so we only have one objective in this lesson and that's to talk about nodal analysis now in your textbook in a lot of textbooks they really like to make the sound like it's something very different than what we've done but really this is just using KCl to find unknown values in a circuit now I will put here this is typically used to find voltages now that does not mean it's the only thing you could that you couldn't find a current with it you absolutely could find a current using nodal analysis depending on the problem but typically this is used to find unknown voltages so let's look at an example here we're just going to dive into examples because there's really not anything else to explain about nodal analysis we're just using KCl now we'll go through this here on this first example relatively slowly here okay so this is the circuit diagram we're looking at I suppose I should label all the components I've got a three amp power supply this is 40 ohms this is 10 ohms this is 5 ohms and this is 50 ohms or 50 watt volts ohms and we're supposed to find here VX now in circuits like this the way they're drawn one thing I don't like about this is that there's no ground or no reference now I don't think we've actually talked about it but most of you are probably familiar with this or have already seen it in your text that this is the symbol for ground now when we get into things in more detail and later courses we'll have to be careful there's there's earth ground there's floating grounds there's an AC ground there's DC grounds there's digital grounds there's all sorts of different types of grounds and you might say how can that be the case well remember a ground here this is what we've referred to this is just a reference point now typically you know for most circuit analysis we'll be doing will be thinking that ground is just earth ground you know which is zero volts and so what I want to do is apply a ground here on one of these nodes now let's first again just go back and identify how many nodes I have I have 1 2 3 nodes now technically you can put the ground anywhere on those nodes but what you want to think about is what's the one that's gonna make the most sense and if you just look at this circuit well it's gonna meet the most sense to put the ground on the negative end because that's what we're probably used to just with household wiring or batteries or anything like that oh wait the negative side is usually ground and so that's what we'll do here we'll make this node right here ground so I want to emphasize that this is at 0 volts which also then tells me the voltage at this node here and if you think about it for a second the voltage at that node has to be 50 why does it have to be 50 because there is a 50 volt drop by this power supply and I'm going to emphasize this by I'm going to call this node a this node here B and then this node here C so if we're talking about having like a little measurement probe here so like if you were to connect up some probes here you'd probably have like you know a little probe connected up to there and then the black probe you know connected up to here and then this is going to the DMM the digital multimeter you know so if you connected this up to the circuit like you were actually measuring the voltage you course would measure 50 volts but what you're actually doing is you're saying that 50 volts is equal to the voltage at node a minus the voltage at node C because those are the two points of contact and since the positive is at node a and the negatives at node C what we are saying is well that's the voltage drop we're measuring and so we get VA minus VC but since we applied ground down here on the negative we have 50 volts is equal to VA minus zero and so we get 50 volts is equal to VA now we're supposed to find VX here well VX again following this idea of a voltage drop VX is going to be the voltage at node B minus the voltage at node C and I'm gonna erase these here just so that my diagram doesn't get too messy here which again this would just be VB and VC is zero because we've labeled that as ground so the voltage I'm looking for is just the voltage at node be now nodal analysis remember is where we look at currents we know that if I sum the currents leaving a node that's equal to zero or the current entering a node is equal to the current leaving a node okay well let's go ahead and draw some currents in here on our circuit diagram to kind of illustrate this so I've got node B well first off let's just emphasize I have three amps coming into that node now whenever I draw currents leaving currents going through resistors I always draw them leaving the node and I'll explain why in a second here we'll see why but I'm gonna draw and I've got three more currents because remember this whole line up here is node B and so I'll label this I 1 I 2 and then I 3 so from KCl we know that I 1 plus I 2 plus I 3 has to equal 3 amps because I 1 plus I 2 and I 3 is what's leaving the node 3 amps is what's entering the node now this is where we use Ohm's law again and I'm just going to emphasize it here by drawing a resistor off to the side on what we're going to be using here so if we have some current going through this and we have some plus/minus voltage across it and I'm going to label these points as simply just 1 and 2 for my terminals 1 and 2 well V is equal to the voltage at 1 minus the voltage at 2 because it's the plus side minus the minus side and then we also have from Ohm's law that V is equal to R times I so setting those two equal to each other you get that R times I is equal to V 1 minus V 2 I is equal to v1 minus v2 over R so this is actually going to be very useful in fact we're going to use this kind of equation here for nodal analysis for almost every single current in our problems here so let's go back to this problem here and so we have I 1 I 2 and I 3 well we're going to use this idea here that well it's going to be I 1 would be the voltage at this node which is B minus the voltage at this node which is VA which is 50 volts divided by the resistance because that's just using Ohm's law like we have here so I 1 is going to be VB minus 50 volts divided by 5 ohms now doing a similar thing we get I 2 would be V B minus now let's think about this I 2 is going through this 10 ohm resistor its VB minus and of course what we've labeled that node as ground so it's VB minus 0 over 10 ohms i3 going to be the same thing VB minus 0 over 40 now I'm going to substitute those all into this equation here so I'm going to take all of these and substitute it into this equation now when we get familiar with nodal analysis we'll actually be able to write this equation that we have right here down first without even bothering writing the other equation and I'll leave the units off just for now and of course if you subtract 0 that's the same thing as just having the number there so VB over 10 plus VB over 40 and this is equal to 3 I've got just a single equation with one unknown that I need to solve here and the easiest way I think to solve this is just to get rid of the fractions and think about multiplying everything by a common denominator at least common denominator that I can see here would be 40 so 40 divided by 5 is 8 so I have 8 times VB minus 50 plus 40 divided by 10 is 4 4 VB for some reason I keep wanting to make a lowercase B plus then 40/40 of course is 1 so VB then you don't want to forget you have to do 3 times 40 which is 120 and now I've got just a single equation with one unknown that I just go through and finish solving this I will go ahead and solve it in this case I get 8 VB minus 8 times 50 would be 400 plus 4 VB plus VB equals 120 and so that would be 12 13 DB is equal to 5 20 and then VB is equal to 5 20 divided by 13 let me do that on my calculator real quickly and this gets me 40 volts and now remember what we were looking for though is VX but then we have right here that VX is simply equal to vb and so we have here that VX is equal to 40 volts now I as I said I really kind of went through this example much slower than it's needed to be but I wanted to make sure that for this first example we can all clearly see every step on where all of our equations come from now again as I've said one you get familiar with this we will often skip writing this equation here and move directly into writing this equation but I want to do one more quick example that's very similar to this one and see how we would go about doing this one here so let me draw the circuit here so this is going to be very similar to the problem we did just previously and in this case we're asked to find V 1 and V 2 now we can actually answer one of these questions straight away and sometimes we do this in problems to make sure that you understand all the concepts we've talked about and if you look at this circuit for a second you can figure out what V 2 is immediately with no work and the reason is is that if I look here I've got this node here and this node here so this 10 ohm is in parallel with the 20 volt supply but what do we know about components in parallel we should recall that components in parallel have the same voltage drop so v2 is simply 20 volts so there's nothing really too challenging about finding v2 sometimes people make problems more difficult than they need to be now we do want to find v1 here and I should again kind of label I like to label all my nodes here and I like to using letters as you can see from the previous one I'll label these a B and C I've only got three nodes and remember we want to make one of these ground and again the one that makes the most sense is the bottom one to make ground here so I'm gonna make this node C ground which makes this again zero volts now I will still go ahead and draw the currents here and if I'm looking for v1 we should apply KCl at node a and if we think about it the reason we'd apply KCl node a is that's of course where v1 is located now again drawing currents here I know that I've got three amps coming into the node and then I have a current i1 and i current i2 that are leaving those nodes again the new labeling really doesn't make that much difference so I will go ahead and still write down the node equation that we'd get or the KCl equation you get three is equal to i1 plus i2 and now that we have this KCl equation 3 is equal to i1 plus i2 we're gonna do the same thing we did with Ohm's law before where I would have then for I 1 3 is going to be equal to the B or not V B I'm sorry I was wrong note here VA minus 0 divided by 5 ohms and then I too would be VA and it's technically minus VB I'll just write it down for now divided by 20 ohms but of course what do we have here if we go back to the circuit here we should see that VB is equal to 20 volts because again we made note C ground so we know there has to be 20 volt drop across the power supply and so we wind up with an equation that is 3 is equal to VA divided by 5 plus VA minus 20 over 20 now we want to make sure we apply algebra correctly so I multiply both sides by 20 and I get 63 times 20 is 60 and then I have 20 divided by 5 which is 4 so I get 4 VA plus 20 times 20 is of course 1 so I'd have VA minus 20 so we'd wind up with 80 is equal to 5 VA and we wind up then finally with VA equaling 16 volts and again looking at my circuit we can see here that VA and C being ground that v1 whoops v1 is simply equal to VA so v1 is equal to 16 volts and we're done now some of the problems may be in the homework are going to ask also for power and things of that nature but now that we have these voltages here finding the power across all the resistors is going to be easy for instance if we wanted the power dissipated by 20 ohm resistor well the voltage across the 20 ohm resistor you can actually technically do it in any order you can make it plus or minus whichever way you want and to illustrate I'm gonna go here make it plus minus that way because of the way we drew I too so if I did that it would be VA minus VB which would be 16 minus 20 which of course is negative 4 volts now we don't have to worry about the fact that it's negative 4 because we're gonna use the special power equation here for us the fact that power dissipated by a resistor so the power of the 20 ohm resistor we're going to use that it's V squared over R so this would be negative 4 squared over 20 ohms which is of course 16 over 20 and 16 divided by 20 is 0.8 watts or you could write that as 800 milliwatts but notice we still get positive because we said the resistors always have to dissipate so it should be positive but that's because we're squaring the voltage now if I said that if I had defined my plus-minus differently on this resistor here like made this side the plus and this side the - we would have just gotten positive 4 volts and so you'd have 4 positive 4 squared but that's still 16 so you'd get the same answer either way so sometimes we're asked to find additional information besides just finding these unknown voltages