kof's rules and complex DC circuit is going to be the topic of this lesson in my new General Physics playlist which when complete will cover a full year of University algebra based physics in this lesson we're going to find out that not all circuits can have all the resistors uh kind of reduced down into a single equivalent resistance uh these more complex circuits we're going to have to use what are called kirov rules uh to come up with a system of equations to solve for the different currents that are flowing throughout the circuit my name is Chad and welcome to Chad's prep where my goal is to take the stress out of learning science now if you're new to the channel we've got comprehensive playlists for General chemistry organic chemistry General Physics and high school chemistry and on Chads prep.com you'll find premium Master courses for the same that include study guides and a ton of practice you'll also find comprehensive prep courses for the DAT the MCAT and the oat so we stated in the intro that some circuits are so complex that the combination of their resistors can't be reduced down into a single equival resistance well how do you recognize that you might have one of those complex circuits well the single biggest indicator is that you've got more than one battery in the circuit like this setup right here good chance you're going to need to use kof's rules uh and set up a system of equations to solve for in such a case all right so Kirk's got two rules first one is the junction rule second one's going to be the loop Rule and the junction rule simply says that any current flowing into a junction is going to equal the current flowing out so anything flowing in flows out so to speak so if I've got a junction here and I've got a junction here maybe we deal with this one right here and maybe current's flowing in right here so then it's got to flow out here and here and if I've got a total of say 10 amps flowing in here then I'd have to have a total of 10 amps flowing out here and here or you know maybe it's currents flowing in in from both of these to a total of you know 6 amps and 4 amps then 10 amps here would have to flow out but the total current flowing in equals the total current flowing out and here's the deal we're going to define the current in the direction it's flowing and we're going to try and do it correctly and the directions and all but sometimes we're going to get it wrong sometimes it's not intuitively obvious which way the current's flowing what's great though is that if you set up the system of equations we're about to using kof's rules correctly uh the math is going to work out anyways if you choose the wrong direction for the current then when you solve for it you'll end up getting a negative number and the fact that when you solve for a current it comes out negative it means if you defin it to go one way it was really going the other way and you just guessed wrong when you were setting up the circuit no big deal kirkoff second rule is the loop Rule and the loop rule simply says that the voltage increases on any closed loop of a circuit are going to be perfectly matched by voltage decreases so voltage increases voltage decreases when do we encounter those well when you cross a battery you're going to get either a voltage increase or decrease if you cross it if I Define my current as going around this way then when it crosses the battery it's going from negative terminal to positive terminal that corresponds to a voltage increase but if I Define my current in such a way that actually I go the other direction and I go from positive terminal to negative terminal that's going to correspond to a voltage decrease now typically when a a current crosses a resistor you're going to get a voltage decrease a voltage drop across that resistor now here's the deal if you look at the way you've defined your current so if you cross a resistor in the same direction as the current you've defined that's going to be a voltage drop so but if you're going around a loop and actually as you encounter that particular particular uh resistor in the loop if you're actually encountering it in the opposite direction from the way you defined your current then you're actually going to experience a voltage increase instead so technically at a battery or a resistor you could get a voltage increase or decrease depending on how you define things now again normally we think of crossing a battery as a voltage increase but again most of the time we try to conceptually set it up to where we're going uh the current is going from negative terminal to positive terminal to get that but if you do the opposite it would be negative instead of a positive same thing you know if I Define my current right here as coming off this uh battery here from negative to positive and coming through this 8 Ohm resistor in this Direction that's going to be a voltage drop across this resistor because I'm encountering it in the direction uh of my current but let's see I decide to you know make one big enclosed loop around the big loop right here and as I go around the loop I go around it this way well the current I defined a second ago was going the opposite way I'm just happening to go around this Loop in a certain direction it could be clockwise or counterclockwise but once you choose your direction you stick with it and because I chose in this case counterclockwise whereas the current in that part of the circuit was going the opposite direction then instead of a voltage decrease for resistor it's going to be a voltage increase instead so but the idea is that by by the time you factor in all the batteries and all the resistors any voltage increases are going to be perfectly matched by voltage decreases so that the overall sum is zero between them all right so these are kof's rules so and notice in a circuit like this uh we've got a junction here and a junction here and even though there's two Junctions we'll find out there's only going to be one Junction rule because it's going to be the same sets of three currents that are flowing in and out of this Junction as this one if there were more Junctions we might end up with more than one Junction rule but it's pretty common just to have a single Junction rule but we have three different Loop rules we're going to set up because we have a small Loop right here a small Loop right here and a large loop right here giving us three possible Loop rules that we could set up so we're going to set up this system I've done four equations in order to solve for some currents so the question simply gives you this diagram here and says find the current through each of the resistors in the following diagram so now we've got to Define some currents and again we want to do this kind of conceptually and try and get it right but even if we get it wrong we still get it right with the math all right so I'm going to go start at the biggest battery I got 20 volts and I'm going to let him kind of govern which way the current would go and so in this case the current I'm going to have it flow from uh negative to positive terminal flowing through here and notice it's not just from here at the battery out but this whole part of the circuit is one common current flowing all the way through from this Junction all the way to this Junction and let's just call that i1 so we could have called it anything but it's common to use i1 I2 I3 that sort of thing all right next I'm going to go to my other battery so in this case I would normally see 16 volts I'm like oh from negative to positive that's likely to flow this way so but in this case again from Junction to Junction it's not just from here to here it's all the way around that particular part of the loop from this Junction to this Junction and we'll call that I2 and now we can see that there's current flowing into this Junction right here from both of these places so if you decided to make the other current look like this you would have a big problem on your hands because you'd have current flowing from three different ways into this Junction and never flowing out and and so conceptually you wouldn't want to do that now again the math might still work out but conceptually you definitely wouldn't want to do that so in this case we're going to make that right there the current flowing the other side and I really should again in similar fashion make it throughout that whole part of the wire and we'll call that I3 and so now we've defined the currents now you could have defined it any way but you're still going to come out with three different currents you could have defined here and which one was i1 and which one was I2 and which one was I3 you might have defined it differently and in fact you might have defined the directions wrong as well maybe you defined i1 going around this way just for the fun of it so in all likelihood it probably flows this way but if it doesn't then at the end when I solve for i1 it'll come out negative and it'll let me know that I chose wrong so even if you choose wrong the math still works out in the end all right so now we're going to set up a system of equations we'll start with the junction rule if I can spell Junction that would be great so and the junction rule here just says pick a junction we can pick this one or we can pick this one so if I pick this one I can see that i1 and I2 flow in I3 flows out and so the junction rule would simply say that i1 + I2 equal I3 if You' have picked the other Junction it would have worked the same way because now it's I3 flowing in and i1 and two flowing out and you still would have ended up with I3 equaling i1 + I2 which is mathematically the same thing we've got right here okay so now we'll deal with the loop rules and again we've got three possible Loops we can set up here so and in this case I'm going to call this little one over here Loop one this one over here Loop two uh and then the big loop all the way around is going to be loop three so and I can go around these Loops in a clockwise fashion or a counterclockwise fashion is totally my call it doesn't really matter so but again I like to often try to make sense of things and so in this little loop loop number one right here again the currents like in the way I've defined it to be flowing around this way and so that's the way I'm going to go around it and I'm going to start with the battery so and that battery again as a voltage increase as we go from negative to positive terminal so that's going to be positive 20 volts so and then the first thing we counter next is this resistor right here so in that 2 ohm resistor is going to experience a voltage drop as the current goes through it if you recall ohms law says that Delta V equals ir and that's how we get the voltage drop across resist and in this case we're en counting encountering that resistor in the same direction as the current we draw on the diagram I3 and when that happens again that's when you get the voltage drop and so in this case it's going to be I which is I3 * R which is 2 ohms and so it's going to be I3 * 2 ohms and I'm going to write that backwards it's a voltage drop so minus so 2 ohms * I3 and I'm going to leave off the units here they're all going to be in ohms so but I'm going to leave it off it'll make the math uh a little easier to solve for in the end if we leave those units off easy to get lost in the weeds with all the all the units on there so uh next thing we're going to count on the same Loop is now the 4 ohm resistor and again we're going to hit that 4 ohm resistor it's current i1 that's flowing through it and we're hitting it in the same direction as i1 travels so it's going to be another voltage drop and so we're going to have minus so again i1 * 4 Ohms which I'm going to write is minus 4 i1 and that's it we've made a complete Loop now back to the battery and again all the voltage increases when added to the voltage decreases sum to zero that's Kirk Hof's Loop rule all right I'm going to write this a little bit different in a little bit but that's suffice today for now that's what the loop rule is going to say all right Loop Number Two here and and this one here I'm again I'm going to start at the battery and start with 16 volts so in fact let's just write it right beneath so 16 volts so next thing we're going to encounter after the battery notice because I went from NE terminal to positive terminal that's why I'm calling it 16 volts in this one case I'm going around clockwise so then we're going to hit the 8 Ohm resistor and we're going to hit it in the direction in which I2 was defined so it's going to be a voltage drop so we'll subtract and it's - 8 * I2 carry on around this Loop and now we're going to hit the 2 ohm resistor in the same direction in which I3 is defined so it's going to be another voltage drop and so - 2 * I3 equal 0 so and notice I say equal Z because that's the last thing we encounter before reaching the battery again in that Loop and then finally now we've got to do the big loop and again once you're doing a loop it doesn't actually matter where you start I like starting from a battery you could like starting from a resistor it's your call but you do got to do the complete Loop for kirkoff Loop rule to apply and so for the big loop I'm going to start once again at the 20 volt battery it's the biggest battery so it's kind of the thing governing which the way current flows the more more than anything else so to speak and so I'm going to do it in a counterclockwise fashion all the way around and so first thing it's positive 20 volts from negative to positive terminal next thing we're going to encounter is that 8 Ohm resistor but notice since we're going around counterclockwise but I2 is going in the opposite direction and so in this case when you encounter it uh in such fashion in one of your Loop rules if if you're encountering it in the opposite direction in which the current's defined instead of a voltage decrease you're going to get a voltage increase instead so I'm not going to subtract here so I'm going to add eight time I2 8 ohms time I2 same thing here with the battery so when we're going to counter this battery here we're actually going from positive terminal to negative terminal it's you only get that potential increase when you go from negative to positive if you're going from positive to negative you actually get a voltage decrease so that's going to be minus 16 volts keep going around the loop and now we're going to encounter the 4 ohm resistor so but we will encounter it in the direction in which the current flowing through it is defined i1 so it will be once again a voltage drop and so we're going to get Min - 4 i1 and then we're back to our battery so this all sums up to zero so here's our system of equations so in this case we actually have three unknowns with four equations with three unknowns you need at least three equations so apparently we've probably got a redundant equation somewhere in here we might figure out where that is we might not but we've got more than enough information in order to solve this sometimes what they'll do to you is they'll actually come back and and give you one more variable they'll take one of these resistors or one of these voltage sources batteries uh and and not give you the value for it so that way you get four unknowns in four equations well this one's going to be a little bit easier in fact I chose all the numbers here to make the math really nice you're probably going to encounter some problems where the math is not so nice FYI okay so we're going to set this up just a little bit different from here so I'm going to rewrite each one of these equations I'm going to do a couple different things so I'm going to move the total volts to the right hand side of the equation so and then I'm going to try and make the total volts positive where I can as well as we'll see so like this first equation here I'm going to rewrite that I'm going to write so NE -2 I3 so I'm going write i1 I2 and I3 in order as well as you'll see so actually I'm going to start with -4 i1 I'm going to leave a space because there's no term with I2 in it so and then I'm going to move on to I3 in this case it's going to be -2 I3 and if I subtract 20 volts from both sides it's going to equal -20 volts and I'm doing this for a reason that won't be immediately obvious but it will be very obvious at the end of the lesson now one thing to note I would prefer to have positive 20 volts over here so it the math will work out either way it's just my preference uh and so in this case all of these are negative I can multiply the entire equation by negative one which is therefore going to make this positive 20 Vols positive 2 I3 and positive 4 I and I could just erase the positive signs I don't actually have to be there and it's equivalent to the same equation we had just a second ago all right second equation here I'm going to do the same thing so in this case there's no i1 term but there's a Min - 8 I2 there's a min-2 I3 so and then if I subtract 16 volts to the other side it's going to be 16 volts and we're into the same situation here and again I would prefer to have this being a positive number over here preferentially so I'm going to multiply the entire equation by1 which makes both of these positive as well and notice I left space over here because there's no i1 term in that equation and then finally the last one we've got uh -4 i1 we got Plus 8 I2 so we've got no I3 term and then 20 volts - 16 volts would be four volts and then subtracting over the other side would be -4 volt but again I prefer to have that positive if possible so I'm going to make that positive I multiply the whole equation by1 which makes this positive but it actually makes this one right here negative cool and we've got three Loop rule equations right here set up the way I want them set up so and then we've got this Junction rule one right here and I'm going to rewrite him as well and we'll see why I do in just a second uh again towards the end of the lesson there's an easier way to solve this as we'll see so but if I rewrite this one more time the the uh the junction rule would read i1 + I2 minus I3 is equal to zero I just subtracted the I3 back to the other side so the again the currents were all in exactly the same places and ultimately what I'm setting up here is to solve this with a matrix and even if you haven't had Matrix algebra I'm going to show you how you can take advantage of M Matrix algebra with your calculator to solve these now we're first going to solve it without Matrix algebra and then we're going to go and do it with the Matrix and you're be like why don't we just do it with the Matrix every time and that's a great question why don't you do it with the Matrix every time uh once you get comfortable using your your calculator to do so all right so we've got a system of four equations here with three unknowns and we want to solve for them so how are we going to go about doing this so and and again if we're not solving this with a matrix it turns out you've got a system of substitutions or linear combinations or something of that sort to figure this out uh and in this case no two of these equations have the same two variables that you can reduce it down to one variable and solve for one so equation one's got i1 and three next equation is two and three 1 and two one two and three and so you have a problem here so one thing we might do uh notice this is 1 and three and this is 2 and three we might take this is already solved for I3 on the junction Rule and substitute in i1 + I2 right in for it and then this first equation would only have i1 and I2 the third equation only has i1 and I2 and then we could do some sort of linear combination or substitution between those two to solve for it as well so a little bit of work here to solve these not the easiest thing in the world but that's the approach I'm going to take I'm going to substitute in to that first one in for I3 I'm going to substitute in i1 plus I2 from The Junction rule it's not the only way to solve this there's a lot of different ways we can approach this that's just the way I'm going to approach it all right so we're going to have 4 i1 plus 2 times and substitute in for again uh for I3 is going to be i1 + I2 so and then going back that's going to equal 20 volts all right so if we expand that out just a little bit that's going to be 4 i1 + 2 i1 + 2 I2 = 20 volts then I'll combine these as well and get in fact I'm just going to do it real quick and make that 6 i1 move that back over too so plus 2 I2 equals 20 volts okay so now we've got a new equation that only has i1 and I2 in it we had a another equation back over here that only had i1 and I2 in it I'm going to rewrite it and so here we had 4 i1 plus or sorry minus 8 I2 equal to 4 volts and this is a little easier to handle now because we've got the same two variables in two equations we could do another set of substitutions if we solve one of these for i1 or I2 or we could do my preference which is some sort of linear combination here so if I take a look at these uh if I want to try make the 6 i1 and the 4 i1 cancel I'm going have to multiply both equations by some sort of factor so and I could do that so but I can see with the i2s so this is already Min - 8 I2 this is 2 I2 and if I take and multiply this first equation by 4 I would get Plus 8 I2 and then I can just simply add the two equations together to make that fall out and so in this case if I do multiply that by 4 I'm going to get 24 i1 + 8 I2 equals and 4 * 20 Vols would be 80 Vol and then I'm going to combine those and so in this case 4 i1 + 24 i1 is going to be 28 i1 but notice the i2s cancel negative 8 I2 plus I2 zero and then here 4 + 80 is 84 volts and then I'll divide by 28 and we'll see that i1 equals let our calculator do some work but I chose the problem so I know this is going to work out too but 84 / 28 is exactly three and in this case that's 3 amps or more properly 3.0 amps with two sigfigs so now we know that i1 is 3 amps and this problem just got significantly easier because once we know i1 we can go back to our original equations or these versions of it right here and plug it in for i1 here to solve for I3 here or plug it in for i1 here to solve for I2 here and we'll get the other two currents in Rapid succession so one other thing to note because having this one variable solve for makes the rest of the problem fairly straightforward sometimes on a test these are really long to solve for so sometimes on a test they'll give you either i1 or I2 or I3 so that you can actually solve it in the time allotted for the test to make it that much faster so from here we're going to work uh on getting those other two and so in this case if we put i1 in against 3 amps right here we'd have 4 * 3 m would be 12 volts we'd subtract that from the other side so and 20 - 12 would be 8 volt and so 2 I3 would equal 8 volts divide by two and we see that I3 is going to equal 4 amps all right so going over to I3 I3 is right here and that equals 4 amps and from here we actually don't have to go to any other other loop rules we technically go straight to the junction rule which again you can think of it as a junction rule in just another equation but if you understand again what it means any current flowing in has to equal the current flowing out well from this Junction there's only one current flowing out and it's 4 amps so far I've got three amps flowing in how much more do I need well I need another one amp and I2 must equal 1 amp and again that just flows from the equation as well we'd have in this case 3 amps plus I2 = 4 amps I2 must equal 1 amp great and now we've solved for all three currents life is good so and sometimes instead of simply asking for the currents they might ask for the voltage drop across a particular resistor notice for the 8 Ohm resistor it's 8 times that 1 amp and the voltage drop would be 8 volts so if I said what's the voltage drop across the 4 ohm resistor well in this case it's going to be 4 Ohms * 3 amps would be 12vt drop so on and so forth notice we can also verify our answers and stuff like that to make sure they work pick a loop if we go back to Loop one here 20 volt increase followed by 4 * so 4 amps * 2 ohms 8 volt decrease 4 Ohms * 3 amp 12vt decrease and an 8vt and 12vt decrease is a total of 20 volt decrease to match the 20 volt increase and you could go back and verify that with any one of those Loops all right one more thing to do here and again that's going back to the Matrix algebra that would make this problem significantly easier than solving a system of equations with a series of substitutions so if we go back I wrote this in this fashion again on purpose because I'm setting the Matrix up for you so turns out when you set a matrix up you've got your variables and I got i1 I2 and I3 these could be X Y and Z so anything of the sort and it's common to set it up to where you've got the variables on the left hand side so and then any numbers with no variables so on the right hand side of the equation and so in this case we've got a a matrix with four equations and each of the equations is going to have four terms now you might be like Well Chad this one only has three terms in fact this one in fact they all only have three terms well when you've got a a variable missing from an equation you put a zero for that entry in The Matrix so it turns out for I2 right here I'll put it in a different color for I2 in the first entry I'm going to put a zero so but you only put the coefficient in front of the variable as well so we're not ever going to enter i1 or I2 or I3 we're just going to put in the coefficient 4 0 2 and then 20 will be the first line in our Matrix and the next one the first term is going to be zero cuz there's no y1 and then 8 2 and 16 so in the third row of the matrix it's going to be 48 you got to factor in that negative so and then 0er for I3 and then four and then for the last one will be 1 1 - 1 and 0 so let's write that out again we've got 4 0 2 and 20 we've got 0 8 2 and then 16 we've got four uh8 zero and four and I'm running out of room here you know let's let's just erase this and make give ourselves just a little bit of room there so and finally last one 1 1 1 0 and that's our Matrix right there that's what we're going to plug into our calculator so and you just got to know how to do just a little little bit of Matrix algebra with your calculator and I'm going to use the most popular calculator out there this one's rather old but the ti 83 is still probably the most abundant calculator on planet Earth uh so I'm going to show you how to use this to solve this Matrix and solve for both i1 I2 and I3 all at the same time all right so here's your t83 I first got to turn it on obviously we'll clear that out so in this case you want to set up your Matrix first you have to Define your Matrix and put all the inputs in which are right next to the calculator here on the screen uh and then you've got to go back and solve it which is straightforward in your calculator as well now there's a matrix button right here I press it it gives me some options so what I want to do is take one of these predefined matrices and actually edit it so here it's just set on a by default and I'm just going to go back in and edit a and hit enter so and in this case I actually want it to have uh four rows and four columns so it's going to be a 4x4 not a 1 by one all right so now it's a 4x4 and then I'm going to start putting the terms in so the first one was a four enter 0 enter 2 Enter 20 enter we're on to row two zero is the first term enter Then eight enter two enter 16 enter four enter on the third row now -8 enter zero enter four enter onto the fourth row one enter one enter -1 enter 0o enter now the whole thing is defined and now we're going to quit out of there so I'm going to hit second function and quit here we'll get out of there we'll go back to Matrix one more time so and now we want to do some math with that Matrix so and we want to use What's called the reduced row Echelon form R RF so when you reduce a matrix down to its reduced row Echelon form you've effectively solved for every variable as we'll see I'm going hit enter right there so and then I want to hit the Matrix button again and I want it to do it to Matrix a so I'm going hit enter there for Matrix a and then we just hit enter and notice we've got 1 03 for the first row 0 1 0 1 0 014 0000 0 0 Let's Take a look at how this gives us the ultimate solution to i1 I2 and I3 so here's our Matrix in reduced row Echelon form and before we go back and show you that it directly gives us the values for i1 I2 and I3 I just want to remind you what each row in The Matrix originally comes from and it's from the equation here so here we had 4 i1 + 0 I2 + 2 I3 = 20 volts and that's where the four the zero the two and the 20 for that first row came from so these represent again the coefficients in front of each of the variables and this is the answer on the other side of the equal sign and so we can see from this first one we've got 1 i1 + 0 I2 plus 0 I3 = 3 well effectively it just means that 1 i1 = 3 and that's what I meant by directly it just gives you the value of i1 the second one it's 0 i1 and 0 I3 it's I2 equaling one and then the last one here is I3 equaling 4 and you see that those batch perfectly the values we got i1 was equal to 3 I2 was equal to 1 and I3 was equal to 4 right off the Matrix now maybe you've had a matrix algebra class and you know how to convert a matrix like this and know all the steps to get it into reduced row Echelon form manually uh but you don't need to the calculator would do all the work for you should be told you don't have to have a whole lot of background in Matrix algebra at all to just uh kind of use it the way I've shown you on your TI calculator if you found this lesson helpful consider giving it a like happy studying