Hi, welcome to another Fundamentals Friday video. This is a follow-on to a previous video I did on Kirchhoff's current law and Kirchhoff's voltage law. so click here if you haven't seen that because that was a build-up video to what we're going to do in today's video, which is have a look at some basic DC circuit theorems, specifically Nodal Analysis, mesh analysis, and superposition theorem as well. And we're going to actually apply Kirchoff's voltage law and Kirchhoff's current Law, which we learn in the previous video.

to analyze a real basic circuit like this. Doesn't get much simpler than this, but you'll find there's actually a little bit of math involved little bit of cleverness in how you actually solve this. So what we're going to do here is we're going to solve this basic circuit in three different ways: Nodal Analysis, mesh analysis, and superposition theorem, and hopefully come up with the same answer for all three three different methods to analyze it. First one we're going to take a look at is Nodal Analysis now.

Nodal Analysis uses KCl Kirchoff's current law that we looked at in the previous video to actually solve this value. and what we want to solve is the current through this resistor here. So what we've got is a basic red circuit with three resistors and two voltage sources. Here, this is one a year like your more classic textbook circuits they give you when you're learning nodal and mesh analysis and Kirchhoff's current laws and things like that and solving them and it's a little bit tricky.

it's actually not obvious at first glance, aren't by all means go away right now if you don't know about Kirchhoff's current laws and try and solve this and see what you get without using a circuit simulator. Yes, we can just whack this into a circuit simulator and calculate the current down r2 here. Easy peasy, right. Done in a couple of minutes.

But really, we're trying to learn fundamental electronics here. It's a fundamental Theory Nodal Analysis and mesh analysis that enables really deeper, rich mathematical treatment and calculation of it. Lots Ton of stuff in electronics, but we're going to look at a basic circuit today so that you can understand how these techniques work. In this case, Nodal Analysis Oh Fun, right? So what? We've got two different voltages Just to mix it up a little bit.

This one's 10 volts positive up the top. This one's 1 volt. I've labeled them E1 and E2. so there's two voltage sources and three resistors like this.

We want to determine the current through R2. We've got three different value resistors here I've just chosen them different values because we can. and basically we've got a node here. hence the name Nodal Analysis And this is important to understand: We've got one node in this circuit I've labeled at A and you'll see why in a minute.

Now We could actually use this same nodal analysis technique to analyze any number of nodes in a circuit. But today, we're going to simplify it to one because there's already going to be enough involved. Is actually solving just one node here and you might think Dave we've got two nodes of one down here. Well, no, because what we've got is when doing nodal analysis, we need a reference point.

So we're going to call it ground like that. We're just going to put the ground, see what we're going to say. This is our reference point so we don't have to worry about this node. We really want to solve this node up here because what we want is like the voltage at node A relative to our reference point.

Here, this point is exactly the same, so there's no point analyzing that node. So basically I only got one node in this circuit that's of particular interest to us and we want to solve for now. of course, in nodal analysis, we can actually choose any reference point we want. It's pretty much arbitrary, but hey, we're just going to pick this.

It makes it easy and familiar to you. Now, this node is a junction, is it not? We learnt in the previous video Kirchoff's current law that the current entry into a junction must equal to the current leaving the junction or in other words, the algebraic sum of the currents at a particular Junction is equal to zero. and we're going to use that formula we learnt last time to analyze all the equation that we learned last time to analyze this thing. Now, by convention, although you don't have to do this by convention, we are going to assume that all currents are leaving the node like this.

There's no current flowing in and yet that's not possible in practice. Okay, but but trust me, it'll work out in the end because technically we don't know. Without, you know, analyzing the circuit, we don't know which way the currents are flowing depends on the voltages that they could be flowing in and out. We don't know.

So we're choosing an arbitrary reference point. And as I said, by convention, nodal analysis just have all currents leaving the node like this. So we'll draw Direction arrows. Now, we'll start off by deriving the equations for each current, which I've labeled I 1, I 2, and I 3 here.

So we're fairly consistent. R 1 equals I 1, etc. so you know it just keeps it nice and tidy. So we start by deriving our equation.

Okay, so I1 Well, what's I1 equal to? Let's go back to Ohm's law. We know I 1 is flowing through our 1. So how do you calculate the current through a resistor voltage and the resistance right? So we can go V ie. because the current is flowing A is more positive.

We're talking about conventional current flow. A is more positive than this point over here. So we go V Ay minus this voltage here, which is actually E 1. We could have called it like node B if we're doing you know a multi node analysis and things like that.

but in this case it's E1. So VA minus E1 that is the voltage difference across the resistor. the voltage drop across the resistor there and divided by V on our I equals V on our Ohm's law. r12 Easy And we can just plug in the numbers that we know.

R 1 is 10 Ohms V 1 is in this case 1 volt and we drop the unit's here. It just makes it easy. trust me and so Ll. Now our equation is VI minus 1 divided by 10.

That is the equation for I 1. Beauty. Two more to go. So let's do I - what is it? Wow, it's pretty easy.

Remember, it's always a reference to back to this reference point: V A The voltage at Point A What's the differential voltage just like we've got here instead of V1? Look, it's ground that's connected directly. So VI its - nothing. So it's just VI on Aa - easy Ohm's law and I forgot that's actually our two there and we plug in the number that we know we know our two. We still don't know what VA is.

We have no idea it'll come out in the wash and so it's VA on 20 and I three. Well pause the video and try and do it yourself. It's easy peasy. Look, it's going to be once again VA Our point.

Look how current is flowing out. So VA is the more positive side according to the current that we've chosen arbitrarily so. VA -. well, the voltage on this side to get the differential voltage across the resistor.

In this case, it's not anyone like before, it's E to So it's minus E -. It's exactly the same formula that we got before over Ah, Three, Two Easy. And then we can just plug in the numbers that we know we've got. V Ay minus E 2 is 10 volts.

Drop the units over R 330. Ohms. Bingo! We now have the equations for all three of our currents. Beautiful! Now we apply Kirchhoff's current law.

Remember Kirchoff's current law. The algebraic sum of the currents at a junction equals to zero. There's our currents I1 I2 I3 equals 0. Bingo.

We just write it down I 1. Plus because it's the algebraic sum. Well, it's the sum plus I 2 plus I 3 equals 0. That is our Kirchhoff's current law equation for this node.

So that's called our nodal equation and we know all these value well. We know the equations for I 1, I 2 on I 3. So we just plug them in. There we go VA minus 1 on 10.

that's I 1 and then I plus I 2 plus I 3 equals 0. Now all we got to do is solve the nodal equation and our value VA A voltage at that node will pop out and that's the idea. Whole idea of nodal analysis. We can calculate the voltage at a particular node even though I think I said right back at the start.

we want to actually calculate the current I2 down here. Well, nodal analysis is actually calculating voltages, but once we get the a year, Bingo is just relative to earth Ohm's law VA on r2 that'll give us our current. So this looks a little bit tricky. and if you're not good at algebra, then well, you might just whack it into a calculator.

or Wolfram Alpha or something like that and just you know, VA will just pop out. But hey, we'll do it simply. We'll just like expand this out. So VA minus 1 on 10 can be written as VA on 10 minus 1 on 10.

So you just expand that term out and then VA on 20 still remains VA on 20 and do the same expansion here. VA on 30 minus 10 on 30. So just expand that out equal to 0. That just makes it a bit easier to reduce it down and get the value of VA.

At least that's how I do it. And then we can just further rearrange this and expand it out so that we just look at so there's no divisions in there. We're just looking at additions and multiplications here. This is just the way I happen to do it.

So VA on 10 is 0.1 times VA You just bring the 10 up. minus one on 10 Point 1 plus so VA on 20 now becomes 0.05 so one on twentieth times VA so 0.05 VA You typically don't show the multiplication sign in there. you just go point oh five VA is Typically some people put the little dot in there whichever way you want to do it. I'm just going to leave it like that and then this term here can become 0.33 VA Minus 10 on 30 is 0.33 3 equals zero.

Once again, it's looking pretty easy now. So what we do Now this is all basic math. Um, you know, probably familiar with. If you've done any sort of you know, high school type math, you should know all this.

Then we can I gather our like terms so we can actually just ignore the brackets. You can take those out. Our - not 0.33 comes over. The other side of the equal sign becomes plus point 3 3 3.

The - not point one comes over and becomes plus not point one over here. and then we can actually group these terms together for VA point one V AO V VA m point 3 3 3 VA becomes nought point 1 plus 0.05 plus 0.03 3 3 times. So now we've only got the single VA equals that. We can now easily solve a VA So it's simple, we just add all these up that becomes naught Point 1 8 3, 3.

Take it to the other side. Divide it. This is point 4 4 Foot. Just do that.

VA equals point 4 3, 3, 3 Divided by point 1 8, 3 3 it equals two Point Three Six Three Six Repeater. actually our vaults. Bingo Tada. We've just sold with just a nodal analysis to solve for Node A using Kirchhoff's current law Beauty.

So as we originally asked what is the current through R2 here, what is I2? Well, Ohm's law. Because we've already done our nodal analysis Beauty, we know what VA is a VA I am I to Ohm's law equals VA on are two point two, three three six three six repeater on 20 nor point one one eight one O, nor point one one Eight one eight repeater. Actually amps. That's it.

Beauty. So hopefully that wasn't too hard and you follow through and it looks good. but it's easy. Kirchoff's current law and we just did.

Some button derives some basic equations for the various currents. Used our Kirchhoff's current law equals to zero. You see how I promised it would be powerful to analyze this sort of circuit and we went through and just plugged in the numbers we derived and we got our answer. We worked out what VA here is and we can.

from that we can work out anything else and as I said, we can do that for any number of nodes. You would just repeat this process for all the different nodes and then you'd end up with actually some quite complex equations where you're going to have to do some matrices, the you know determinants, and things like that to actually get your final answer. It's a bit more messy, but there you go. That's the sort of working you can see.

It's not that hard once you actually sit through and go and do it. But yeah, it does look ugly. I got to admit it's easier just to type it into Wolfram Alpha or use your a formula solver and you calculate a bit. we're learning so you know an X.

But at Noodle Analysis, let's go on to Mesh analysis and use exactly the same circuit and see if we can get the value of current through R2 again like we got before we want. Remember, we're looking for the answer. Point One One Eight, One Eight Repite. That's the answer we want to get.

Let's see if we can repeat it using Mesh analysis. I'm pretty confident in the basic laws of engineering and I was just testing you because I got something wrong. This I had point three, three, three. Here it should be 0.03 three and yeah, thanks.

There you go. Fixed Yeah, I was just testing it and if you're wondering what uses this in the real world, well, you know that circuit simulator you take for granted. It just produces magical results. How can analyze a circuit with hundreds of nodes in it and do it? Does it at each time step.

How does it know what the voltages and currents are? I'll give you one guess. Now next up, we have yet another DC Circuit theorem. This one's called Mesh Analysis. And just like Nodal analysis before, this one uses Kirchhoff's laws it.

But instead of using Kirchhoff's current law like we use last time, you'll notice I've changed it to Kvl Kirchoff's voltage law. That's what's used in mesh analysis. Now, the difference here is that Nodal analysis is what you would typically use to calculate a voltage at a particular node or junction within a circuit. But if you want to calculate a current like as what our original question was, then mesh analysis might be a better technique to use because we're going to look at calculating the current through I2 this time.

and that's what mesh analysis is good at calculating currents in a circuit. We don't care about nodes. In fact, I haven't even labeled this node node A It doesn't actually matter and we don't need any circuit a reference point like we did last time. Because we're not calculating any reference voltages like this, we're looking at loop currents.

So what is a mesh? What is mesh analysis all about? It's a little bit confusing. Stick with me. a Mesh is an individual loop within a circuit. So for example, E 1 R 1 and R 2.

Here if we have a current goes around like this, that is a mesh and likewise I draw it in a different color. You'll see why in a minute we can have another loop or around here like this. This is also a mesh. Now we've also got another loop around the outside here, but that is not a mesh and this is pretty critical.

You can't have a mesh with inside a mesh. so unless you just sort of like the smallest loop possible within a circuit. so this one has, this circuit has two meshes and will I actually solve these individually and then bingo out will pop our answer for the current through R2 at the end. And yes, mesh analysis is more fun than a barrel of monkeys.

Let's go. First of all, we need to label these currents here. So I'm going to label this I 1 and I'm going to label this one I 2. And just like we did in nodal analysis and analysis, we're going to get some equations for these two currents and then solve them exactly.

You know? Basically the same technique is what we did in nodal, except we're solving currents now instead of solving our node voltages. Now Mission Alysus is sometimes known as loop analysis or loop equation analysis or something like that. You can see why, because it's to do with solving current loops through a circuit. Now, what is Kirchhoff's voltage law? If you remember from the previous video, it's the algebraic sum of the voltages around a closed loop must equal to zero.

Bingo! What can? You're probably guessing what our equations going to look like Now, just like nodal analysis, our current directions that we've drawn in here are arbitrary. They can be any direction as long as you're consistent. But by convention, when you're doing mesh analysis like this, you should use a clockwise current flow like this. That's why I've drawn them in going clockwise and this is conventional current flow.

You always do conventional, not electron current flow. But as I said, you can actually do it in the opposite direction. You can use electron current flow if you look one and it'll you know all the numbers will come out in the wash. But this is the convention.

Now, this is the magic part about mesh analysis. We've assumed that the current is going in a clockwise direction like this. Now, it could actually be flowing in the other direction. The current through our one might not be flowing that way.

it could be flowing that way. It and our two. Likewise, look, you'll see that I 1 is actually flowing down our 2 whereas I2 is flowing up our two. This can't happen, You kind of account, flowing down and up right.

It's impossible. And yes, that's true, But this is just for the purposes of mathematical analysis and this is the magic part. You'll see in the end. how that? if we've assumed the wrong direction for the current flow, it'll come out as a negative answer at the end.

And that actually tells us something. Gives us information about our circuit. Wait and see. It's magic.

So we're going to choose this starting point down. here. We're going to derive our equation for the various voltages and voltage. voltage is generated and the voltage drops in this particular loop I 1 here.

Okay, and as I said, it must equal 0. That's Kirchhoff's voltage law. Ok, so start out with a 1 here. So E1 will start out at this point.

Ok II 1. It's positive up here. So it's going. The current flow is going from negative up to positive.

So that means it is a positive. Voltage is generating a voltage in the circuit. Ok, so we don't go B 1. It's positive.

You don't have to put the positive in there. It's just not negative. Ok, so it's actually generating a voltage. and then we have a look R 1 here.

What have we got? It's a voltage drops are actually going to have a positive voltage here and a negative voltage here as opposed to this one. We went from negative to positive. So it's A it's generating a voltage so it's positive. This one is going from positive to negative because resistors drop voltage.

Okay, we know because it's a resistor. You know how resistors work. They don't generate voltage. They actually drop voltage when you pass current through.

So it's a voltage drop. So it's actually -. And what is the voltage Drop on here? Easy. It's R1, You got it times I1 Ohm's law Voltage equals I times R or R times I Am going to put it in first but it doesn't matter.

You'll see why in a minute and then well. we've got another one. It flows down through R2 Here this is going to be positive. This is going to be negative here.

So we've got another voltage drop. So it's - ah - in this case and the current is I1. So that's our voltage drop and we're back to our point here. But we're not finished yet.

Look, I - is also flowing through here. So we have to take I to endure count. And what is I I - is going from a negative to a positive, right? So in this case it's negative to a positive. It's like it's generating a voltage because it's going in the opposite direction to I1 here.

Now here's the tricky bit and you have to stick with me with this. You can see that I 2 is flowing in this direction. so just like I 1 flowed in this direction and we have a positive here and a negative here. The same thing is going to happen here and I'm going to draw it in blue because it's of caused by I2 year.

So going to have across R2 positive and negative like that and you'll notice that it's actually the opposite polarity. But we're writing our term for I1 here. So what effect does I - because it's interacting on I1, What effect does it have relative to the direction of I1? Well if I one you remember it's going from negative to positive in this case because it's been influenced by I2 I1 relatives. So just picture I 1 coming around here going on this side.

Here, it's going from a negative to a positive. What does that mean? Just like here, it's going from a negative to a positive. It's a positive voltage is generating a voltage in a circuit even though it's a resistor. The effect of I 2 flowing in this direction which which which we've chosen are Vitara Li Remember, flowing in this direction is causing a voltage to be induced into the circuit.

So it's a positive voltage just like as if it was actually a little battery, a little power supply. They're generating that voltage so it's going to be positive. Ah, two again times, not I1. It's I2 like that.

Bingo! And that is our equation equals to zero. Like that, you can see how the color coding really helps you identify which terms are relative and cause by which particular currents. So we've actually got four terms in our equation there. Even though you might think with at first glance, I 1 is only being influenced by three particular parts in the circuit, you're forgetting that I 2 is having an influence as well.

And if we had, if when this was a bigger circuit and we had another mesh up here like this and a current circulating around here, we'd have a fifth term on here. Doing the same thing, we'd have to take into account I 3 up here and it'll be our I 1 times I 3. And that would be another positive term in here because the current would be flowing in a clockwise direction from negative to positive like this. Now let's do the equation from I - once again, we have to start a choose a starting point and I can choose down here.

but I'm actually going to choose up here and go down and do the voltage first. Just so our terms are like in the same order here. it doesn't matter it, you know it makes no difference. it's just I want it to be a bit neater.

So I'm going to start with this point up here. So our current is flowing in a clockwise direction. so this is our starting point our current is actually flowing from. Look, battery is positive to negative.

It's different to what we had over here is flowing from negative to positive. So it was actually producing voltage in the circuit. Now it's flowing from positive to negative just like it was here. So this battery is effectively working like a voltage drop based on the arbitary current direction that we chose.

So you guessed it, it's negative e to Just Like It's a drop. Brilliant. So you've got to be so careful doing these things, you can easily miss that and go. Oh, it's a voltage.

It generates voltage in a circuit. But no, it can be a drop depends on the current direction. and once you do this a few times, you it - you know you'll get to know and love this technique. And trust me, it'll all come out in the end.

It's brilliant. So what's our next term? Now next time. Okay, we've done this point here. Our current.

Now we're looking at our -. so it's flowing from positive to negative. It's a drop. Okay, so it's negative.

R - I - Okay, so that's our voltage across there. And then it goes through our three like this. So once again, it's positive. Negative Like that.

it's a drop because it's just a resistor. So we've got drop drop drop. Everything's dropping gone. Where are the voltage is being generated? Wait for it.

Okay, so we've got our three this time times: I -. But aha, just like before, we have a fourth term which I'll draw in red because I won now interacts with I - Over here Now we've got to look at the direction that it does that. Once again, this is flowing around like this. It's positive negative.

So it was a drop before in the previous equation, whereas at R2 it was a drop. it was negative. But now you can see that relative to I - a suit. Just imagine this blue arrow going around on this side.

Now it's going from negative to positive. Uh-huh It's a voltage generated. Brilliant! So that's a plus. So I1 is actually generating a voltage into the I2 loop equation here.

Fantastic! I Love this! Ah, so we're looking up. so what does it do? R2 plus r2 times I1 because it's I one doing the interacting and then Da equals to zero. We have our two loop equations. Now all we have to do is solve them.

So just like before in the nodal analysis, we've got some no ones in this circle. We know what the resistor values are. We know what the voltage values are. so we can plug those into our equation here.

I've moved the one we just derived down here because I need some space to actually do this. So what we look at that ee 1 E 1 is 1 volt. Okay, it's positive one volt and then - our one is 10. We don't know what I one is yet.

it's going to come out in the wash. So 10 I 1 minus R 2 which is 20 ohms times I 1. Still don't know what I 1 is and then we can go + R 2 which is 20 again and we don't know what I 2 is yet. Once again, it will come out in the wash.

So Bingo! We can now reduce that and solve it. And just like before, we can leave it like that and just shove it into our equation solver. how solve it whatever method you prefer but can actually reduce that a bit further. we can just group like terms yet again 1.

- in this case we've got 2 I 1 terms here so we can just 10 - 20 like that so we can go - 30 I 1 and then we can just go plus 20 I 2 and that equals 0. And likewise with this one down here we can go - E 2 which is 10 minus R 2 which is 20 I 2 minus R 3 which is 30. Ohms. Once again we take out the units I - and then we go + R 2 which is 20 and then I 1 which we don't know yet and bingo equals to 0.

So I've reduced that one as well. Just gather the like terms 20 and 30 there and Bingo! We now have our two equations which we need to solve. Now let's take a look at these. You can see that they contain two unknown terms I 1 and I 2, I 2 and I 1 there and that's a bit tricky.

You know usually you might solve that with by determinants with the matrix R for example, you might as I said, plug it into you of the formula solver on your calculator which does the same thing. you can actually expand it out and do some things and try and solve it manually and I probably don't have the space to do here. so let's solve this the modern way. shall we go to the internet? We'll use Wolfram Alpha to actually uh, plug these equations in and we'll get the answer for I 1 and I 2 our to unknown terms.

No, a lot of people say that's cheating but this is not a math video. I'm not going to show you how to solve for two unknowns in these equations. you know, do it. However, you know, whatever floats your boat.

let's go to the Internet Ok so here we go. We got a sheet like any modern student does and we're going to go to Wolfram Alpha. But you could do this formula solver on your calculator or you can do it in your head. or you can do a pencil and paper however you want to do it.

whatever method you want to do to solve for two unknowns. So we can actually enter the equation here I've entered the first one in here and then we can actually enter the second one by going a comma like that and then typing in our second equation and it'll automatically know that we've got terms in there now I can't label them I 1 and I 2 because I think it thinks they're complex numbers or something like that. So I 1 is going to be A and I 2 is going to be B. So you'll notice that we've got two unknowns in there.

so let's just press Enter Magic happens. Magic happens. Wait for it to know we've got our answers as actual our fractions because it tries to give you an exact form. but you can actually go to approximate form here.

And bingo, there's our two answers: A which is I 1 equals negative point 1 3 6, 36 X, and B which is I 2 is approximately equal to what negative 0.25 4, 5 5 amps. Let's go back to the whiteboard. So Bob's your uncle. We've solved our two unknowns I1 and I2, so we know everything in this circuit.

Now, we know the voltages, we know the resistances, and we know the currents flowing through each component. Well, we kind of do. And this is the magic I was telling you about. Before you remember how we started out going clockwise, we assumed that the current was looping around clockwise, but we got an answer both of them.

I 1 is negative. Our answer popped out of our equation as negative and so did I - what does that mean? It means that we chose the wrong direction. both of these currents actually flowing anti-clockwise like that. So what does that mean you for R2 here, for example.

So I, one, for example, is flowing. We assumed it was flowing down R2 like that. but we got a negative result. So I1 is actually flowing up like that through R2 and R2 Here, that's also a negative.

We assumed it was flowing up, but it's actually flowing down because it's negative. So it's flowing down R2 like that. Now, we can actually work out the current through R2 d2 So what do we do now to actually calculate the current through R2? Well, we've got to subtract one current from the other because they're flowing in opposite directions. They're still flowing in opposite directions because they're both negative.

So one is actually our I 1 is actually going up there I T's going down there. So we've got to subtract them to cancel them out. Now here's where we can actually drop the sign and we know that the larger value point 2, 5, 4, 5, 5 amps is flowing down. So L that's larger value absolute than the other one.

So our final current is going to be flowing down. So we just subtract the smaller value point 1, 3, 6, 3, 6 from 2 point 5, 4, 5, 5 and we get what do we get? Point 1 1, 8, 1 No. And it should actually be 1 8 repeater. It's exactly the same as what we got last time.

Winner winner chicken dinner High-five Wahoo! So you Kirchoff's current law Kirchhoff's voltage law they hold. we got the same answer using two different techniques: nodal analysis and mesh analysis. Got exactly the same answer for the current through R2 and freaking tastic. That was easy, wasn't it? Piece of cake? No worries.

But now there's actually a third method we can use to calculate through R2. It doesn't use Kirchhoff's voltage or current laws, but I thought I'd show you anyway. Let's see if we can get the exact same answer yet again. And I can't really leave this one out because although it doesn't really have anything to do with Kirchoff's voltage law, Kirchhoff's current law, it is one of the basic DC circuit solving theorems and it's called the superposition theorem.

the superposition technique whatever you want to call it and it's a bit of a mouthful, but this is what it basically states. The current in any element is the sum of currents produced by each source acting independently whilst the other sources are replaced by their internal resistances. Bibbity, bibbity, bibbity. It might be easier if I just show you.

please excuse the crude if the model didn't have time to build it to scale or to paint it. Now what it basically means: if we want to solve the current through R2 down here, we can do that by the sum of the currents produced by each source acting independently. So what we do is we can. Just if we start with E one, we take out a two and we replace it by its internal resistance, which is a short circuit.

And you should know that from your basic circuit theory, a power supply in ideal power supply is zero internal resistance and ideal current source is open infinite our internal resistance. So we replace it by that and now you should be able to do this. Anyone can do this right? This is you can now calculate the current through here, but uh-huh we then have to. It's the sum.

Remember, So now we have to replace this one with a short-circuit put this one back in here, and then we calculate the current through there. and then we sum them is the sum of the currents produced by each source acting independently. So you can see how this is much easier. We won't end up with any weird equations and generate any weird equations like we had two with nodal and mesh analysis.

It's basically just Ohm's law and current divider stuff. Very simple. So let's go through the exact same example we had last time and see if we come out the same answer you think we will. I'm pretty confident once again, engineering for the wind.

So all the way over here. I've got the original circuit that we wanted to solve our two voltages with our three resistors Exactly the same as before. I've read, you're on it here because we need to solve this twice because we've got two sources. And by the way, the superposition theorem of course only applies if you've got more than one source.

You gotta have multiple sources and they've got to be linear as well. So let's solve for E one. So what do we have to do? We have to replace E two. We have to replace in this case, the other source we have.

but if we had more than one source, we would have to replace. If we had more than two sources, we'd have to replace all the others with their internal resistances. And as I said, a power supply, a voltage, a battery, or whatever it is, it has a zero internal resistance. So we replace it with a short-circuit.

If it was a current source in our circuit, we would actually replace that with and open. It would just be open circuit. So I'm going to replace all the other sources and now it just becomes a simple question. But what I've labeled here is I've got our I T which is total the total amount of current coming from this source and then I r2 down here.

So now we want to derive an equation for the total current first. So now what we've got to do is derive an equation for our total current here and that's easy. It's just Ohm's law. Look, I T equals what does current equal voltage divided by resistance? So in this case, E 1 Ok is on top divided by our resistance.

which is going to be the total resistance of our circuit which is R 1 in series with R 2 and R 3 in parallel. Easy R 1 plus R 2 in parallel with R 3. There we go. And if you haven't seen those two lines before, that's just a common way to express parallel.

And of course, you can solve your parallel resistors any way you like. If you're lucky, you might have a calculator that has a parallel key. As far as I know, there's only two calculators on the market that has a the casio FX 61f and my own micro calc so you haven't got one of those. Could do it the old-fashioned way I choose to do it r2 times r3 over r2 plus r3 you can do the one over version if you like.

So there you go. I've just expanded that equation. although I've plugged in the numbers that we have so we have all of our resistor values, we have our voltage e1 which I forgot to write up there before. Bingo! Our answer is not point O four, five, four five repeater.

actually I Like that. That gives us a bit of confidence because we know that our solution is a repeater. so you know that gives me the warm fuzzies. Now we know our total current flowing here.

How do we calculate our current flowing down our - Well, along with Ohm's law, some of the basic stuff you should learn is the voltage divider equation and also the current divider equation. Very similar to the voltage divider equation. In this case, I 1 is equal to I T and then you use your current divider because some goes down here, some goes down there. In this case, it's the opposite resistor.

We want our three like that over r2 plus R3 that is a current divider. easy. And it's simple. We just plug in our no ones because everything is knowing none of this unknown.

It's solving equations for unknown rubbish that we did before and we pop out with an answer of 0.02 7 - 7 Repeater. Once again, another repeater. Brilliant. So then we do exactly the same thing for E2.

We replace E1 with its internal resistance, which is a short circuit and we solve for IR - yet again, it's exactly the same way. I Won't bore you with all the details. Exactly the same equation we generate: I We calculate I T First the total current coming from the battery and then we use our current divider equation - once again, calculate IR - I had that labeled a one before that was oopsie. mental brain fart from the previous one and we'll call it I R to a and this is IR to be.

And Bingo! it comes out to an answer of 0.09 The point yet 0.09 O9o. Once again, a repeater. Brilliant. feeling pretty confident.

So we've got our two different currents here for two IR twos. so we now have to get the algebraic sum. Once again, we have to take signs into account. In this case, it just so happens that they're both positive for what flowing down like that.

So there's no negative or whatever, but it could have been depending on the circuit that you're actually analyzing. So we take those two values, whack those into the equation. just the algebraic sum to get our final value down. I r2 which is what we're trying to get here.

So it's point O to 7 to 7 plus point Zero Nine, Oh nine. Oh What do we get? Ha Magic Zakah The same as before and the time before that we're not. So there you have it. Sorry about the length of this, but I Wanted to go through step by step in detail I Hope you enjoyed these series of two videos here.

one showing what Kirchhoff's current law and Kirchhoff's voltage law is about and then actually applying it and applying three different circuit. DC Circuit analysis techniques, nodal analysis, mesh analysis, and superposition theorem here to solve exactly the same circuit. We get the same answer three different ways and you think, well, what's the difference? Which one should you use or you can probably saw here. you know this superposition in this particular instance was the easiest.

Probably the easiest one to do because we didn't have to. You know, get some weird algebra with. You know, unknowns and stuff like that. It was just basic Ohm's law and current divider.

So that happened to be the easiest case here. but that's not always the case. Nodal analysis as you saw you would use if you want to calculate a voltage in a circuit. And as I said, that's a popular technique used in ER Spice circuit simulators and things like that.

They use various versions of nodal analysis and they also do some mesh, but it's it's easier to do nodal analysis. Mesh analysis are easier if you have a lot of sources and you want to calculate currents and things like that. So yeah, choose whichever one is appropriate for your particular circuit. But there you go.

That's really interesting. So this is really basic, fundamental stuff. So fundamental. it should be taught directly after our Ohm's law.

and typically is you learn Ohm's law. Then you learn what you know: voltage dividers, current dividers, what a voltage source is, what a current source is. and then you learn Kirchhoff's current law, voltage law. And then you learn your basic DC circuit theorems for solving circuits.

And as I said, you know, look, these are often academic. You know examples and things like that, and sometimes you don't. You can probably spend your whole career and never have to do nodal or mesh analysis or something like that. But hey, it is a fundamental technique which is so important, just the mathematics of it goes into actually solving a whole bunch of other stuff.

as I said circuit simulators. All that sort of stuff wouldn't work without stuff like this. So math full win. Even if you don't like math, it's pretty easy.

Wasn't that hard at all. So there you go You now an expert Kirchhoff's laws and no mesh and circuit position theorem come over play. If you like that, please give it a big thumb. Thumbs up.

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