Recording of BioEng Lecture Series: Session 1



This learning series provides a comprehensive introduction to the fundamentals of bioengineering and electronics for medical professionals. Through a series of lectures, this series will go over the basics of voltage, current, resistors, meters, Kirchoff's laws, NovoLog methods of analysis and current voltage providers. Expert tutor Danny, who went through bioengineering last year, will guide attendees through their learning journey and provide practical exercises to help them nail down the techniques. Themedic bioengineering series is the perfect resource for medical professionals who want to learn the basics of bioengineering and electronics.
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MedED is delighted to announce the first instalment of the BioEng Lecture Series!

Session 1: Fundamentals of BioMedical Engineering by Danny Cowen

6-8pm run time, Mon 24th Oct

Learning objectives

Learning Objectives: 1. Identify the difference between voltage and current in a circuit 2. Describe the concept of positive and negative current 3. Explain the importance of measuring voltage and current 4. Discuss the application of Ohm's law and how it can be used to calculate the total resistance of a circuit 5. Understand the difference between analog and digital signals
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The following transcript was generated automatically from the content and has not been checked or corrected manually.

on live. And then let's wait for perfect. It's gone red. So that means everything from now on is going to be recorded. Um, so you can go ahead and give your introduction and start, and I'll invite everyone to the stage. They can interact with you. Perfect. Right now. Yeah. Okay, cool. So, everyone, welcome to the first lecture in Meds Bioengineering series. Uh, I'm really happy that we have a tone out of a few people. At least, um, it's pretty good. It would be a good lecture. Basically, I saw when I thought about this lecture series I thought would be good because, um, Medicare doesn't really have We didn't have anything for bioengineering, but I had all the other bs CS and I remember doing bio last year and being sort of it felt quite isolated from the rest of the medical school like there wasn't really much support. And even in the bio engineering department, um, there wasn't like, a society, that medicine, that we do tutorials. So I thought we'd give it a go. Um, having said that, bear with us just because, um, I'm doing things on my ipad as well be doing examples on there and like getting that to interact with medal and teams as well. You might see, I've got teams link up just here. It's a bit chaotic, so things might be a little bit like, a little bit slow. Just bear with me. I am going to try and get some interaction in if I remember to, um, I can't actually see your names so I can't call you out. Unfortunately, But if I ask you a question, please do respond. Um, and you will just be going through examples like the fundamentals of lectures. 123 basically in a nice chill manner. Um, this is these. These lectures are like the bread and butter, the first three of fundamentals and a lot of electronics as well. So it really is important to sort of nailed down the technique. Um, and there is a technique which is ideal. There is a structure to all of the madness days, bioengineering and electronic, especially, which is what we're going to delve into, um so that further ado Let's go. I'm assuming you can also your screen. We'll see my screen. If you can't please, please, please yell out. Because I can't hear or I can't I can't see your chat. Um, so yeah, right. So again, welcome. My name is Danny, By the way. I did buy engineering last year. Um, here's what. The issue is sort of going to like so again, basic electronics. At the very start, we have some of the laws. So Kirchoff laws, uh, then we'll go into the NovoLog method superposition. Uh, and then we'll go into current voltage providers, and we'll be practicing through as well. Okay, so basic electronic. Um, so voltage is obviously a crucial, um, concept to understand. It's probably best described on this this diagram here. Yeah, this one here. Um, basically, voltage is what we say. A difference in potential energy. You might see it written as potential difference between two parts of the circuit. And it pretty much means, um if there's if you look at this diagram, uh, there's going to be energy differences at this point, and then the current flows, and then everything happens through the component. And then there's going to be an energy difference here at this point and this point, we'll have two different energies. Okay? And that's basically voltages it just measures the two different energies. It is potential difference. Um so different. Don't get voltage confusing. Don't let it get confusing. It's It's quite simple concept, just potential difference. 11 number minus another number. And that's the voltage that has been consumed. If you will component just along the lines of denoting things, this is how we draw. The voltage has been dropped. It's a bit weird because if we imagine current going this way, so from left to right, um, which the arrow is showing, let me draw that. There's been a voltage dropped across component. We actually draw the arrow the other way. And if you see the two arrows, the current voltage going the same way, just as a little little rule of thumb something that's going to be a negative sign involved somewhere. So that's just something to bear in mind. Um, all of these things really as well that they are important to understand, but I would emphasize that it comes down to technique you can get by not completely understanding how electrons work and like the flow of electrons and just by practicing your technique in the voltage and superposition and things like that. Okay? But it does help to have an intuition of these things, Which is why I'm teaching you so current is the flow of electrons. Um, so this really is, like, the way electrons flow. Basically pretty straightforward. Doesn't really need anything more to be said about it, because it is so simple. And then you can see here, this area, the notes, current flow so currently flowing from the left to the right through the electrical component, which could be a resistant or something like that. And so we said that as current flows across this component, there's going to be a drop in energy. Right, So we know that there's a higher voltage, this left part here and a low voltage here. So we're going to lose out some of that energy because this electrical component have resistance. And that resistance, you can think of it basically spend or sucks up some voltage. Um and yeah, and that was January. That's how I remembered what resistance was. And it was really helpful to just have that intuition about it. So something important to know is that current will flow from a higher voltage to a low voltage. And if it does, then it's called a positive current. Um, and a good way to think of that would be like in biology. In a little biology, for example, you have your diffusion gradients. You have a really high concentration of particles, and it would spread out to go to low concentrations. That's pretty much what current does. It goes from a high high voltage to a low voltage, and that's a positive current, right? So you guys know that concept of high too low being a positive gradient. Perfect. Um, the negative current. We'll go from low to high voltage. Um, I don't really worry about this is it's like So even now, I don't completely understand it, but when you are doing circuit diagrams, you don't need to understand it just like this. Very beautiful. Do you understand how, um so, yeah, don't worry too much about negative current. It's just a minor sign, and it means that current will go the other way. But we'll encounter all of this. It's much more important to apply it than to understand it. Um, so remember, measuring things we're going to talk. We're going to talk about about meters and meters. So a volt meter we put in parallel with any component. So say we want to measure the voltage dropped across this lamp, right? We put a node or a connector for a wire of our volt meter here and then the wire here. Okay. And that will basically be the higher point here. Sorry. This is the positive node. So this will be the higher voltage because it's closer to the positive node in the cell. This is higher, higher voltage. Then there'll be a voltage dropped across the lamp as it dissipates energy in the form of light and heat as well. Um, and then there'll be a low voltage here, and this volt meter is just going to measure the difference between the two values. Really, really simple stuff. And 4 m have to be in parallel, I think, Doctor. Good. I want you to know that vote meters have infinite resistance as well. If you know what resistance is, and you know that a high resistance means that less current goes through, then that's good, because it means that all the current will go to the lab and onto the volt meter which is good. You don't want the volt meter to interfere in your circuit. You just want to measure, um, and contrasting to that. We have the meter, and it just measures the current flowing through a certain point. So that will be current flowing all the way across this circuit like this, literally in this fashion from positive or negative. And the meter has zero resistance, so it's basically just a wire. It doesn't. It doesn't, um, doesn't suck up any voltage. Doesn't use any voltage ideally, um hum, which is what you guys have to worry about. And it just measures How much current is passing through this particular point? Um, like I said for me to put in parallel and you can even see I've done a nice little diagram for you. I'm going to just see you in a little bit. Um, you have your vomiter put in parallel. Okay, um but this is again, this is all just understanding stuff, and you don't really get tested on this, but it's just nice to have this intuition of health as well. You will be tested on this. This is generally the bread and butter of all of electronics. This is owned law. I'm sure you know by now, just in case you don't run through it, it basically just says that the voltage across two points, uh, between two points is equal to the current. The pulses through that component times by the resistance of that component. So if I wanted to work out the voltage dropped across the resistor, I could, uh, just multiply the current that passes through the resistor and times by the resistance. And that would be the voltage job, easy stuff. And then it makes me sad that you guys and my year as well didn't really get any, you know, teaching past the precession learning. So I put this in just to illustrate the difference between analogue and digital. So are you are doing Those are typically digital because you just turn a pin on and you like, right, digital right high. And that just flicked it on like that, or you turn it low, which is off. And so because it can have only two states, we say it's digital. Okay, so high or low? One or zero to the same thing. If that pin is on, then it's high if it's low and then it's off and that represents a zero. And digital is obviously only two values as well. Um, so when you set your clock sta digital, you have a M and P M. Because there's only two values. Digitals, only two values and the dye in digital makes sense. Um, analog, you don't really have to worry about too much unless you're going to do the instrumentation module. But even still is really cool is again a lot of what makes up electronic. So it's good fun to use. Um, yeah, analog is basically continuous fluctuation of voltages between two values. So it's like a sine wave, basically voltages. And if that was applied to the lady just for your conceptualization, you might see the led fade because they're just going to go up and then down and up and down, okay, but minor. So getting into the laws of resistors this again is quite basic, but it's just important to to write it down for certain analysis. If I wanted to identify the total resistance, let's see if this work you know, in this soccer, then I'm just gonna use this. Actually, I do in this circuit here, then that's just a means of adding up. Are one or two or three the values of those resistors and that's it. Keep it simple. Don't get confused that Israel is. And then if I wanted to identify the total resistance of this circuit here you see, there are three resistors in powder, right? And the way we would do this is by using this formula. So it will be one over the first resistor, plus one over the second resistor, plus one over the third resistor. Okay, And that will give you your answer in one over the total resistance so you'll get an answer Will be, will be in this form. Yeah, you'll do. One over are one plus one of our two. And then you need to basically one over all of that. So it looks like this 1/1 plus one over to, and then that will be in the form one over our top. And then we want to take this value. We're going to do 1/1 over our top, and that will give us the total resistance. Makes sense. Cool. I'm going to take the silence as a yes as well, because if you guys do have any questions throughout, please just get out. Um, I'll be happy to answer it, but yeah, cool. So another general rule of circuit is that if the current split then reset, the circuit is parallel. Um, and this is illustrated on this slide, As you can see, um, so this this is not parallel circuit, another current split. And if you follow the little red thing, it just goes completely around continuously. There's no current splitting, but here, when we go from from here from this battery, we'll get to this point. But then current going to split it, it's going to go down. And it's going to go this way as well. And so because current does split, this is a parallel circuit. That's it. Nothing too confusing. Okay. And then again, it's quite small thing not necessary to worry about. I think we did get a question on this in our progress test. Actually, maybe it was in our mark, but it's just capacities. General rule is that when you're trying to sum up the overall capacity to the circuit, um, it's just the opposite math to resistor. So he said that for resistance and series, it's just one or two or three. All added together really nice and easy. No room for confusion. Um, but for capacities, um, if they're in parallel, then that's when we do that. So if the resistance and serious, then it's just all of them ended up their capacity is in parallel. Then it's just all of them added up. Okay, so it will just flip the resistant laws. Probably the easiest way to understand it. Um, just as a understanding point again, a capacitor is something that will store charge and then discharge. Okay, So, um yeah, a good example of pasta would be something. Um, that could act as like, a temporary battery in the circuit. They could charge up, and then it can discharge something like that. Um, but yeah. So, just for completion, the overall capacity of a circuit where all the capacities are in Syria's is followed by this one over C equals one C one plus one C two plus one, oversee three. Obviously, this is the exact same as resistance in parallel. So again it flips. This looks like resistance in parallel. Specifically this this equation here. Yeah, but then it flips because these capacities are in serious. So just remember that resistance and capacities when it comes to math complete opposite. Okay. And then, just as a side note, you might see capacity is down like this. Just, uh, differentiate them from batteries. So in my diagram, I drew here, as in the capacity has two straight lines, Right? And then the battery also has two straight lines. So just in case you get some of that's worse than I am and has really close closely length lines, Um, some people draw one line, the capacity is curved, and that's actually fine as well. But again, minor. So cut off of all these law, Um, it's quite straightforward. There are 22 of Kirchoff laws that you need to know. The first is the voltage law, and it's basically describes the in the loop. Um, all of the voltages? Well, some to zero. I'm going to demonstrate a few examples a sec. Um, but I just wanted to know that kcl is it's less used. It's easier to remember, in my opinion, just that all voltages in a circuit will in a in a loop, we'll add to zero. I think that's quite quite straightforward to remember, but kcl coaches current law is much, much, much more commonly used. Okay, so yeah, this is coach off voltage law. Um, let's jump into an example straight away. So I'm just gonna change change slides. I hope you guys can see that again. If you can. Please shut out. Um and yeah, so let's prove it. So we want to prove that all of the resistance is the resistance across this resistor across this resistor and across this resistor will equal zero. Yeah, the voltages across all three of those, including this one, will be equal zero. And typically the formula for that would be 12. So 12 volt minus, Let's call us are one called Cozaar to call. These are three. So the 12 minus the voltage across our one minus the voltage across our two minus the voltage across our three equals zero. And that's really what we're kind of aiming for, um, in this example. But I'm just going to prove it to you anyway. So to to find out how much voltage, um, is consumed by each of those resistors we need to use only. Okay, so I'm going to use the because I are. But just as a side note, quickly, if you ever see this, it just means the voltage across our one and the same for this voltage across our two. And that's just, uh, like electronic notation. Quite straightforward. But just as a little side note, because there are some some things that can just sort of get quite confusing. So just to take them off as you go cool. So the equal ir that's your own law. Um, and I want to find out the current throughout the circuit. And the way to do that means that I need to know the total resistance across the circuit. So I'm just gonna organize this to find current first. So current is equal to be divided by our Does anyone have any? Any problems with that? The current is equal to be done by by our If you have any problems or any questions, please, please, please get out. Okay. Um cool. So current is equal to the overall, which means we need to find, uh, total are which I'm just gonna call Cigna are someone Please shout out how I would go about doing that. Using the resistance is that we found out. How would I find a total resistance in the circuit if any of my bio and mentees here, by the way, you're you're getting at it now. We'll put it in the chat. You just add them. Yes. Fantastic. Thank you, ma'am. Good. You just add them. So it's really just going to be our one. Plus are two plus are three. Okay. Which means that's just going to be one plus two homes plus three homes just from these values here. Yeah, and that's just going to be six zones. Great. So we know our total voltage across the circuit. We now got a little resistance, so current is going to be equal to I was going to be equal to 12 divided by six, which is too amps. Great. Now we can sub homes log into this equation. I'm just going to bring it down here. We can send it in. We're using this. Whoops. We're using this, and the way we're going to do that is 12 minus. So we know that the was ir so current across our one times by the resistance of our one. Does anyone not see how that's the same as this? These two? The exact same? Because that that is quite fundamental to understand before you move on, Does anyone not understand if they if they if they don't then just yell out, we can go into in a bit more detail. But otherwise it is just something in IR for the Yeah, cool. So it's the exact same as exactly for the other ones as well. So minus R two times by our two minus, I are three times by our three and actually zero. So let's start everything in. So we know that current doesn't split in the circuit. It's just gonna give you the positive terminal across all of them. Okay. Which means that all of these resistance have the same current, which means that I are one is equal to I l t which is equal to two or three. Okay, so they're all equal. So I'm gonna say two times by the resistance of our one, which we know is one minus to for the current times by the resistance of our two, which I put two minus to the current again. Times by the resistance of three, which is three simplifying that down you get 12 minus two, minus 46. And then that week was zero. That's approved for Cady. Okay, you don't obviously need to know that proof, but I just thought I'd do it just for your understanding. Um, kbl literally just means that all the voltage is in the loop with some to zero. Okay, which we've demonstrated here. Fantastic. Righty. Oh, next. So that's the current law is probably the one you need to fully understand. Because, as I said, I don't really use KBR Last year, that case I was pretty pretty important. You won't? Yeah, It's probably quite difficult to get a first if you don't like nail down kcl. Thankfully, it's pretty simple. Um, and I've got a little diagram here for you guys to. So, um, Kay Ciel basically states that the voltage entering a node will be equal to the voltage leaving a node. And I've illustrated that here for you, So Whoops. Yeah. So this current coming in current coming in from my battery here, and I'll just call that I one okay, and then that's going to split into I three and I too. And we know how to write that down as an equation using kcl. So kcl states that the voltage entering a node is equal to the voltage leaving a node. And I've drawn arrows on this and they're quite arbitrary. I mean, realistically, current will flow from positive two. Negative. So this one coming out of the positive makes sense. Um, so I want is there. But then I just said that it splits this way. So going out of this node, you can see the arrow points away from this black dot here, right? So that means it's going away. And then there's another arrow going away here. Does anyone have any questions about arrow direction? Because I feel like that was a source of confusion in our year. Does everyone understand why I put the arrows the way they are? Technically, it doesn't really matter. In all honesty, it will still add up to the same thing. But it just like intuitively, makes sense to me that current is going to be coming out the positive note of the battery, and it's going to split and go in two directions back towards the negative note. That's probably a real wide stick by, Um, just when you're doing the circuit analysis, the current goes out of the positive node. Sometimes you might get a quirky kind of left field question where current will come out with the negative mode. Um, and that's that's fine. It will just be There will be a negative in that question somewhere. Um, yeah. So that's that's the current law. And it is best understood when you're using no voltage method. Ultimately everything that I taught you so far before this, um, well, just become an intuition, like, you just sort of feel for it in the question and know exactly what to do once you've done a few. So I don't really I don't really worry if you don't completely understand it now, but, yeah, the best way to get good that these questions is just to do them. Which brings me to my next point no voltage method. Now there are two methods that they teach you for circuit analysis because obviously, like with this circuit here, the resistant one that we just did like That's easy because it's all in serious. But when you get something like this. I'm not entirely sure how I would approach this if I didn't have no voltage superposition. Um, something to note as well. You do need to know how to do superposition and no voltage. Um, like in our exam. We got asked. Perform this by the superposition method. Or, um, like you can get asked as well performed this and analyze the circuit via the note voltage method. So if you don't, no one likes to lose marks you straight out. Well, but, um, you'll probably be better at one than the other. And that's fine. I was better on the voltage because it's much more methodical. Um, and I just referred it. It took less time. It was less confusing to me. So that's what we're gonna test out first. Okay, now, this is the same example. This is the first example we got to. I'm not sure if if you guys had it as well, but this is the first example we got taught, and it didn't really make sense to me the first time. I learned it in all honestly. And the more I looked at it, the less sense it made. So we're going to take this one really, really slow. Um, just so that everyone here completely understand it. And again, if you guys have any questions, don't understand anything. Just a mute. That's fine. It's less of a lecture than it is like, uh, editorial. So if you do have any questions, then just pop up and ask like, I'll be happy to answer them, um, and take it really, really slow. So, yeah, let's jump in. So the first step of no voltage um, you need to do is choose where your ground goes, Okay? Choose ground placement. Um, and there is some are to this, like if you wanted to put ground here, I mean, you could, but it's it's not the best place is it would be quite confusing. That's probably something I do if I wanted to test myself before an exam, if I run out of questions, but it's not something I would do to learn. Does anyone know that the most optimal place would be to to place ground? Does anyone wanna have a guess will make it easier? Is it a Is it a, B, C or D? Anyone? One in four chance I told you that's not a three close, not see. Not quite. It's actually D. And the reason for that is because you kind of at least in my head, I want to keep all of your charges, um, quite close together. So we know that current is probably going to be coming out of the circuit out the battery in that way, right? Just because that's a positive note and the current dose and positive negative, which means it's very likely that current will come out there, which is fine. Now, if you wanted to go from positive or negative, and that means that negative can't really be on this side, it could be and you definitely could do that. And you can solve question by doing that. But if you're learning the voltage and want to, like, completely grasp it, um, I personally would put here, I would always put my ground on a node. So when I say no, I mean like this black dot this black dot here, this black dot here and this black dot here always a node, um, and ideally, as close to the negative terminal of the battery as possible. Okay, that's like just just to make life easier for you. Anyway, I think intuitively in my head the negative terminal of the battery is zero voltage. Because all of the voltage would have been dissipated. Would have been, like, sucked up by the registers by this point, anyway. Like if you go back, Excuse me. If we go back to to here, there's been the voltage drop. Then we just proved it with KBR, right? We lost voltage across the first resistor. First resistor, the second resistor, the third resistor. And then we took all of those voltages away from everybody. 12. So by the time we got past our third resistor, there was nothing. There's no voltage there. Okay? So that intuitively makes sense in my head that before you get to just before the negative terminal, the battery that's going to be zero voltage. I hope that makes sense. It makes sense to me, and it comes in practice. All of this comes to practice. So if you don't understand it, but I'm doing it first, shot out and ask, um, the second, don't worry. It really, really does just come in practice. Truth Ground placement closest to negative terminal great. Um, so we've got the ground, and then we want to start labeling. Are nodes okay? So arbitrarily like it doesn't matter. I'm just gonna say this is no day. This is going to be no be. And this is going to be no see label nodes. And now we've done that. The next step is to perform Casey on these nodes. And the best way to do that would just be first off by inspection. So in med school, you do inspection at the end of the bed, right? First way to to do this would be by inspection. So a has has a current coming out here. This is a current source for you guys that don't know which means it doesn't produce any voltage. Just this is current. Um, and obviously, the arrow is telling us that the current is passing out here. Okay, that's fine. Sounds good. And then I'm just going to say that if current going that way and it's going in that way, then here's ground. So I probably guess that current in life, if this is a real circuit, would go that way up and then let's just say we'll go this way. Same way down towards towards ground or going to the right. Okay. So I can now do a equation in Kcl for know day, and I'll teach you guys how to do it first. So we said, Let's just reiterate what kcl is Kcl means that at a node current in in people's current out really fundamental like, yeah, I have that nailed down the understanding. Everyone understands that. I hope if you don't get out, um, that that is just that's like one of those things you need to take for granted. Like you just get told it You just have to accept it without kind of thinking about it too much. Um, but yeah, so that's That's no day and I'm going to call this. I'm going to say this is I one. Just because it's this current here that's being produced is the same current as this one here. Nothing's changed because there's no resistance or anything like that or no split in the circuit. Um, and then this current here, this one is going to go through our one, um, and it's still going to be, uh, the same current. Want to go through the resistor. Um, so I'm going to call this current IR one and then this current up here, I'm just going to call my are four because it goes through the fourth resistor, which is called our four okay, straightforward stuff. So I'm not going to for my first kcl equation, and I'm going to say that the current going in So which arrows are going in to know day only? I want cool. So I one equals the current going out now which current are going out of the nose and point it away from the node, which arrows pointing away from the node. And that's going to be your eye are one, and I are four. Um, I would also say that I haven't I'm not looking at solutions for these, so just bear with me because I'm thinking about it too. Um, but just to give you, like, an organic sort of demonstration of my thought process for circuit analysis, Yes. So this is Casey often a day Fine. That's perfect. That's an ideal kcl equation. And it's good. I hope that makes sense to everyone. If anyone has questions, then just yell out. Um, and we can simplify that a little bit further, actually. So we already know the value for my one. Uh, and it's it's 100 million hoops. Does someone to yell out what 100 million doses in amps for? For a point? No. Okay, it's 0.1 and you'll get used to the the units as well, just just putting questions to get used to it. And that's going to equal. I are one plus, I are four. So just copying that down from above, the substituting in the value which we know from here. Okay, cool. So something I'm going to do again is use the equals IR, and I'm just going to reorganize it. So we have in terms of we have current in terms of voltage and resistance, and we're going to come back to that a bit. But I just want to perform my circuit analysis for B and my know analysis for B and C as well. So is going to be quite straightforward KCl be so we can just put the arrows and make sure to be consistent. So you see the arrow that's going from left to right here, this one for our one Yeah, I'm highlighting in a lot of black ink. That's gonna be the same current, um, going in. So it needs to go in the same direction because that is where you get confused. Okay, then we can probably guess that if ground is here and current flows towards ground, usually then there's going to be an arrow going that way. And then I would also assume that or just guess it's in. It's arbitrary. You can put the arrows whichever way you want, but be consistent in this case here. But then, if I'm choosing, for example, this one next, what's going to be this arrow? I can do whatever I want because I haven't assigned one to this like strip yet. Okay, so again, current flows from positive negative. So I'm going to assume it's going to come up this way and go up here. So that's what my arrows going to do as well. So I'm going to go like that. I'm going to say there are two current going in Perfect, right? So, Casey Oh, for B, using current in equals, current out. Does anyone wanna? If I said this is, I are, too, by the way. This is IR one, and this is I are three. Does anyone want to form? I KCl equation for me. Just shut out. Uh, I see him. Who's in there? Oh, yeah. Guys, I can't see the chat. Unfortunately, Um, okay. Memory. Do you want to yell out? Hello? Can you hear me? Hi. I'm sorry. It's a bit noisy. Uh, it's, um we have IR Mm. I are three equal. I are one plus. I are too. Fantastic. That's great. Thanks, Mary. Fantastic. So current going out equals current going in perfect ideal. Um, that's great. Thanks. January. So I'm just going to move on to K C O S. C. Because this this is where it gets a little bit tricky. So we do have a current going from from the self and the battery, and then we're gonna have one here, too, and then we're gonna have a current going this way to We can just say that because again, from positive or negative, um, and then across were just there arbitrary, right? This arrow here So there will probably be one here just to stay consistent, but something that is going to be quite difficult is when we get to our analysis. So let's just say that this let's just call this eye zero and then this is I want and this will be I 22. So let's write it out. So current going and equals current going out. I'm going to say that I zero equals eye one. Plus, I are two, and now we're going to use our own floor. Okay, so let's start with I are too. We're going to say if I just actually, I'm just going to rewrite this and rewrite it as I are. Two plus one equals zero. I'm just writing that out. Awesome. So let's start analyzing my L2. In fact, what I'm going to do is I'm going to spell my screen just like keeping on the questions. Great. I hope you guys can still see that, but it's just so I can keep it on the sac as well. Um, cool. So I are too. We know the current is equal to voltage difference or voltage divided by resistance. Right? So if I wanted to write this out to be the voltage across our two divided by the resistance of arty right and we said at the very start of this lecture, that voltage is equal to a potential difference. The higher value minus the smaller value. And, um, we can literally do that math, like, right now, if you want to, which is in fact, what we're going to do. So that can be written as the voltage of notes. See? So this note here, Yeah, we can just say that Just a notation. The voltage that is at that node. Um, we're just going to call VC, and then we'll have a V B as well for the voltage. That's that. That node. We have a V A to, um, But for the moment, let's just stick with the C so v c minus VB minus V be divided by the value of our two, which is five ks. It says there. Um, but we don't need to worry about that just yet. Just symbolic purposes. So v c minus vb of our two. And the reason we do it as V C minus V B and not V B minus fvc is because we follow the direction of current. Okay, so we can see the current goes this way. Current is going to go this way. Yeah, great. Everyone says that. So that pretty much means that we have a higher voltage at C than it be. And we normally take our voltage measurements from the higher minus a smaller um, a k from where the current start, the the note at which the current is exiting minus the note of which the current enters. Which is exactly what's happening here and in words. It doesn't really make much sense, but when you start doing it and you'll see more of it in this example, it will make more sense. Okay, so that's what we can do. That that's fine. That's perfect. That's the correct answer. I'm just gonna struggle with out. Great. So that's why I asked you, I one we already know it's 0.1 amps. Yeah, and then I zero. This is where it gets tough because although this value is zero here because it's connected to ground, right, Um and this value is just VC. There isn't. We could write that out so we could say that I zero would be in owned long term's, um if you follow the direction of current, the current is going this way. Yeah. Following the area that we set, then we can say that it would be zero volts because of ground minus VC, divided by the resistance. But they're actually isn't a resistor in this little branch here. So it's stuck because we can't put it over zero because anything divided by zero is something back? Um, yeah. I don't know what that what that actually is. So we can't do that right. But thankfully, we can still analyze the circuit because we're essentially doing is trouble with this out. If we know that this is ground here, that's zero. And then there's five volts that comes out of nowhere. So plus five volts VC, we know that V C is going to be five volts. The value at C is five volts. Okay, so we don't really need to worry about. Okay, so let's see, um, because we know the equation. And even if we did, we'd still be doing zero minus V C, which is five now. We're sending that in zero minus five, divided by and are which we don't have. So we can't really analyze that part of the circuit. Thankfully, we don't need to. We now know the voltage of the C, and it's five. Does everyone everyone understand that that really does take practice Like, uh, I didn't really understand that concept of you can just assign a note of voltage if there's if there's, um, a voltage source connected to it. Um, when I when I was doing, like, right up until the very end of studying for the exams. Um, so if you don't get it, then don't worry. But yeah, that's essentially what's that's what. I'm just gonna leave. Actually, that's what that does. So we know that D C now is equal to V. C. Is equal to five votes. There's zero votes here, zero votes here because it's connected to ground. Um, and so we add on five volts here, plus five. And then we just go here. Which means that the C is five volts. Hope that makes sense. It is a bit of a you know, it's quite confusing, but that is the easiest way to think of it. There's zero, then you just add five. And because there's nothing else added, that must be five volts. So we have that value. That's fine. It's perfect, Um, which means that we can carry on with our analysis. And like I said, we are taking it really slow. Just so all of you guys can properly understand what is what is going on? Um, cool. So we're just going to take this equation from up here? I'm just going to steal it. This is our analysis of know day, and we're actually gonna solve everything in the bones, little wise, um, for either one and four so we can get it in terms of get each current value in terms of voltage and resistance. Okay, So ir one, um, in terms of voting resistance is going to look like the the voltage dropped across our one divided by the value of are one plus. It's just that that plus, um, those are the over our four, and that just means voltage dropped. Does anyone want to guess how else I could write down to the Like what? I'm going to put on that top top part of the equation. If anyone knows looking at the circuit we're analyzing know day, um and I'm going to sub in something for this. This note here, this branch no, no worries. It's quite difficult, but you will get it. So we said the voltage was again a voltage difference, right? And it follows the direction of current will take. We'll take the highest voltage so v a, follow the current and then subtract V be from that. Okay, that's literally just find rather current about the current arrow exit, which is via here, and then follow it along and then subtract v be from that. And that's what we're gonna do for the rest of this. And that's basically no voltage. Once you understand that concept, you're flying. So we're going to say that V A minus vb over are one plus. Now for our four, it's going to be V eight. So here, following the current arrow all across get the ground, which also has a value of zero, which is really nice. So v a minus zero divided by our four, and that is our whoops. That's an advantage for case er and we can just something values and simplify that, um as and when that's what we're gonna do. So 0.1 equals O V. A minus. B B is still unknown. We only know V C and B CS, five volts. Right. So we have that here. Um, divided by our one of our one value is 500 mg plus V A. So I'm just getting rid of the minor zero. Um v a divided by, uh 44 is 1500 so many times everything by 1500 to simplify it. So 0.1 times 500 that's 150 equals. I'm good. That's three. So three times v A minus V B plus v a. I'm just going to keep simplifying. That's a three V A minus three B B plus v a I'm going to simplify again. Four v A minus three d b and that is your super super simplified down. Um, answer for K C l A. I'm just gonna do the same with be, and then we'll have our answer. Okay. So, K c R b, we already have our, um, equation for K c B. And it's this one here, so I'm just going to copy that down. Great. Let me just do this. That's fine. Cool. So we have our case. I'll be here. I'm just going to do the exact same as I did before. So v Delta V over three equals Delta V over, uh, one plus Delta V over there too. So it is really just following the same structure of once. You have your case. You know, in terms of current which we had here, then you just suck in each the overall time for that. Okay, Um, cool. So if you have got to be over our three going to stop that in with some more equations with Delta V. So if you look at the current off by our three, it's going to be if we follow the current arrow, find where the current arrow exits from. So that's V B. Right? So that's going to be V B. Then subtract from where the current arrow goes into, which is ground. Here, Subtract that bridge is zero over the value of our three, which is 10-K. Um, then just do the same for one. So V a minus vb over our want 500 plus five minus B B over. Okay, Does anyone not understand why I've used five instead of the see? Because that's that's basically I'm doing this is. I could have written this as V C as well, and we didn't right, But I just stopped in from before with the C equals five volts. Okay, so, five. And I'm going to time everything by 10,000 just to get rid of our denominators. So be be minus zero times 10,000 is just B B minus zero, which is just be be, um, and then let's do something. And then that's 20 V A minus B B plus to five minus. Be. Yeah. So just going to simplify that down quickly? It turns out everything be, and I'm just gonna keep simplifying it until we get a really, really simplified answer. So minus 10, I'm just going to move to 10 over equals minus 23 day minus 21 day minus 23 v be lost. 20. Be a It's too. You just Yeah, Yeah, cool. And then this is, uh, second equation. And so, for the keynote of you probably see that we have now two equations to unknowns, and that is classic. Um, so the tetanus equation, which is exactly what we have to do. And then we have all of our answers Okay, so let's copy that. Bring it down here. Great. So, to get I like time, I'm going to aim to find 20. Via in this equation, which means I need two times. This one by 54 v A will turn it to 20 a day. Okay, so 1 50 times five and 7. 50 four V eight times five is 20 v A and then minus three V b times five is minus 15. Maybe contact. And then I'm going to add my equations together. We'll subtract my equations. So that will go 7 15 minus 10 minus minus 10 is 7. 60 and then minus 15. BP minus minus 23 8. So a baby and then we can find the be by just saying that 7. 60 to write about eight. It's 95 votes. Great. And then to find me A It's just the matter of subbing in, however, be value into this well into any of the symptoms equations. Let's just use this one, because the closest and simplify it down, which is what we're going to do. I hope everyone's okay. But I'm just gonna do offscreen really quickly. Um, it'll be 7 50 plus 15 times by 95. Divided by 20. Yeah, which is one of 8.75 volts. Great. And then because the voltages can defer depending on where you put ground, um, Doctor Google might ask you in your exam to find the current, which would be constant, irrespective of where you put your ground. Okay, so I'm just gonna calculate one of the current, uh, across one of the resistors. But you would do this for whichever one you're being asked to find out whether it was one eye or two or three. I'm actually going to ask one of you guys, um, tha find it for me. So you have your B B and your values. I want you to find me Final One, and Or if you could just tell me what the equation would look like they would put values into, um, but don't worry. If not, but have a crack. Uh, I wanna do you want to have a crap? No. Okay. I'll speed up just to get through this one. So I are one is going to be equal to be over our the voltage difference. Okay, so we take again Where does the current come out of it Comes out of the A. So we're going to do V A minus vb over the resistor value, which is 500. So via is 95 minus 108.75. Uh, be be over 500. Which means that I are one is equal to 95 minus one Oh, 8.75 divided by 500 which is minus 0.0 to 75, 275 amps, Which is correct. Um, and then that's no voltage done. And I'm going to speed up the next example just so you can see, um, like, just so you can make the links in your head as to how how it's done properly. And I'm just more speed. But that's it. Does that make sense for everyone? Everyone understand that? I hope so. Um, there's no thumbs up that I can see. I don't even think metal has a thumbs up feature, but yes, that's no voltage. Um, I hope that was easy to understand. Let me just flash back to our slides, and then, yeah, I'm going to go straight into another example. Some more quickly. Um, this one's actually from your supplementary questions on blackboard. If you still use black board. Domestic doesn't but bioengineering. My. So if you go to your electronic resources on blackboard, there are supplementary questions, uh, answers as well, and I just I've got this from there. That's a really good, really good resource for extra questions. They're kind of quite challenging. Like this one looks kind of daunting just because of how big it is. And there's multiple voltage sources, etcetera. But it's not too bad, so I'm going to start rattling through this one. So step one is find your ground. And so I'm just gonna use my logic from before and say, Where is the negative tunnel of the battery here? So that's where my ground is going to be a step to his label nose. So that's why I'm going to do next. I'm going to say that this is not a day I'm going to call this one to be. I'm going to call this one. Let's see, I can actually already eliminate one of the notes, and so I know the voltages, and it's because see is directly connected to ground, which means that we see is zero. Okay, so I'm just going to say from the offset that we see is zero, which is true. Um, cool. Let me just move this one down just to get out of the way, because I'm going to need that space. Yeah, Okay. Okay. So we've got a B c. And I'm just going to stop this one again so we can keep the sock on one side of the screen and we're going to go a little bit of a faster pace. Okay, So kcl at a I'm gonna say the arrows, they're giving you the arrow direction. So this is by our four. This is our three, and then this is going to be our five. Okay, So current seven equals current out. The current is going in, and I are four. Plus, I are three. Plus, I are five. What know equals ir five. Because I are five is coming out. I are four, and I are three going in now. I'm going to sub in my V e r equals I values. Okay, So, Final Four, I'm not gonna ask you guys to, uh I guess what I have for would be because it's a bit of a trick question. Really. You don't need to do if in any of your nodes, any of your branches. Branches. I just mean connections between two nodes. You have a current source like you do here. You already know the the current because it's only coming out that source, right? Do you ever see that in the branch? I like here. Then you know the current list of that gland. So I already know I to will be high to being here. That's going to be five milliamps. And I already know that I are four is going to be 10 million because they told me. And that's just how it works. And it it's just one of those things again that you just need to take for granted and not necessarily ask questions about but just getting your head. If there's a current source in the branch, then that's it. That is the current going through that branch. Okay, so by our four, you know is 10 million, 10 million times in and is 0.1. I'm actually going to check out 10 divided by 1000. Yeah, 0.1 arms. Great. I are three. We don't know, but we know it's going to be zero because of this here minus following the current arrow minus V A. Okay, so zero minus V a over five k for the value of the resistor. Excuse me. Equals are five so again following the current Where does IR five come out of it Comes out of the A. Where does it go into? It's going to go into V B. Yeah, yeah. Um so it can be be a minus B b divided by the resistant really, which is also five k so many times everything by 5000 just to get rid of that. So 50 plus minus v A. Because obviously the zero will just eradicated um, equals V A minus Phoebe. Then I'm gonna add, um, I'll be a to that side to be a mild CP, and you see how quick that was. That was That's That's one of the notable equations. They're easily done. We just need to DVD and resolve this. So let's do exactly that kcl be, um Right. So this one, this one's a bit funky. We actually don't have any labeled current going out. If if you guys can see that we have a high five and that's going to come in, yeah, I'll too and that's going to come in And they told us that I won is also going to come in now. That is less ideal because it's, uh, one of the things in electronics that can confuse you. I feel like there's a lot of those. There's a lot of things that aren't like one very small thing in the question that will just stop you, and that will be it, and you can understand the entire rest of the question. But if you don't understand that one thing, that's it. Thankfully, I mentioned before the arrow directions arbitrary. We actually haven't assigned a single arrow to this note yet, so I'm going to be naughty and flip it, and I'm just gonna start minus I one, okay, because it goes in the direction of what it would go into if that makes sense. So this is no be, by the way Kasey would be. I'm going to say we have IR five going in. We just change it to black ir five going in. We're gonna have ir to going in. But we already know what that is because, like I said, if you have a current source of the branch, that's it. So you have plus five milliamps five million chance is 0.0 five. Okay, 5 1000. Yeah. Um, that's going in equals minus I. Oh, one. Does that make sense to everyone? If it doesn't make sense to check me, it should make sense. And if it doesn't, then again, just please yell out. Great. That makes sense, then. Fantastic. Brilliant. Okay, so let's start subbing in our voltages. Um, so we can get to the final solution. So about five, that's just going to be again. Where is the current for IR five? Come out of it comes out of V a Soviet A minus. Where does it go into V B? Divided by the resistance of that part of the circuit. It's only got one resistance, so it's going to be that value five K plus your current value equals minus ir one. Now, if you look at the current direction, it's going to be going from V B to see which one it was zero and then divide that by the resistance, which is two K. So we have our equation. Okay, Sounds good. Um, I'm going to type everything up by 10,000 just to get rid of all of our denominators. So this is going to be a minus to be be. Plus, it's just straight up. Five 50 15 equals minus five vb. I've just found that by 10,000 again and then simplified the negatives. Okay, Hopefully not too confusing, so I'm going to move everything around. So we have everything in terms of the M V B. So we have 50 equals. Oh, actually, let's go back. Where is my, uh, three. Yeah, I have three. We need to do this again. I've forgotten there's a voltage source here that we didn't take into account. So by all three won't be zero minus v a. If we follow the current, we have zero. But then we add two volts. Okay, so it's actually going to be zero plus two or just to, if you want to write it out for you would be zero class too. So I'm just going to read you this quickly. It's going to be a so two minus K five k and then three times the up by five k started sticking to my V A. So that's just going to be 52. 52 equals two games on the B B. Yep, that checks out. Okay, good. Um, and then moving this over. So 50 equals minus five db. Moving over minus three B B minus to be a then I'm just going to be real negative. So just so you have post in terms minus five, it's time. Great. Okay, um, so now we have two occasions, okay? Which means that we can just do again are single test equation magic, and this will come up. So, uh, let me just bring this one down so you guys can see, um cool. So I'm just going to add, uh, subtract, um, so attractive one from two. Because we have to be a and R T v a. And you have two positives and you subtract so 52 minus 15 minus minus 50 is 102 equals minus V B minus three. PBS minus four V b And then TV A comes from the TV A So one oh two divided by four is equal to minus 25.5 volts, and that's equal to be be. And then we can just start that in 12 years of age. Okay, so we can say 52. Yeah, 52 equals two times or 52 plus V b equals to be a so I'm just rearranging equation, too. So I'm going to see that 52 minus 25.5 equals T v a 15 minus 25.5 divided by two So v a e equals 13.25. And then we would just sobbing hour be be, you know, be a into equations and we can find our current values. So let's just do that quickly for Let's do it for our five. So our five is going to be a my V b again following the current divided by our five, which is 5000. We know those values, so we're going to say 13.25 minus minus 25. 25. Divided by 15,000 equals 13.5 minus minus 25.5 by 5000 equals 7.75 times 10 minus three. Great. And that's your current. That's literally no voltage is easy to fly through. Then once you have the technique like a rattle down, you've done it a few times. It's easy. There's so much offensive position. But you need to know both, unfortunately, but that's no voltage. I hope that's clear. Um, it really is one of the easier ones to understand. Okay, so moving on, we have voltage device as well, just something to quickly mention they do try to catch up with wordings. Um, and basically so voltage divider. It basically describes how voltages going to be split across to resistors, and it does intuitively makes sense. Like I said, the bottom a larger sister is going to use up more voltage than the lower stuff. And you check that when you get your equation out and your value for the voltage across lost across the other one. So you are. One was 10,000 owns and are too on here was one own. You can bet that are. One is going to suck up more voltage than our tea. If our two is one of our one is 10,000 homes and that that's your sanity check. That's how you make sure that your answers, right? Just like intuitive without any math, Uh, bigger, bigger resistance. We use more voltage. Easy. Um, like I said, it's just a ratio of how the voltage used. So let's quickly do an example. And we want to find the voltage the costs are for So to do this, we need to first find what this total voltages. Um, and that's across our one or two and three. So I'm going to write this down like this and these two lines here, that means, um, in parallel. So our one is in parallel with our two, which is in parallel with all three. And remember, if you want to find total resistance, then we just put 1/3 K. So one over our one, um plus one of our two just three k plus one of our three, which again is three k. I'll just write the formula, get you guys to be what three equals one over our top equal. Which one of our talk. So what we basically have is one over 1000 equal one over on top. So if you do one over 1/1, over 1000 equal 1/1 over the top. That's being equal, Um, 1000, which is equal to our top. I hope that's clear. It's quite straightforward Mass, which is so it is clear. So our topic is now 1000, and I'm going to redo that second. Great. So this is 1000 homes and this is 1500 mg. Okay, And so your formula for voltage divider is here. So if I want to work out how much voltage is dropped across our one, then I use this equation. I basically say, Well, with this particular statement Whoops. With this, I'm putting our one on top and then adding two. So it's literally just a ratio. But you guys have done ratios before the existing medicine, like you're running aldosterone ratios. Um, so that's basically saying how much bigger is our one compared to our to and then sign? Voltage, Uh, are one. If it's bigger, some more voltage are one. If it's bigger and that's it, it's really, really easy. So I'm going to use this formula, and I know that is a five vote voltage source. So if I want to find the ER for one, this is my fault. And I'm just going to say this is our tea. Then my equation is going to look like the N times by our four. Divided by are four plus R t and again I'm trying to find the voltage dropped across our four. Remember? That's how you you draw the voltage arrow so current will be going this way out positive. But then voltage has the arrow going the opposite with the current. So again, current going to come around like that like that. And then you'll see the current faces that way, if one of them is if they're both positive in positive direction. Soviet is five times that by our four, which is 1500. And I'm gonna, uh, on the bottom 1500 plus 1000. Cool. So that is basically just five times by 1500 over 2500 or five times by 3/5 which is just going to have three volts. So the voltage dropped across our four is three volts and then by deduction, the only vault is left in the circuit or by KB. Oh, I should say, is two volts and that's going to be a signed Artie. So that's two volts. And then that's three volts. And that makes sense both in terms of KBL and in terms of what we said earlier. Bigger resistor sucks up more voltage. And in this case, that's happening. The big sister is indeed sucking up more voltage. Perfect voltage providers. Easy. Um, yeah. Does anyone have any questions about my goal today either? Let me just see if I can move slides. Great. Yeah, so that's the voltage divider. Then there's obviously current divider as well, just for completion. Um, and it's the opposite theory. So remember what I said. That would be, like, intuitive. We found out bigger. Um, resistors will use more voltage. Um, it's the opposite. In current divider, you're going to get less current flowing through big resistors. Um, which is why so water doesn't have more resistance. Um, but then obviously, why? It will have have more water. So that's why you get short circuits if you spill water on your electronics because all the current one instead divert to the thing with the smallest resistance. So current will be more current going through, um, resistors. That are smaller. That makes sense. Current takes the path of least resistance. Think of current as smart. It takes part of the least resistance. Um, and that's what happens. So this is the formula again. We have a similar sex before, but instead there's a current source. And interestingly, we actually have every one of our current, the current that goes across. We want to work out the current that goes across this resistor. Then you actually have to sum up one or two or three, which you just say is R T um, and put that on the top. So when he worked out the voltage, that would be going across our X, we put Rx on the top of the fraction, but instead, in current divider, we put, um, Artie on top. So I want to find the current that goes across here, and I'm going to put the other resistance early. And that's because, like we said, current is inversely proportional to the resistance of a, um, resistor. We have more current with less resistance, more current going through resistance with smaller values there inversely proportional, and it makes sense. Um, if this rt is bigger than the time by source. Then our ex will be bigger, right? Because the other resistance big so Rx the current going through the small resistance will be larger. Part of the least resistance makes sense. Cool. So let's do an example. I'm going to use this formula as well. Let me just move this again. Okay, so we said that we want to find the current through our three. Okay. Um so, interestingly, we actually we can do this by intuition. Um, some of again the chemo that you might have spread this. Okay, so we know that current is going to go across here. They can't split at this point. You can't split before here, but here it does split, and it's probably going to do something like that at this node. Then, at this note you got coming in from here, it's probably going to split up that we want to work out the ratio which is split across our one or two and are three. But something interesting to notice is that, um, through all the same values, so we actually don't really need the current divider equation on this particular instance. Instead, what we can do is just work out the total resistance in the dark. It cabinet, the current and then just divided by three. Because each of the resistors are one or two or three, they're all going to get the same amount of current because they're valued. Now, if I do enough really massive, let's say that they were like 10-K and 10-K, um and are one was one K. Then there would be more current going through our one because current take the possibility of distance makes sense. Good. All right, so let's just quickly do list. So our one is in parallel with our tea, you know, And that's in parallel with our three. And that's going to be the one at the top so we can do. We can say 1/3 K plus one of the three K plus 1/3 K 1 over the top three of 3000. All right, so you have the same thing the same seconds before one over everything, so I'll top equals 1000 three drawers. Okay, that's 1500. That's 1000. So you're one of our camper. Total resistance in the socket to it's in these two resistors, and now it's serious. We've done more up in there in serious. Okay, Um So what we can do now is just which is called this are These are for what we call this rt. So are four. Plus r t equals 2500. So 1500 plus 1000. And that is the total resistance to the circuit. We can then use own law and to work out the current can rearrange to find the current, and we're gonna put five divided by 2500 and that is going to be equal to two million, um, or 0.2 abs. Okay. And then he said, We know that now there's going to be two million ants. There is going to split equally threefold. So literally every single resistor of these three are one or two and three. We're going to receive the same. It's just gonna be two divided by three million's 2000 million. And that is the current two or three. Okay, so that's clear. Please. Please remember, the intuition of current divider, um, current take the path of least resistance. Larger resistors suck up more voltage. It makes sense. Like just get that in your head and you will never get any of these questions from promise you. Um, that's the key point, right? Okay, so this is the last last point now, um, and it's on superposition. So superstition is probably the thing that I hate the most in this course. It's cool to understand that if there are all linear components, um, in a in an example, in the circuit, then you can take things out and add them all together, and we'll give you the same with the voltage, but it takes longer. In my opinion, one of my friends prefer using it. I prefer no voltage. I think the voltage is is just better helped understanding of kcl. Um, with this one, As you see, you need to be a reorganizing the circuit and resistors. But the key points of superposition are that when you do it, you basically analyze the circuit with only one of the power sources. The power source has been current or voltage source. So let me look at this. We're going to analyze the second test. Um, in the first one, uh, we'll remove the voltage source and have only the current source. And then the second one, we'll remove the current source and have only the voltage source. And then we'll have the current in each of the resistors. Add them together at the end from each of the two iterations done, and they'll give us the same as the voltage examples needed for cool. So, um yeah, another really important thing is to be consistent with signs, because current are going to go in different ways in the two, um, iterations that we do. And if you mess it up, then you won't get the right answer. So you need to be conscious of, but negative. Because if the current gets flipped, then it has a negative in front of it. Okay, cool. Let's crack on the example, Um, and hopefully finish a little bit early as well. So just move this down, right? Okay. So first thing first, I am going to remove my voltage source all my current. So it really doesn't matter which one you do first, um, let's say I'll remove my current source first. So when we remove current source and do superposition, um, it basically becomes a an open circuit, which means there's nothing going through it. So I'm just going to redraw the circuit. But without the currency was and it would look something like that. But there's no current source, so this is open. So there's no current going through there that way. So I'm just not even going to bother with that, because that note is now completely relevant. Okay, great. So that's our for the one. This is our three that's up to you, and this is still my five volts source. Now we will start working out current, and eventually we're gonna have to use current divider because we know that there's going to be a current go through there and it passes up to our two, and then it's going to split, which means we need to know the total current of the circuit, and we've seen that before. We just like we did earlier in voltage our current dividers. We just simplify all existence down into a serious circuit, get the total resistance to the circuit and then new vehicles. I are because we know the have our total resistance. Just find the current, So let's do exactly that So, uh, I'm going to redraw this circuit again because it's quite hard to say, and it takes practice. It's quite hard to see how you can read wrong resistors. Um, and we did have a question on that on our exam. That was quite difficult. It was absolute chaos. There were resistant everywhere. And you have to simplify it down. Um, and yeah. So I would practice the kind of questions. Um, just putting it down so we can we can see. I'm going to read your work for you guys, so you can see if you get ready, this happens. And then this happened. So this is still too still the same thing. This is still our three, but I just moved our one up to sit right next to our four. Literally the exact same circuit. I just read on it, Okay. Exact same circuit. Every component is still exactly where it would be. Um, but now they're next to each other and in my head, like it just makes it so much intuitively easier that they're next to each other and so I can just push them together, is what I'm gonna do. I'm gonna simplify even more and say this is now are one plus are four. Yeah, this is our three. This is not true. Does everyone see that? If you don't, then like, yeah, shout out again Because, like, I can't see the chat. But, uh, I hope that makes sense. The next step is going to be to identify it. Um, which resistors are in parallel? Which one? C again. So it really is just simplifying it down. Resistors. Um so if you don't have a knife for that, you can see why you prefer no voltage as a process of position. But so I have are one plus are four, and that is clearly in parallel with our three. Easy, easy. So that's what we're going to simplify it down to. I'm just going to do the math as well. So one over our three plus one over are one plus four because now we're taking that as a single resistor equals one over X, a r t. Now I've done these questions like a lot before. It does like you just work. Um, practice it, getting in the eye for it. But essentially, just you write the entire thing as 1/1 over our three plus one over one plus four equals, Artie. So you just basically do one over that equals one over that, and this one simplifies down to our tea. When you do that, I got to make sense. Um, yeah. So this is our new our new value for our tea. So I'm just going to simplify that down even more to have not coming up. We have a new anti there, and then this is our two. Great. And that's a serious circuit again. So it's going to be our d plus are too. And that's five volts, I believe. Yeah, cold. Great. That's a circuit literally. That's a socket. And that is a total distance in the circuit as well, which is fantastic. So what I'm going to do is I'm going to suck in the values for resistance so we can get a resistance value. So they're saying R t r. The total resistance is equal to 1/1 over our three. Looking at this on my left are three years of 10,000 plus 1/1, which is 500 plus are four, which is 1500 and then all of that. Plus are two. Which one is 5000? I'm just gonna shove all of that into my calculator. And that leaves me with 6.6 recurring kilotons. Okay, so 6666 homes recurring? Um, yeah. So that's what the resistance we already know of the five volts. So you can get what we're gonna do. We're going to rearrange under law again, and we're going to put 5/6, 666.7. And that is the current part of the cycle, which is 7.5 times 10 minus four pumps. Oh, 0.75 milliamps. But we're just going to leave it in, and it's just for simplicity at the moment. Great. Okay, that's fantastic. So we have have got our current, uh, for this operation, Remember? We were still our first iteration. We've we've not analyzed it with the voltage sort of moved yet. So we now want to work out the current across each of our, um, each of our resistance. So I have one. I are two. There are three everything like that. We just wanna some all of that, because we'll be something. It, um at the very end. So following this circuit on the left. Now I'm going over to the left. A judge? I'm going to say the voltage comes out of the battery, it goes up this way, and then it's going to go that way. And then we can probably say it's going to get it that way as well. And then it's gonna put around that way. Okay, so I know that this current here is equal to this current here. Yeah, we went out the current circuit, so if you go down, you might make it easier to visualize This is the current going through the circuit. This one, this is I. And then as we go up, um, simplify our circuit. This is still high. Which means that because I have a split yet L2 is just gonna be straight away. I Okay, so I are two equals I, which is 7.5 times 10 2 months for cramps. And this is like the fine stage now of this. It's original superposition. Okay, um, we can also see that are currently going to split. Now, if we know our current going to split. That means we're gonna do current provider to see how it feels. Okay, um, which is exactly what we're gonna do. So I want to work out. I are three. And to do that, I need to know what I two is which I have. I need to know what the value for our three years, which I also have three schools, 10,000. And then I need to know according to this is great. This diagram here, what are one plus are four is so are one plus are four equals what? 500 plus 1500. So that's 2000. Okay, so I want to calculate, um, exactly how much current is going through our three. Now, remember what we said we were doing current divider? I'm not going to put our three on the top, and I'm going to put our one plus off four on the top. So I'm gonna do this. Are one plus are four over are one plus four plus our three because it's a fresh in times bye bye. Or in other words, times by 7.5 times 10 minus four. And that is going to be my other three. So if we create that, you can say that I have three is going to be equal to bear with me. It's going to stop everything into the calculator. 2000 Divide by a 2000 plus 10,000 times by 7.5 and 10 is the most for, So our three has only 10,000, and then the resistor above it has a value of 2000. This is 2000. This is 10,000. So intuitively I know that are three current is going to probably be quite small compared to our brain and our four because our three is huge compared to them. It's five times bigger than 2000, right? Which means that because it's bigger current, it's going to go to the path of least resistance. There's going to be less current going through our three, so L3 should be quite small. It is. It's 1.25 times 10 to minus four and cool. Something important to know is that we've set the arrows going this way. So this way, um, for our tea, which is fine, but we just need to keep an eye on all of them because when we do the second iteration, that might change, and that's gonna increase negative. But that's okay. Um, so we want to work out how much current now goes through the one in our four. Um, so this is we said this is I want which was our 7.5, This one on the right here. Whoops. Yeah. So that's why one we've worked out what I are three is just now it's 1.5 times at the most four hands, so we need to work out what I are one plus, I are forward. We can do that by current provider again. We could just instead, um, put our three on the top here, um, which we could do or we could use case, er so you know, the eye. One current going in equals I are three current going out. Plus, I are one plus. Plus, are four current going out. We know I want, you know, our our three. So I'm going to do one minus. I are three equals I are one plus four. Just Katie. All right, Um, and so you know that I one is 7.5 times 10 minus four amps to subtract 1.25 times 10 to a minus four pumps from that, and that will give us a value of 746.25 times 10 to minus four times. And that is I are one plus four. Let's see, are one plus four. In other words, that is this I are one plus or four, and that's 6.25 this value here. Okay, So, as you can see, the current is not going to split when we're going this way. Which means the value for the current of IR one is also equal to the value of the current. For I are four. So I'm just going to go up here to illustrate that. Um, so this is our three. This area here. Um, in fact, I'm trying to get a little closer. That's o L three ir one. Um, Plus, I often would that be the same value? Um, so that is 6.25 times 10 to five times 10 to the minus four. Fantastic. That's done. So we now have the same values, and I are one equals. I are four because it doesn't split. It's just gonna go around here like that. So both are one, and I forget the exact same current value. Uh, value is equal to 6.25 times 10 to minus four. So we have that we have our I are three value. It's 1.25 times 10 to the minus four aunts, and then we have a high arch to value. It is 7.5 times 10 to 1 is four, and that's done. First iteration of superstition completed next is going to be re drawing the circuit. Um, but instead of removing the council's, we're not going to remove the voltage source. Okay? Now, when we remove the voltage source, we keep our current source. When we remove the voltage source, we actually don't sure we don't have an open circuit. So I mean a second. So before this was empty and I just completely disregarded it because there would be no no current. There's no connection, right? But when it's just a rule, when you do superposition, you maintain your your current. You maintain your wire for your voltage source so it becomes a short circuit, and I'll say, Sure, it literally just means there's no there's no resistors know components. Just a little wire. That's exactly what happens. So I'm just going to redraw it. Um, this is now iteration to I'm just gonna have to use as well as we can come back to them that we need. Great. Um, so, yeah, this is going to be a lot of simplification again, So I'm actually going to redraw, uh, the socket, and it's going to look something the best. Now this is Yeah, this is a bit confusing. I'm not gonna lie. Like, if I was in, like, take back to October last year. Um, in bio and genuinely not chance I could do this question, Not a chance. So if you can and you're looking at it like how it's just done that like don't worry. Um, I've literally just brought this wire down, like, basically just shortened it. Um, this is the exact same target. Okay? So don't stress if you if you don't know what I've done, the exact same thing. Um but it just takes practice to be able to like to move circuits around like that and see when resistors are in parallel and in series so one. So we can see that these two are two and three are parallel. It's a bit weird because they're in the middle of a circuit, but they are nonetheless and parallel, so we can say that one over there too. Plus one at our three equals one over t. In other words, 1/1 over are two plus 1/3 e r. T. So we can simplify that down. Okay, that's exactly what we're gonna do. Do that. That's fantastic. So this is now four. This is now our t. Let me call it R T one. Uh, this is one. Okay, next contestant five we can see that are one and our t one are in serious, which is great, so we can easily simplify. I love it when stuff some serious because it's so much more straightforward and paddle. So I'm going to simplify that even more. Going to draw the big sister up here and then coming up That was gonna have the one resistor. The value of that is just going to be our one plus r t one. This is a four and now you can see there are four and this other resist that are one plus 51 also. Well, which means that we're going to end up with something that our total resistance it looks like this. Oh, in fact, we don't need to worry about that. We don't need to worry about that at all. So if we if we did that, we could simplify it further. But because we're going to be covering the current going through components, we already have our current source. So you just start cracking away. Basically, that 0.1 and we have our full value. It's 1500 homes are one class are t one. Let's just find out the one it's one over all over are two plus one over our three. So I'm just sending in my r T one value, which is this down into are one plus lt one, which was simplified earlier. Um, and I'm just gonna soften the value to that almost there. Guys, I know it's on Monday evening, but we're almost there. So thanks for paying attention. So far, one over our to 5000 plus one over are three. There's one over 10,000. Um, and that I'm just going to set that equal to r T two, but it's a little t t t two. Yeah, r t two and then just add them because, uh, yeah, we're just adding them because previously there in series. So I'm going to run that through my calculator and that comes out to 1 to 12. Mhm, Yeah, that's going to come out to 3833.3 bones. That's R T t great. So now it's going to be a case of current divider. Um, just bear with me, because it does it seem to be a Dre. We're just finding out the current spit. But you can appreciate this current there, and it's going to split go in that direction, that direction. Yeah, Cool. That's what you need to understand so far. So I'm just going to work out the current goes through our four first. So I are for and to do that with your current flyer. So the know a current is 0.1 and 0.1 times it by. If you're finding I are four, we're going to put our one or 32 sorry, a new value and that is just this here. We're going to put that on top, and then we're gonna add and find the ratio. Remember, our society check more current should flow through the smaller resistor. So in this case, we have our t two Which way, Which way is which? Has the value of 3800 then we're going to have, um uh four, which has 1500. So in theory, are four should be, um, conducting more current than our tea too. And that's how we know how we know that we're on the right track. So let's do that. I'm just going to plug in the value for that. It's gonna be 0.1 times by our t two, which is 3833.3 divided by 3833.3 Plus are four 2500. Um, and that is zero point, actually. You know, I'm going to use just gonna sub in the original value so we don't get running hours, which is a good practice. Usually get into, for example, to one of the, uh I think I'm just typing everything out. Plus more than 500. Yeah. Perfect. It gives me 0.0 71875 lbs And using K C. L. We know this current going out this. Come. Yeah. Okay, that's number one. Yes, we know it's current 01 ounce. We just worked out with this current whoops. This current is this value, so we can do kcl. Um And it's going to be 0.1 minus, uh, are four, which we just calculated. So I r t two equals 0.1 minus 0.718 75. And that is 0.0 to 8125 amps. Okay, so we know our current. We got 0.1 coming out here. We've got IR for their for and then we're gonna have Bell Ti to going there. Now, we don't need to worry about calculating a new eye are four because it's going to go around the whole way. But I are teachers going to go through both of these resistors. I are one, and I are, and it's going to go through both resistors. So are one and our t one. Right. So we know are ones current because it's going to be that. So I are. One is going to be equal to the same thing. 0.0 to 8125. Fantastic. Then R t one, it's gonna be the same thing, right? Cool. But you can see this is the same circle. We're just working back in operation. Um, so we have our i r t two there, and then it's going to split this way this way, uh, across our to and are three. Yeah, you can appreciate that. Uh, so we're just gonna use current divider again? Uh, I'm almost there now, so I'm just going to copy this so I can bring it down. So I want to work out. I are too. Um So I'm going to put the current, which is the one which is 0.0 to 8125. So this current here times by going to work out, I are too. I put I put the resistor values are three on top. That's 10,000 over 10,000. Plus are two, which is 5000. So 0.0 to 81 five times by equals Said I are to 0.1875 amps and then by kcl just quickly, Um, that's going to be 0.0. So 0.0 to 8125 minus 0.1875. So you can see that I've just used case er and I said that if this current IR one goes in and I know now, this current then to find this I'm just gonna subtract this current are two from our one. Because I are too is a fraction of our one. And then that will give me how much current is left for our three. Right? And the value is you might have to stop. That is 9.375 times 10 to the minus three. Okay, so we now have ir one ir 23 and I are four for the second one, which means we're done. We just need to add up now. And that's superstition done. But we need to run into a problem or we're going to run into a problem. And that problem is going to be hours may not be consistent across the two by grams, which really annoying. But once you do it once, then you realize that you need to keep checking. So I'm going to compare our first iteration for arrows in our second iteration for arrows. Okay, so we said that does that does that. Great. So when we remove the current source, let's have a look at my arm. One first. So across here, across here, Whoops. We can see the arroz. So the current arrows to these red things across in different directions. Yeah, you can see that. These are pointing towards the right on the right hand side of my screen and then on the left hand side of my screen, they're pointing towards the left. Uh, so that means we're gonna have to introduce a negative. And this is when we start something up because we have our values. So if you want to be consistent, we do need to use a negative, because the current is basically flipped. So if I want to find true, I are one. I need to sum up with my IR one value from example one on the left and my ir one value. My example on the right. Okay. And that is going to be a case of Yeah. Where is my first one. Let's have a look. Uh huh. Right. So let's just copy this down so fast. It's racing. Great. It's our first situation number one. And so yeah, so I are one is going on to the left on the first iteration, and then to the right of the second iteration. You guys can see that, um, which means that we're gonna have to have a negative sign when we actually complete I want, which is what we're gonna do now. Um, and the way that we can stay consistent in doing this is just by Well, it's after it. It doesn't really matter. As long as you throw a negative in there, Then then you're being consistent. I would personally, um, put and negative across the second the first one, Although it doesn't matter. Just comes flip like you want to mark. It just means your current flipped. Um, so 6.25 times 10 to minus 43 to minus the second value. Where's my second value? 10.0 to a 25 0.0 to 8125. And that is going to equal 6.25 times four or five minus 0.0 to 75 amps. That's ir one eye or two. Um, we need to check that We've been consistent signs across both examples, so we haven't, as you can see, right? So in this example, the arrow in this example on the left arrow is going upwards. But then in the example on the right of the diet is going down. So I'm just going to say that I are, too. And the second one was negative, actually, no, we need to be consistent. If we've assigned the second was negative, then we should do the same for this one. So I are too. First value was 1.25 times 10 to the most four. And that's just from here. Minus a second value is 0.0 1875. Yeah. So 1.25 to 24 minus 0.1875 0.5. Yeah. Not equal 0.18625 minus minus that. Okay, I are three is equal to, So we need to compare signs fast. So we say that in this example we relieve the current in the current PSA was in. We say the current across our three is going this way. So it's going this way. Which means that on the circuit above, where we going this way, we may consider is consistent on this side. It's going from left to right, Yeah, which means we can just add them. So at 1.25 times 10 to the minus four plus 9.375 times 10 to minus three know, give us 75. That's pretty 0.1 Oh, 625. And then finally, for I are four, we just do the same. So we compare the arrows going the same way, we'll be consistent. It looks like we buy The arrow is going up from the current source. And then when we remove the current source, it was still going up, so we can just lower values. Fantastic. So 6.25 times 10 minus four plus 0.71875. Your joints here are the one 6.2 bottles of the first one and 0.71875 718 75. Yeah, yeah, plus 6.25 times the most four. And that gives you your 0.7 to 5. That's it. That's super position. You can see that. It it seems like it's a bit more error prone. Um, which is why I prefer not voltage. Honestly, like it is a bit more confusing. It does take more time as well. But you need to know it because you could get asked in the exam, um, to do it. Which is why it's so important. Um, yeah, so that's That's the first of the metal lectures for bio and I hope you guys enjoyed. Does anyone have any questions? Know Also, it's approved, you guys that it works and we got the same value. This is by our one using superposition. And then if we go back up to the first example, we calculated either one. We got that. So it works, But I prefer the voltage. Yes, that's it. That's everything. Enjoying Monday, Guys, if you have any questions, then please feel free to email me. Um, whenever, uh Or, like, send me a message. Whatever is easiest for you guys. Hope that was useful. Um, I'm gonna stop sharing. Not see. Yeah, thanks. Everyone for coming. Um, keep an eye out for the other Lectures are going to be coming up. Um, and then if you have any questions, you can ask now where you can just come in the mail. Mm. Yeah. I'm just gonna find the feedback for