10 - Intro to Mesh Current Circuit Analysis (EE Circuits)

hello and welcome to the section of the circuit analysis tutor and in this section we're going to begin to talk about the next major solution technique of circuits that's super super important and that is called the mesh current method alright so in the previous several sections we've talked about the node voltage method and I think you'll agree that the node voltage method lets you solve more complicated types of circuit than regular solution techniques like Kirchhoff's laws with fewer equations so basically less work on your part but there's some work involved in understanding the node voltage method how to write the equations and also how to make sure that you actually have enough equations to solve the entire circuit to find the currents everywhere alright so the node voltage method super super important the mesh current method ranks right up with the node voltage method in terms of importance so it's it's not really a matter of you just needing to remember one or the other one of these types of methods you must know both of them because what you're going to find out obviously on your test you'll be on your exams you'll be asked you know solve circuit by by a certain method but beyond that you're going to find out as we work more problems that some circuits are going to lend themselves by the way that they're drawn to naturally be easy to solve with node voltage equations some circuits are gonna find out by the way they're John and how they're constructed are going to naturally lend themselves to being easily solved with the mesh current method so it's not so much that one of these methods is better than the other and you can just pick and choose every problem I'm going to do node voltage or mesh current it's a matter of having both of these these methods in your in your toolbox so to speak so that when the situation arises you can pull out the proper method that makes the most sense whatever you're most comfortable with for that problem and get and get the answer now I can also tell you up front though that in my opinion I think mesh current is actually a little bit easier to deal with then node voltage and I'll try to explain why that is as I as I go forward so big picture we've learned node voltage super-important definitely don't flush that knowledge because you're gonna have to use it on your test and unreal you know practical so that's mesh current ranks right there up there with node voltage you're gonna need to know it it's gonna lend itself it's gonna make your life easier solving some of these problems as we go alright so again one more point of comparison for the node voltage method if you remember even though we called it the node voltage method what we were really doing at every essential node that we were kind of looking at we were writing an equation and basically that equation was pretending that the current was leaving the node and we're basically summing the current leaving the node so basically it was kind of like a modified kerkhof current law is what we were doing in the node voltage method we were going essential node to a central node and for each node we were writing a big long equation that was basically expressing how much current is leaving that node through all of the branches and we were sitting in equal to zero so the sum of the currents leaving a node equal to 0 that's Kirchhoff's current law right so it's basically you can kind of think of node voltage even though the equations look a little bit weird it was basically a modified node modified kerkhof current law designed to be streamlined for those problems so you don't have to have quite so many equations right that's what node voltage really was in a nutshell mesh current in contrast is basically gonna be a modified kerkhof voltage law that's gonna be we're gonna set it up in such a way that it's gonna be rock-solid bulletproof and it's not gonna require as many equations as you would otherwise need to solve but basically it's gonna be a modified cut called voltage law as we walk around the loops in a circuit so again big picture Kirk the big picture for node voltages it's a modified kerkhof current law at the nodes the big picture for the mesh current is it's a modified kerkhof voltage law as we walk around the miche's in the circuit and I'll explain what all that stuff means in a second so before we can really do much it's easier really to explain the mesh current solution method with an actual circuit so I'm going to draw a pretty simple circuit on the board 40 volts ok and this guy is connected to a resistor here I've got a resistor here there's no resistor down below we've got a resistor here nothing down below got a resistor here got a resistor over here there is another source over here that is connected like this and this one over here is twenty volts alright so let's just do some labeling to ohms this is 8 ohms this is 6 ohms this is 6 ohms this is 4 ohms so here we have everything labeled now one more thing I want to label for you this is again given in the problem let's call this current let's just call it I and let's call this the knot the voltage across the 8 ohm resistors V naught and the problem states find I and me not all right now first of all obviously you could use node voltage for this problem right if we did we would just select you know the bottom node here as the reference node we have an essential node here and we have an essential node here so we would basically be writing equations at these nodes we know how to pop up we would label the node voltages we would go up we would subtract we would divide we would have all these node voltage equations and then we would find the node voltage here and the node voltage here which would allow us basically to find everything else on the problem so I'm trying to point out to you that mesh current node voltage it's not like you you have to use one or the other or one of them's more appropriate than another as far as exclusivity you can almost always serve solve every problem with a mesh current or a node voltage it's just an alternative method there so what you need to do for every single mesh current problem is first identify what are the meshes what you really need to do to identify the meshes is pretend that your circuit it's a cookie cutter you know those things that that when you bake cookies you stab it into the cookie dough and it makes a dumb dinosaur or something like that for your cookie dough you could stamp them out like that that's we're gonna call that basically a mesh when you look if you were to pretend that this were a giant cookie cutter then you would make a cookie with this guy right here you make a cookie with this guy right here and you make a cookie with this guy here so if this were a cookie cutter right then you would basically make three cookies every time you push it into the dough right each little individual square in there is what we call a mesh all right so this guy has three meshes this is mesh one this is mesh two this is mesh three so basically any time you look at a circuit no matter how complicated it is it's the number of meshes you have is the number of windows I guess you can think of it as windows also window panes or something the number of individual rectangular panes and I called them rectangular or square really it doesn't matter you can have you know if I could have a resistor going from here to here that would divide this into two more meshes one mesh would be up here the upper triangle the lower mesh would be down here so it's basically literally like the cookie cutter analogy everywhere you would make a cookie that is your mesh so we have three meshes it's very easy with the mesh current method to realize that the number of equations you need for the mesh current method is just equal to the number of meshes you have here we have three meshes in this circuit so we're going to have three equations to deal with it and to solve our circuit so we have to do some labeling much like the node voltage method let's label our meshes so what we need to do not only have we identified our meshes we need to label the mesh currents so what I'm gonna do and what almost every book does is do it like this I'm gonna draw like a little semicircular arc and inside of it I'm gonna put ia this is mesh current ia we oriented clockwise like this let me finish and I'll do a little bit of talking this is mesh current I so B this is measure current I sub C so you you have to label them something so almost always or at least in all of my problems I'm gonna label a mesh a mesh a B and mesh of C so this is gonna be called mesh current a mesh current B mesh current C the goal of the mesh currents a solution technique is just for you to be able to actually calculate the values of these mesh currents ie IB and IC so we're gonna write some equations in a minute they're all gonna be in terms of ia IB and IC and you're gonna solve for ia IB and IC and you know those mesh currents you can calculate everything else on this in the circuit much like the node voltage method we label the node voltages then we wrote equations dealing with the node voltages and once we calculated the node voltages we could calculate anything else in the circuit same thing is true here it's just we're trying to find these mesh currents now a couple things I want to point out to you you can do it however you like but I recommend that you since you're learning for me that you do it the same way I'm doing it I think it's easier but you might have a you know some epiphany that makes no sense to you to do stew it some other way when I label my mesh currents I always write them clockwise like the hands of a clock so here's the arrow going this way ia IB IC it doesn't really matter you could label all of your meshes going the other way if you like and call them ia IB and IC and as long as you follow the sign conventions properly you'll still get the right answer I label it clock lines I suggest you do the same thing now what we need to do is write a mesh current equation for this mesh and then a mesh current equation for this mesh and then a mesh current equation for this mesh so we're gonna have I told you three equations one for each mesh and then you can solve the problem alright so let's go ahead and do that so let's write the mesh current equation or mesh a and I literally write it on my paper like this mesh a and then I write the equation you know mess with it simplify and then I write mesh beat and I do it for me mesh C and so on and then you solve your system of equations I told you a minute ago that node voltage was basically a modified kerkhof current law at each node mesh current is a modified kerkhof voltage law so what we're gonna do is we're going to walk around each mesh soph right now we're working on mesh a right only this little square only the square we don't care about anything else in the circuit we're gonna walk around this guy and basically we're gonna do Karkov voltage law we're gonna sum the voltage drops as we go around that circle and set it equal to zero because that's what car club voltage law is right you can take any path in a circuit as long as you come back to the same spot the sum of the voltage drops as you go around have got to be equal to zero we've talked about that a great length and we talked about kerkhof voltage law so the only thing that you need to be careful about just like anything with circuit analysis is the sign convention right so the sign convention so how does the sign convention work whenever you are traveling around a path in a circuit and you come across a a voltage drop if you go from plus to minus and we've been talking about this for a long time so this is nothing new when you go from plus to minus like through a resistor in the direction of the voltage drop we count it as positive when we're writing our kirk off voltage log equations or our mesh current equations when we encounter a voltage rise from negative to positive we always treat that as a negative you have to have some kind of sign convention it doesn't really matter how you do it it's just that a voltage rise has to be one sign and a voltage drop has to be the opposite sign so that you can keep track of everything so in this case if we start here and we go around this way the first thing we encounter is a voltage rise it is it is going from negative to positive through the source so that is treated as a negative in our mesh current equation so we literally write negative 40 because we're summing voltage drops this is 40 volts it's negative because of the way it's oriented all right now we get up here to the 2 ohm resistor now we're again we're traveling around like this and one thing I want to say before we get any farther is that this eye that's labeled here this is what we're trying to calculate and this V nas what we're trying to calculate much like them at the node voltage method you don't mess with that stuff the labels on the drawings until the final step of the answer in other words don't try to write your mesh currents in terms of I and V naught it's just gonna confuse you write your mesh currents ia IB IC calculate the answers and then at the very end go back and figure out what I must be and what v naught must be so you can kind of ignore those labels for now here in the beginning only worry about what you have you have yourself labeled which is the mesh currents so we're getting up to the two home resistor now when we label a mesh current like this ia what we're basically saying is that there's some current ia we don't know what it is it's circulating in this mesh right and there's some IB that's kind of circulating in this mesh and there's some icy kind of circulating in this match there so forget right now how that could make physical sense I'll explain it all later but if you just pretend with me for a second that this is circulating once you get up to the two ohm resistor if there really is a current ia going this way then there's going to be a voltage drop plus minus because it's going through the two ohm resistor plus - all right so you're going in the direction of the mesh current clockwise plus minus so that means you treat this as a plus because the voltage drop is gonna go from plus to minus it's gonna be a drop what would the voltage drop be across this well the currents ia then it's going to be ia times two because V is equal to IR okay so if that current is ia in this mesh then the voltage drop across this resistor is IR and it's oriented with a plus sign because of the direction then we talked about okay now so far it's been pretty easy negative forty because of the sign the direction that we're going and then it's positive because of the voltage drop across this resistor I times R now when we get to the eighth ohms we have to be a little careful and this is kind of a magic of the mesh current method right here what is the current we want to find the voltage drop across this resistor obviously and we know by the way that we're going down like this so we need to figure out what would be the voltage across this resistor well it's gonna be I times R right I talked about what is the current in that leg what is the current in that leg with the 8 ohm resistor well you might say well I'm doing a mesh current so this is the current ia so it's gonna be ia times 8 plus ia times 8 but that's not really right because notice right next door we have an another mesh current IB that we're saying is circulating this direction so what you really have if you think about it if you follow this current here ia in this leg if you can pretend for a second I sub a is going down like this because look at the direction of the arrow it's going down this leg ib if you look at the direction of the arrow is going up so what you have and it's all fictitious okay this is not really happening this is just a solution technique you ia IB and IC now we're talking about a resistor that's on the border of two meshes so ia is going down based on the way we drew it and IB is going up in the very same leg so they're fighting each other ia is going down IB is going up so what's the current in that in that actual branch well it's not ia or IB it is ia minus I B this is the current you know fictitiously right now in this resistor because we're saying that ia should be going down we're saying that IB should be going up so if we take a minus IB that's the net current flowing down okay if we take this current and multiply it by eight that's I times R that's the voltage drop across that resistor so we've taken care of the 40 volt source we've taken care of the voltage drop across this we've taken care of the voltage drop across this so we set it equal to zero this is basically a kerkhof voltage law we have summed the voltage drops arc around this mesh and we've basically set it equal to zero the magic of the mesh current method is that anywhere that there's a boundary like this you find out what the current actually is by subtracting the mesh currents that border it okay now also I want to make sure that you understand it's very important we subtracted ia minus IB we didn't do it the other way we didn't do I B minus ia because since we're walking around the mesh in this direction we want there to be a voltage drop plus to minus across this 8 ohm resistor if there's a voltage forget about V naught what I've labeled here in red that has nothing to do with I'm talking about I'm saying that as we walk around if I want to put a plus sign here okay then there needs to be a voltage drop plus 2 minus so I can go through a voltage drop in order for there to be a voltage drop plus 2 minus then there has to be a net current flowing down through this leg so I'm assuming in this case there's a net current going down therefore it must be ia - whatever IB is the first current that I pick one of my subtraction is going to basically imply what direction I'm hoping this current is flowing in so I'm saying it's going down let's continue the problem listen do the other mesh currents solve the problem and then as we do more and more of these problems you'll get a lot of practice with how to write the mesh current equations and I think you'll see a little bit more about how we have we proceed so let's go ahead and simplify this this is going to be negative 4t plus 2 times I a plus over here eight times I a minus eight times IB is equal to zero so I a is here 8i a is here so I'm gonna have 10 I a is here minus 8i B is here I have a negative 4t let me move it to the other side of the equal sign like this 10 I a minus ni B is equal to 40 and let me just go ahead and put like a little asterisk there because that's gonna be an important equation basically we're gonna solve a system of equations just like we always do and this is the equation and notice that we have mesh currents ia IB and IC so we're gonna need three equations for three unknowns this is the first equation we're gonna have another equation for mesh current B another equation for mesh current C and then we gonna be able to solve the system for for these variables here which will then solve the whole problem another thing I'd like to point out to you is this is one reason why I think the mesh current method is actually a little bit easier to deal with because notice in the node voltage method because you were going up to the node and subtracting and dividing by resistor you always had a lot of division going on you always had all kinds of stuff divided by numbers and then you had to clear the fractions and or deal with decimals or something like that mesh current is actually easier because there's no there's no big fractions everywhere to clear out so once you write your mesh current equation now is very easy to distribute in and simplify your terms you want to write it in terms of the same ordering of the variables ia IB and IC for each equation so you can do your matrix method okay so the next let me switch colors to purple now we're going to deal with mesh current B okay ramesh current B so so here we go let's write mesh like this so let's start kind of at the corner we need to go through the 8 ohm resistor then through the 6 ohm resistor then through the 6 ohm resistor here and then we're done all right so what is the voltage drop across the 8 ohm resistor all right well it's IR what is the current I through this resistor so again you now what you've done before is totally separate throw it away you leave it for later now you're only focused on the middle mesh you don't really care so much about the other ones too much anyway so the way you write this since you're trying to write the voltage drop across this resistor is you want to write it as I be minus I a times 8 make sure you understand this the current through the 8 ohm resistor is gonna be in this case IB minus ia and the reason we write it is IB minus ia is because I want to assume since I'm walking around it the mesh this way I want to assume that the actual net current in this leg is actually flowing up so if I'm going to assume it's flowing up it's gonna give me a voltage drop plus 2 minus which is gonna let me treat it as a positive sign because remember when you're doing your kerkoff voltage law when you encounter a voltage drop from plus to minus you treat it as a positive sign so I'm gonna pretend I have plus to minus here so I do IB minus ia because IB is going up ia is going down notice that this is exactly backwards from the way we wrote it in the original mesh basically each mesh current equation you work on you're basically going to pretend you have positive voltage drops everywhere you can and you write your current equations in terms of that so it doesn't matter that in the previous equation I was pretending the current was going down in this resistor and in this equation I'm pretending the currents going up it doesn't matter each equation you write in terms of how you would like to pretending that you have a positive voltage drop through all of these resistors basically ok so when we get up to the 6 ohm resistor it's just going to be IB times 6 so I'm gonna do it as I B times 6 it's positive because this mesh current is flowing like this positive-negative I'm gonna have an implied positive - negative which I'm going through is a voltage drop so I get a positive sign of a mesh current equation and then now when I get to the 6 ohm resistor again I want to pretend the actual current and this legs going down because I want to pretend to have a voltage drop plus 2 minus so that when I go through it I can treat it with a positive sign so if I'm pretending the currents down then the way I write this current is IB minus IC times 6 ib is going down IC is going up fighting it so the net current down is IB minus IC and that's everything in the whole loop or in the whole mesh so we we set that equal to zero okay so now that we've got that done we can distribute it so we have eight times IB minus eight times I a plus six times IB plus six times IB minus six times IC is equal to zero and now we collect term so all the IAS we have this is the only ia I have so negative 8i a and IB I have a six and A six which is going to give me a 12 and then a twelve plus eights gonna give me twenty so I have a twenty I'm B I have ia and IB all taking care of IC is right here negative six I see and there's no other constant term so we just leave it equal to zero and this is the second mesh current equation that we're gonna basically use to solve for our system of equations okay notice that we're writing our final equations in the same form I well ia IB IC this one was ia IB of course there's no I C so you just leave it there we're writing it in the same ordering so that we can easily dump it into a matrix all right so let's continue on and do the mesh equation for I sub C and then once we're done with it will solve the problem will come back and reinforce how we wrote our mesh current equations so let's go back up to I sub C and write a mesh current equation starting from this corner going through the six own through the forum through the 20 volt you know back to where we started so if we're going through the six on resistor up like we're gonna pretend and assume we have a voltage drop from plus to minus right so that we can write it in a positive sense in our equation so if we assume that then we're gonna assume I sub C is bigger I subsea is going up IB is coming down so we treat it as IC minus IB that's the current multiplied by the six because I times R are six homes so again we treat it as IC minus IB because that's gonna be the net current flowing up then we get to the four ohm resistor the only current up here is I sub C we have an implied plus minus because we're going in the direction of the current flow so we just say plus I sub C times four okay I so C times four is going through there and then finally we get through there to the 20 volt notice that when we go through we're going from a plus to a minus and anytime you go plus to minus we carry that as a positive sign in our equations so plus 20 is equal to zero all right so that is the third and final mesh current equation so I'll say mesh C like this now let's distribute and what we have here is six I sub C minus six times 2i b plus or I sub C plus 20 is equal to zero so we don't have any I sub A's anywhere so we can't do anything with that so we have a negative 6i be I sub C as 6 plus 4 is 10 10 whoops not an equal sign we have a plus sign here plus 10 times I sub C now we have a plus 20 let's move it to the other side of the equal sign like this so we have a negative 6i B plus 10i C is equal to negative 20 and we're gonna put an asterisk by this so let's take stock of where we're actually at and I'll go over again one final time before we close the lesson how we arrived at our mesh current equations but for now let's just say we arrived at this messy mesh equation which we simplified to this this messy mesh equation which we got to this and then this mesh equation which we simplified to this so we have three equations we have three unknowns and you can solve these three equations any way you like you can put them into a calculator if you have a solver if you're comfortable doing that it's honestly just as easy just as fast to just do a matrix method quickly I'll show you how I do it on the board takes you an extra two minutes mate maybe and then you have a full record of everything you were doing as far as salt showing your work on your paper if you make a mistake solving it you're gonna get most credit for the problem if you dump these equations into a solver in your calculator that's fantastic if you get the right answer great if you type them in wrong and you get the wrong answer your professor is not going to give you as much credit because they're not they're not gonna know what the heck you were doing really it's the reason so if I were you I mean there's ain't many ways to do it but if I were you I would solve this guy with a matrix method so and it just takes a few lines to do so what we know about matrix math is that we can write an equation as a times X is equal to B where X is basically the unknowns in this case there I am I see okay so the way you want to write it the matrix a is the coefficient matrix of everything on the left hand side of the equation so for mesh a it's ten negative eight and zero because there's no I C so we write it 10 negative 8 and 0 for mesh B it's negative 8 20 and negative 6 so we say negative 8 20 and negative 6 and for mesh C there's no I a so we write a 0 then a negative 6 then a 10 0 negative 6 10 you need to make sure and put these zeros ends in places where the variables aren't present and then we multiply this by whatever we're solving for ia IB IC and that's equal to the right hand side which is the B matrix so it's gonna be 40 then 0 then negative 20 so it's gonna be 40 0 negative 20 this is the matrix form of the system of equations if you think about it if you might actually multiply this matrix it'll be multiplied this direction times this and that's going to equal 40 this direction times this is going to equal zero this direction times this is going to equal this this reproduces the systems of equations that we're actually trying to solve so we know when we write a when we write a matrix equation like this that the answer which is what we're trying to find is ia IB and IC this is what we're trying to find this column vector here so the way we do that is we take the inverse of matrix a multiplied by B this is an equation you can do anything to both sides take the inverse of a multiply that's gonna wipe out this matrix of course we do it to both sides so the inverse of a times the B matrix is gonna give you what we what we're trying to find okay so this is how I write it on my paper you need to type this matrix in to into your calculator and take the inverse of it and I think any professor is gonna be totally fine with you doing an inverse of a 3x3 matrix when you do it you're gonna get the following zero point one six four zero point zero eight zero zero point zero four eight zero point zero eight zero zero point one zero zero zero point zero six zero zero point zero four eight zero point zero six zero zero point one three six okay so that is this and then obviously you still have to multiply by this column vector over here which is 40 zero negative 20 so forty zero negative twenty like this so you type this a matrix in your calculator you calculate the inverse type this column vector N and then you multiply it oriented the way I have here and what you're gonna then find is ia IB IC which is a column vector that we care about is going to be equal to five point six two point zero negative zero point eight so this means that ia is five point six am ib is 2.0 hams and IC is negative zero point eight amps so we found the mesh currents and I've told you more than once that once you find the mesh currents we've actually solved the circuit you don't have anything you can find anything else once you know the mesh currents so our original problem didn't actually ask us find the mesh currents it said what is the value of the current I here and what is the value of the voltage drop across the eight ohm resistor we're calling it V naught this guy is actually the easiest to find because this I sub a is circulating quote-unquote in this mesh outside here on the outer boundary the value of the current in this outer boundary is ia and it's in the same direction as the arrow here so it's very easy to say that the current I is equal to ia which is just five point six amps that's what you would circle on your test now let me ask you this how do you find the voltage drop across the eight ohm resistor so what's the current in this leg and also notice our voltage is oriented plus to minus so let's assume for a second that the current is going down which would produce a voltage drop like this how do we find the current as it flows down assuming it does flow down well I sub a circulating down but we also have ISO B which is circulating up so we take ia minus IB so we say the naught is equal to ia minus IB that's the current in the 8 ohm resistor times 8 ohm so V is equal to IR the reason we do the subtraction is because we have two mesh currents that border this resistor one's going down one's going up so we subtract them for a net current going down is down B is up so I am eyeness ib gives us a net current down times the 8 ohm resistor ok I sub a is five point six IC B is 2.0 like this and then so V naught when you subtract this guy's three point six times eight is going to give you twenty eight point eight and what's the unit it's volts so these are the values okay so I need to talk about I would like to talk about several things here because it's our very first mesh current problem the following ones that come are going to be once a faster pace but I'm going to assume that you understand some of these things that I'm about to talk about as we go through the rest of the problem so I'm not going to take time to discuss every little nuance of the solutions of the forthcoming problems but for now I would like to say a few things when we find the mesh currents okay notice in this circuit the mesh current a is circulating over here mesh current C is circulating over here so if I asked you what would be the value of the current flowing down through you know through this and certain going down this direction you would look at the circuit and say well that's gonna be I sub C because I sub C is flowing over here that part makes sense in the IB circuit if I sit in this mesh here if I said what is the current flowing through the six ohm resistor let's say this way then you would say well that's equal to I so B because I've labeled it here it's circulating in this mesh so basically whenever you have the mesh currents you can use those values those those are the values of the current flowing in the circuit in the meshes the only gotcha is if I start asking you for currents that flow in these resistors that border the two meshes like the 8 ohm and the 6 ohm so if I ask you just like I did a minute ago what's the current flowing here well you can't just say it's ia even though I a is going down because you have IB that's fighting it flowing up so you have to subtract them I am IB is up so you subtract them that's the net current gonna be flowing down as I've drawn it here if I say what's the current flowing in the six ohm resistor you can't just say oh it's IB and be done because you have an IB component flowing down in this leg and you have an IC component flowing up in this leg and they're fighting each other so the net current is gonna be ID minus IC flowing through the six ohm resistor so anytime you're finding currents a net currents that are flowing through these resistors that border the meshes you basically always have to subtract the mesh currents to get the current actual current flowing through these elements but everywhere else like here we're not bordering anything else it's just ia flowing through here it's just IA flowing through the 40 volt source it's just I be flowing through the six ohm resistor and it's just I see flowing through the four ohm resistor and through the 20 volt source so that's important to know because a lot of times we'll ask you hey what's the power supply orbed by the you know ain't no resistor and you have to find the current in order to find the power and you need to know that in order to find the current I've got to subtract these adjacent mesh and mesh currents to find the current in that resistor so that's important the other thing I want to say is notice that when you calculate your mesh current answer sometimes you can get a negative result that's fine it's just like node voltage I mean you set it all up based on your drawing and then you solve it and then you get an answer and sometimes the answer is positive that just means that the way you drew it is correct your guess was correct sometimes you might get negative which means it's backwards from the way you actually drew it so what this means is I see really doesn't go this direction i subsea really flows the other way which kind of makes sense because I have a 20 volt source here I expect it to be able to supply some current that would mean the current would be coming out of the source so that really means it flows the other way it doesn't mean you go back and change the drawing or anything it just means it's a negative answer don't you just do it you just leave it like that explaining that the negative sign means that your direction was backwards from the way you draw it so those are two things I wanted to point out sort of like in conclusion to the answer we have now I want to briefly go back over this problem and say that when you're given a mesh current problem you had first identified the meshes we have three of them we know we're gonna need three equations because we have three meshes we write a mesh current clockwise another mesh current clockwise and another mesh current clockwise and for each mesh we go sum the voltages as we go around those loops as you sum the voltages if you encounter a voltage rise you treat it as negative that's why we did this if you encounter if you encounter a voltage drop which is like what we got here when we went through this direction it was a voltage drop we treat it as positive that's why the 20 volts was treated as positive here as you go through resistors you basically in this case we pretend I a of circulating this way so there's a voltage drop here so we treated as that's why we did hi any times to positive when we get to resistors that border adjacent meshes you have to subtract the adjacent mesh currents to find the current flowing here but you basically assume that as you walk through it you're going to encounter a voltage drop you just make that assumption even though it may not be true so you pretend that there's plus minus here and so in order to get that you need to pretend the current is going in the direction that you're walking around it and if you're gonna pretend that that's true it's got to be IA minus ib that's why we did IA minus ib here all right so when we get to the next mesh we're going in this direction so we're going in the opposite direction through the 8 ohm resistor so again we want to pretend that the voltage drop is plus minus in the direction of our child will forget about V naught that is something you deal with at the end of the problem but in the interim we're pretending we have a voltage drop across this from plus to minus that means there needs to be a current going up that means that I need to do ib - I egg so IB - I a times 8 that's I times R when we get to the 6 ohm resistor it's just a voltage drop through it plus 2 - so it's IR and then when we get to this border we go around like this we got to pretend the currents going down so we do I be - I see another way of looking at it basically is whenever you're traveling around your meshes you basically if you get to a border with a resistor that borders two meshes you basically always pretend that the current that's actually flowing in the direction that you're traveling through the resistor that's basically what you do so when we get to here in this mesh we pretend the currents going down when we get over here to this mesh we pretend the currents going down if we're looking at this mesh when we get to the third mesh going the opposite direction through the sixth ohm resistor we pretend that the current is actually flowing up so whichever way you're pretending the currents flowing is how you have to do your subtraction if you forget this little little part of it then you're gonna get sign errors in all of your equations that are being inconsistent with one another so make sure you you you do due diligence as we go around it so I'm gonna give you more practice with that as we do more problems but that is basically how you set up your mesh current equations you simplify them and then solve them the rest of the solution is the same thing that we've been doing all along once we finally got to the answer we said okay this current I is just whatever is equal to that's the answer and then in order to find V naught I have to calculate I times are V naught is drawn in this manner so I have to pretend the current is really going down so I say I a minus IV times the resistance so I times R and then you get the voltage there so hopefully that makes some sense to you I would encourage you solve this problem on your own just go back grab a sheet of paper solve it yourself even though you've seen the solution it's still gonna help you all of these solution techniques for circuits none of them are hard but they all have little gotchas with the sign convention if you if you just kind of just willy-nilly do something whether you don't really understand what you're doing you just kind of writing some stuff down as you go around in the loop you're gonna miss signs and you're gonna get the wrong answer and you're not going to know why so you have to force yourself to understand what I'm talking about when I talk about the sign conventions and how you're doing in so you'll get the answer every time so solve this guy yourself make sure you get the right answer follow me on to the next section we're going to do more and more problems a little bit more complicated a little bit tricky here and there but that bad really overall mesh current is not that bad it's not that hard the equations themselves are easier to deal with than node voltage in my opinion so that's a bull that's a bonus and honestly I like working with mesh current equations better than no voltage method and so as we go through the few the future problems you'll see what I'm talking about so I'm Jason with math tutor DVD comm go practice this guy and and follow me to the next section for more practice with mesh currents

Channel: Math and Science

Views: 137,912

Rating: 4.8817043 out of 5

Keywords: mesh current, mesh, current, circuit, circuit analysis, node voltage, electrical engineering, engineering, circuit analysis lectures, circuit analysis 1, circuit analysis tutorial, circuit analysis techniques, electrical engineering basics, electrical engineering 101, electrical engineering lectures, electrical engineering tutorial, electrical engineering course, mesh current analysis, mesh currents, mesh current analysis supermesh, mesh current example

Id: nC0PJqOJJ28

Length: 41min 36sec (2496 seconds)

Published: Sat Feb 17 2018