Solving KVL & KCL Circuit Examples #1 - How To - Electrical Engineering 101 - Episode 12
Hey, hey today, is very exciting. This is going to be the first time when I use the concepts that I've taught so far to solve and analyze an example circuit. So I'm gonna I'm planning on using KVM and KCL to solve probably the most simple contrived circuit.
You can possibly come up with it's going to be the case of two elements together. So you can't have a circuit with less than two elements. So a single element can never create a circuit. Uh, well, I guess it could if you short its terminals. Together with an ideal conductor, but it's, not very exciting. So let's talk about a two element circuit, and I'm going to use one source and one passive element, just so that something actually happens. So let me draw this out what I've drawn here is an ideal voltage source.
And my rendition of a resistor. There is another way to draw this symbol, but I like to draw it this way. So this is a resistor. And this is an ideal voltage source the symbol with it's a circle with a little plus and minus. So for the. Sake of this problem, I'm going to give us a little of information, so we're going to say that we have a voltage across this voltage source.
And now I'm going to draw these polarities like this. I really don't have to do this. Because the way the source is oriented. Usually this is this implies that the polarity follows the plus and minus on the source, but that's, okay, and I'm going to write v. S, here I'm also going to write r here to represent the resistance. And just so this isn't symbolic and.
That we actually have the feeling of solving a issue I'm going to assign some values or numbers to the source and to this passive element. So for the source I'm going to say, let's say, vs equals 10 volts, and I'm going to say that r equals 5, ohms. Okay. So what does it mean to solve a circuit there's? No right?
Answer, that's, really up to you. It depends on what you want to know in that situation in a lot of classes or textbooks or homework issues when your professor tells you to solve a circuit, what. They really mean is just identify all the unknown quantities that you possibly can, and that usually implies voltages currents powers or maybe even a resistance. If you don't know, the element. So in this case given that we know this source voltage, and we know this resistance I'm going to show you how to solve for the current in both element in both elements, the voltage across each element and the power either sourced or consumed by every element. So that's, typically sufficient we'll say that let's. Identify the voltage and current and power in every element, and then we're satisfied with our knowledge of this circuit.
So to begin, we've got a voltage source. So it's, probably a good idea to try and identify as many voltages as we can in the circuit before we move on. And the first thing you want to notice is the fact that these two elements are connected together via ideal conductors, so there's, two ways. Let me just say, I automatically know the voltage across this resistor and there's, really.There are two ways that I know it.
The first is the fact that both elements both sides of each circuit element are connected by ideal conductors. And if you remember that means that well there's no voltage across an ideal conductor, which means that any two points connected by an ideal conductor share the same voltage at that node meaning that I could equivalently just draw this plus and minus vs or 10 volts directly across the resistor r and that's valid. That's, one way to look at it in an equally. Valid way would be to say in this special case. These two elements are in parallel. And if you remember a consequence of circuit elements being in parallel is the fact that they must share the same voltage. So that's, really just two ways of saying the same thing.
So if I go ahead and write this plus minus, and I write VR for the voltage across this resistor they're, really equal. So the first corollary, I can draw from just taking a look at this circuit is that vs equals VR equals 10, uh, volts. So now that we have a voltage across this resistor, we actually can completely solve for all all the knowledge that we need about this resistance. And the first thing it really well, the only way to do this. And the most simple way to do this is, uh, just to talk about ohm's law. So if you remember from my recent video on it, ohm's law says that v equals I r for a resistance, which means that the voltage across that resistance is equal to the current through that element times this constant of. Proportionality that we call its resistance.
So given that I know the voltage, and I know the resistance of this resistor. I actually automatically know the current through it, and I know it because I equal v over r, or in this case, we'll say, VR over r, which is 10 volts over 5 ohms, which gives us very clearly, two, amps. Okay, great. So I know the current voltage and resistance of this element. And I actually didn't even need to solve for the current, but I also know its power. So we now know the power. We could have solved for it with, uh, just its voltage and its resistance.
Because really if you know its voltage and its resistance, you know, its current, and I could also use its current in conjunction with its resistance to solve for the power, or I could use current and voltage together and solve for the power. But before I do that real quick, I just want to point out that. Now we actually also know the current in the other circuit element. So the is, um, well, I guess it was sort of unfair of. Me to say that these circuit elements were in parallel because in this ultra simplified contrived case, yes, you could say these circuit elements are in parallel. It turns out they're actually also in series.
So if you think about this, this loop, there's only a single path for current to flow. And if you recall how I described the way circuit flows through the way that current flows through circuit elements, uh, you know that all the current that flows through this resistance has to flow through. Uh this voltage source, and so then pause for a second think about what I titled this set of notes. I actually when I said that the voltage across these two elements is the same via the fact that they're in parallel via the fact that they're connected via ideal conductors what I was really doing. There is applying KVM back of the hand, saying that the sum of the voltages across, uh, this loop must ultimately wind up in zero. And now what I'm saying is I'm actually applying KCL.
Um by saying that the current through these two elements must be the same, because if you look at either node this side or this side, any current that flows this way from this resistance, well, it has to go through the source because there's, no other path. And so you can see how I didn't even really need to use the words, KVM and KCL. In this case, I didn't need to step through that analysis of counting voltages or counting currents entering nodes.
Because this is an ultra simplified case. There's its. Only two circuit elements, and each node only has two directions that current can go. So it really is that simple, but it just goes to show that you can apply your intuition in a lot of cases.
And I guess what I'm saying is KVM KCL, the conservation of energy in charge are really intuitive, and you'll start to pick this up and not necessarily need to step through rigorous calculations every time you saw the circuit, um, at least in these simple cases. So just remember that what we're really doing. Here, um with my reasoning and intuition is KVM KCL. And so I know I just jumped off on a tangent there.
But what I was saying is we essentially automatically know that not only do these two elements share the same voltage, but they actually share the same current as well. So given that we know ir, we actually know I s as well. And so we'll say, I r equals I s equals two amps. And now, okay, now that we know the voltages and currents for each element, we can actually calculate the power. So let's go. Ahead and start with the source now, do we remember our convention for currents and voltage polarities when it comes to power? So in this case, I'm going to say that because this current is entering the negative node of this circuit element that it's a negative current, or this is going to create a negative power, which we know because it's a source.
So the current is leaving the positive node entering the negative node I'm going to assign a negative polarity of that current. So the power in the. Source p source equals it's v source, and I'm going to say times minus I source, which equals 10 volts times 2 amps, which is 20 watts. And now we can do the same thing. So this is the check you always want to do remember a KVM measures conservation of energy, KCL, maintains conservation of charge. And then one more thing you can do to double-check all your work.
When you analyze a circuit is is is conservation of power maintained. And so really this conservation of power and energy are really saying, the. Same thing so this is just one more check you can do here. So we know that all the power created by this source or all the power sourced from this element must also be consumed by this element. So there's, no power, disappearing, right. All the power that comes out of here has to go into here. So we know that these two things have to align, and how can we measure this?
Well, there are several ways we can calculate this. It would be super easy because we know, we know, VR and vs are the same and IR and is are. The same so actually by writing this out, um, we've essentially already solved for it.
Uh with the caveat that this must be a positive power. And if you notice the flow of current through the resistor versus its voltage polarity it's, actually the opposite direction. So in this case, I'll say that this has a positive current.
So p r. Oops, p, r, power consumed in the resistor equals v r times, and I'm just going to say, positive, just going to say IR here. So then I'll say 10 volts times. Oh, excuse me. This. Should be minus here we go then we're 2 amps equals. And this is positive, 20 watts. So again, uh, there's, there's, really there's, no such thing as negative energy, there's, no such thing as negative power.
So to speak, this polarity, uh, just refers to the fact that it's being sourced. So what we're really calculating is we're saying the power consumed is negative, which means it's just sourcing power. And in the resistor there's, a positive power consumption, which we expect. And with that.
Being said, there's, okay, so there's, two other ways you could calculate the power in this resistor. You could use the voltage resistance relationship or the current resistance relationship. And so let me just write those out equivalently. Pr equals. And if you remember again, this is just an application of ohm's law.
So we have v equal sir. This is ohm's law, and we know that power is iv. So by substituting in one quantity or the other, we can get this in terms of the resistance. So I could say that pr, Equals VR squared over r, which in this case equals 10 squared over 5, which is 100 over 5, which is 20 or 20 watts, like we already know. And then the final alternate method would say pr equal sir squared times r, which equals. So we have 2 squared times 5, which is 4 times 5 or 20 watts.
And so you can see how that works out no matter what you do. And this should give you great confidence. Right here that the total power as you step around this loop. So that this basically says conservation of power. The total power sourced equals the power consumed and that's a great way you can check, uh that you've done your circuit analysis correctly. I hope this was helpful, um, I'm, trying to decide what I'm going to do in the next video.
If I'm going to do a few more examples. I think what I think that's. A good idea I'm going to start gradually increasing the complexity of these circuits, um and continuing to do the analysis and walk through these calculations, maybe I'll get I'll get to circuits that have. You know, multi-path nodes and many more elements, and we'll start doing rigorous application of KVM and KCL, because you do need to know how to do that and start setting up algebraic systems of equations to solve all these things. So I think that's the direction we're going to go before.
I move on to more advanced material. We'll just get perfect at applying KVM and KCL, and we'll use some other elements like possibly some dependent sources or a current source or something like that. And then. We'll start getting into possibly adding more resistive elements.
So again, I hope this was helpful, let me know if there's anything I can do to improve these videos, going forward, and thanks for watching and have a great day.