 Now we're going to take the same circuit we just analyzed by using series and parallel subsections but we're going to analyze it using Kirchoff's rules. And again for my students this circuit is a little bit more complex than the one you have to do on your own but it should give you the general idea of how to work with this that you should be able to apply to your slightly simpler circuit. And for Kirchoff's rules we want to look at the junctions and the loops. And we're going to start with the right side junction and by that I mean this location over here. Now I'm going to do that by drawing some arrows here and I'm just going to use some different colors to kind of show a few things. So first of all you've got this current coming into the junction and it's actually attached by a wire all the way around so that particular value for the current is not just the one but it's this whole section all the way around. Then we're going to see these other two branches and they'll have different values for the current. So now when I come in and I want to actually draw the current say maybe along the top side I'm going to first go up the branch and then along that side and then down that side. And so this particular section here is our next current. And then finally we have the other branch where we're going to have the current come down this way and then across this direction and then finally back up this way. And this section here is the current we really care about going through this one. So we've got one, two, three currents we care about. And right here at this junction those currents have one coming in and two going outwards. And I'm just going to give them some names right now. I'm going to call the current coming into the junction A, the current coming up along the top B, and the current going down along the bottom C. Now these are not the same A, B and C that I used when I did the subsections of the circuit so don't let that confuse you. Now if I were to look at the left side junction what I see is that I've now got current B and C coming into that junction and current A is going out of that junction where it wraps around to the other side. But if you look at these two equations in this particular case it's really the same equation which means I don't need both of those two junction equations. Only one of them is needed. Now if you have a much more complex circuit then you could actually have three or four or even five or more junctions and you'd have an equation for each junction but in the end at least one of those equations will be extra, not needed. Now we need to look at the loops and there's several loops we could look at. Now just to kind of clean this up a little bit I'm going to go ahead and get rid of the sections I had over here so that we can draw in the loops that we care about. Now in order to do that I'm going to choose a color for my top loop and make it an arrow and do my best here to draw a loop which starts at the battery, goes around the circuit, it goes through the one ohm but then it's going to take this top branch, go along through the two and two and three ohms, down and then around back to the start. And that's going to be one of our paths here. And I'm going to call that the upper path. And that upper path I can write out current or Kirchhoff's loop rule. Now remember or go back and review the other videos here. If I'm going around this loop in the direction I've shown I'm first going over the battery and since I went from negative to positive that's an increase of 20 volts. When I get up here to this resistor it's going to be a voltage drop across the resistor and it's the current times the resistance. The resistance is known as one ohm. The current is just my current A that I had up here. Then when I go through these two resistors again they're going to be drops but it's now current B and my 2 ohm and my 3 ohm. And Kirchhoff's rule says that that has to then be equal to zero. Now I'm going to go ahead and before I simplify this equation I'm going to go ahead and show my other loop. And I'm going to grab a different color here and I'm going to do my best to draw this carefully as well. Coming along, going up, again going through that resistor but now going down, around, up and then back to the battery. So this is my second loop I could take a look at. And for this second loop again I can apply Kirchhoff's loop rule. I have an increase as I move over the battery of 20 volts. Again I have the same drop of minus current A times one ohm. And then again a drop so it's minus but now it's IC times four ohms. And then when I get back around to where I started I have to be back at zero because the voltage changes over the loop balance out where it increases and where it drops off. Now there is a third loop I could do but just like our junctions you can end up having a one which is more than you need. I could come around, go through the two and the three ohms and then go backwards through the four ohms. That's another loop I could do and I could write out the equations for it. In that case I would basically have this part of the equation and then plus this part of the equation. And what happens here is that if I had took this equation and subtracted this equation from it I would get the equation for this small loop. And because it came from those two again it's an extra equation that we don't actually need to solve this problem. So for simplicity I'm going to not write that equation out but if you're curious you can ask me about that one. So now I've got a set of equations to solve. I don't actually have to look back at my circuit although I can to make sure I understand what I'm doing. But I can start working through these equations and I'm going to start by simplifying this top equation here for the loop. And in this case the first simplification I can make is that I can factor the common minus B out and then I've got the two ohms plus the three ohms which is our five ohms. I can then take this equation and solve it for current B. I'm going to move this current over to the other side of the equation and then I'm going to divide through by the five ohms. And that means I've got 20 volts divided by five ohms. I've got minus current A times one ohm divided by five ohms. And when I simplify that 20 divided by five gives me four amps and now I've got minus IA times just 0.2. There's no units on that because the ohms and the ohms cancel each other out. Now having seen all that work and knowing that you can pause and go back I'm going to take out that and just put this as our simplified equation. Then we've got this equation. I'm going to do the same process here. I'm going to go ahead and move IC over to the other side of the equation and then I'm going to divide every term by the four ohms so that I can isolate current C by itself. And when I do that I'm going to get that IC is equal to five amps minus current A times 0.25. And again you guys have the ability to go back and look at the video and pause things. So I'm just going to move that out of the way. And then what we're going to do is we are going to take this equation but I'm going to plug in the values for IB and for IC. And when I do that I get this long equation here. Current A is equal to four amps minus current A times 0.2 plus five amps minus current A times 0.25. I can start organizing like terms in here. And that's going to give me that the four and the five amps come together. The point two and the point two five come together. Continuing to rearrange this I can move the zero point four five over which is going to give me one point four five and my nine. And then finally I can divide the one point four five over giving me six point two one amps. So by working through these equations and it's just algebra at this point I've got my current A is six point two one amps. And so I'm going to actually put that up here as a known value. Give me a second. Once I know that as a value I can take another look at these equations because now I can plug in that value for current A and then simplify it. And I've got a value for current B which I'm going to copy up here. And similarly I can now put that value in down here and find a value for current C. And I know I'm copying and pasting stuff here but you should be able to plug these things into your calculator and confirm that you find the same values. So by analyzing the loops and the junctions I've now found all three currents. The next step in the process is to realize that once I know all of my currents I can find all of my voltages because the voltage for each resistor is going to be the value of the current through that resistor and the resistance of the resistor itself. So now I have the voltages across each one of the resistors as well as the current through each resistor. These are the current values. These are the voltage values for these individual resistors. And since I had already analyzed this particular circuit using subsections of the circuit you can go back and compare it to what I got at the end of that video and see that we have the same answers. Now Kirchhoff's rules are very powerful but it uses a lot of algebra.