 and node 0. We have a voltage source there, with the positive side towards node 1. So that means V1 minus V0 gives us 12 V. That’s our first equation. Next, let’s look at node 2. We see that a dependant voltage source is connected between node 2 and ground. So that means the node voltage of node 2 is 2Vx. That’s the voltage of our dependant source. But notice that Vx is
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actually equal to V1. In other words, the voltage across this 1 kohm resistor is V1. So we can write Vx in terms of V1. That is our second equation. If we look at node 0 or node 1, we can’t write a KCL equation. That’s because we don’t know the current flowing through this 12 v source. So we need to draw a super node. We draw it to cover the 12 V source, node 0 and node 1. Now we can write
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a KCL for this super node. So what currents are leaving and what currents are entering? We have I1 leaving, Ix leaving, I2 leaving, 4Ix leaving, and 2 mA coming in. Let’s multiply everything by negative 1 and then bring the 2m to the other side. Now we can write each current in terms of nodal voltages. I1 is V1 over 1k, Ix is V0 over 1k, and I2 is leaving V0 and going towards V2, so it’s V0 minus V2 over 1k. Lastly, we have this dependant current source. But notice, we already
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established that Ix was V0 over 1k. So we can replace this with that. Let’s multiply everything by 1k and then simplify. This is our third equation. Now we can solve and that’s our answer. Let’s take a look at one last example where we need to find v1, v2, v3, and v4. Looking at the circuit, we see there are 5 nodes, so we will need 4 equations to solve it. We are given the labels for the nodal voltages so we will use those as our nodes. Let’s assume our current directions
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and label them. Now let’s focus on node 1 and 3. We see that between them is a 12 V voltage source, with the positive side towards V1. So we can say V1 minus V3 is equal to 12 V. That’s our first equation. What about node 4 voltage? We see that it’s directly connected to a dependant voltage source and ground. So V4 is equal to 2VA. Where is VA on our circuit? It’s right here, and what can
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we say about this voltage? Well, we see that it’s the voltage between nodes 2 and 3. We are told the positive side is towards V3, so that’s the higher potential side, which means VA is equal to V3 minus V2. Let’s plug that into the previous equation. Now we can simply. That’s equation number 2. let’s shift our focus on to node 2 and write a KCL. Next, we can write these in terms of nodal voltages. So I5 is V1 minus V2 over 6ohms, I2 is V2 minus V4 over 10 ohms, and I4 is
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V3 minus V2 over 8 ohms. Let’s multiply everything by 120, which is the lowest common multiple. Let’s simplify. This is our third equation. We need one more equation and we can’t write any KCL equations for node 1 or node 3 because of this voltage source on top. We have no idea about the current flowing through it. So instead, we need a super node. We can draw it to cover the voltage source and nodes 1 and 3. So now, we can write a KCL for this super node. We have 3 amps leaving,
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I1 leaving, I5 leaving, I4 leaving, I3 leaving and 4 amps entering. Let’s write these currents in terms of nodal voltages. Now we can multiply everything by 24 to get rid of the fractions. Let’s simplify. This is our 4th equation. Now we can solve these equations. If you know how to use a matrix from your linear algebra class, this is the time to use it. Otherwise, you’ll be doing some grueling substitutions or just use an online equation solver, whatever works for you.
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As a final tip, make sure you thoroughly check your negative signs, especially when solving your equations. That should cover the types of problems you will face when it comes to nodal
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