 Suppose we are given a circuit with some resistors and some batteries. We are asked to find out the current through each branch. Current through each of these wires is what we need to figure out. So how do we do this? Well, usually when we have a circuit, the first thing I look for is if we can reduce the circuit by using series and parallel formula. But over here, you can't do that. If you don't trust me, go ahead, give it a try. Reduce the circuit using series or parallel. So in general, what do we do when we can't do that? When we have more complicated circuits, that's where we use Kirchhoff's laws. And so that's what we're going to do over here as well. You use Kirchhoff's laws to figure this out. But before I solve this problem, let me give you a general overview of how to solve any complicated circuit. How do you solve any complicated circuit like this? So think of it mathematically. Since I want to calculate current through each branch, and there are three branches over here, as you can see, color differently, I have three unknowns. And to figure out three unknowns, I need three equations. And it's Kirchhoff's laws that are going to help me build these equations. The first equation can be built using the Kirchhoff's current law, which is applied to nodes where three or more wires get connected. And the remaining two equations can be figured out by using Kirchhoff's voltage law, KVL, which is applied to closed loops, like a loop over here or over here or this big loop over here. Now I'm assuming that you have already practiced Kirchhoff's laws. Otherwise, this video is going to be pretty confusing. So if you aren't familiar with this, I highly recommend to go back and watch videos and do practice on our website on these laws and then come back over here. But if you have, well, then let's go ahead and solve this bad boy. All right. Where do I start? Well, since I want to calculate the current, we can start by assuming some current in each of these branches. So let's say in this yellow branch over here, the current is moving to the right. Let me call that value as I1. And I've colored this over here just to indicate that the current through this entire branch will be the same. So over here will be I1, here will be I1, here will be I1 because the current has nowhere else to go. And you could ask me why am I choosing it to go towards right? Well, I've randomly chosen that. You could have also chosen it to go backwards, for example. You're free, feel free to choose whichever direction you want. Similarly, in this branch, let's assume the current is, I don't know, maybe going upwards. So let's call that current as I2. And again, now here also you are free to choose whatever you want. But since it's so nice, I can see that, you know, there's I1 going in and there is I2 going in. I like to think of it as, okay, these two currents are merging and coming out from here. So if I1 goes in and I2 goes in, then what is the current that comes out? You can say I3, but that I3 has to be I1 plus I2, right? Right? I1 and I2 go in and the current that goes out, comes out over here, must be I1 plus I2, sorry. And in effect, what I've now done is applied KCL. This is Kirchhoff's current law, which says that whatever current enters a node should exactly equal the current exiting the node. And you can see that the same applies here as well. I1 plus I2 come in, comes in over here. I2 goes out, I1 goes out over here. Everything works nice and well. So because I applied KCL already, I only have two variables now, which means I only need two more equations. Does that make sense? So KCL is done. Now to build two more equations, I'll have to use KVL. So let me write that down. So next step would be KVL. And this is where things get usually a little tricky with the signs and everything. But if you, you know, if you do this conceptually, then there will be no problem whatsoever. All right. So how do you apply KVL? You choose a closed loop first. So for example, I can choose this closed loop over here. And the whole idea is you start from some random point and let's say I start from a point somewhere over here. And let me just give that point a name. Let's call it point as A. And then you figure out what the voltage at this point is and whatever that value is. And then you keep track of the changes in the voltage. At every point. And then you come back all the way. If you do that, you will build an equation. And the whole idea behind KVL is energy conservation. Okay. And again, like mentioned before, this is something that we've been explaining great detail in a previous video. So if you need more details, feel free to go back and watch that. And of course, please practice as well. This is definitely requires practice. All right. Having said that, let's quickly start this. So the way I do this is I start at this point and I say, let's say that the voltage over here is VA. Okay. Some voltage, I'm going to call that VA. And now I keep walking. And every time I cross a battery or I cross a resistor, I will figure out what the new voltages. So let's say when I walk, walk, walk, and I cross this battery, notice some voltage gets added or subtracted. So the voltage over here is going to be VA plus or minus something. So now the question is, what is the voltage at this point? Well, it's going to be, first question is, am I going up in potential or am I going down in potential? The way to figure that out, at least for a battery, would be able to look at the terminals. This is the negative terminal of the battery. This is the negative terminal of the battery. This is the positive. Negative means low, positive means high. So I'm going from low to high, meaning when I'm going from here to here, I'm going uphill. It's like climbing a mountain. I'm going uphill. All right. So when I go from here to here, voltage gets added. So the voltage over here must be VA plus something. Plus how much? Two volts. It's given in the battery. It's given over here in the question itself. So the voltage at this point is VA plus two. All right. Let's continue walking, walk all the way until we come back. Now when I, when I cross this resistor, again, there's going to be a potential drop. Remember Ohm's law, whenever a current flows through a resistor, there will be an IR drop. Now the question is, when I go from here to here, am I going up in potential or am I going down in potential? How do I figure it out over here? Well, notice I'm going in the same direction as the current. Current is moving this way. Right. And remember through a resistor, current always flows from higher potential to lower potential, but think of positive charges. They will always move from higher to lower potential. So I can again say this has to be positive this side. And this side has to be negative. And basically I'm saying that this should be higher than this one because current is moving this way. And since I'm also going in the same direction, I'm also walking. I can now say that I'm going down high to low. And so when I come over here, the potential drops by some value. So negative by how much when I go from here to here, potential drops by the whatever is the voltage over here. And that's V equals IR. And I know I is I one, I know R is one. So the potential drop over here is going to be I one times one. And now I am over here. This is the voltage at this point. Does that make sense? The same thing. Continue, you walk, walk, walk, walk, walk, walk, walk. And you keep doing this all the way till you come over here. So great idea for you to pause the video and see if you can just complete and see if you can build the equation. All right. If you're given it a shot, let's continue. So I come over here, now I cross this resistor. This time notice I'm going in the opposite direction of the current. So again, same up the same rule. If the current is going this way, this must be the positive. This must be the negative. But remember, I am walking in the opposite direction. So notice I'm going from low to high. I'm going in the opposite direction. I'm going low to high, meaning I'm going up the potential. All right. And so when I come over here, my potential increases. Oh, let me use the same color. Increases by how much IR, five, two times five. And this is where mistakes are usually made. I mean, when I was studying this, I used to not think about it. Think about whether I'm going up the voltage or down the voltage. I used to mug up in the direction means negative. Opposite direction of the current means positive without even understanding what was going on. And obviously if you try to mug up like this, you are bound to forget it sometime, definitely during the exams. That's what used to happen to me. But now I don't do that. Don't mug up those arbitrary, you know, conventions. Instead, think about it. Think about, are you going up in potential or down in potential? All right. So finally I'm over here. Now I'm going from here to here. When I'm going through a battery, you don't look at the current because a battery is an active device through a battery. Current can flow either ways, can flow from higher to lower potential or from over to higher also because battery is a pump. It can pump charges in the opposite direction. So when you're going through the battery, that's how you don't look at the current. You only look at the positive and the negative terminals. Another place where commonly I should make mistakes. So now when I'm going through a battery, notice this is the negative terminal of the battery. This is the positive terminal of the battery. And so when I go from here to here, again, I'm going up in potential just by looking at the terminals. All right, low to high. And so by the time I'm over here, what's the voltage? What's the potential that has to be plus two. And now I'm over here. This is the voltage at this point, right? But hey, that point has the same voltage as point A because there is no more potential difference. That means that has to equal VA. And ladies and gentlemen, this is how we build equation using KVL. Okay. Now all I have to do is go ahead and, and reduce this. So VA cancels out on both sides. And notice what I'm end up with. I end up with two plus two is four minus I one plus five I two equals zero. One equation down, one more to go to build a second equation. We have to choose a different loop. Maybe we can choose this one or we can choose this big loop as well. Okay. And I want you to give this a shot. So since we want, since I know I want us to get the same equation, let's choose a particular loop. Let's choose this and let's choose this small loop. All right. So let's say we start. Let's do it together. Let's say we start at this point. Let me call that as point B and let's walk in this direction like this. Let's walk this way and come back over here. So I want you to give this a shot. Okay. See if you can build an equation and then let, you know, then let's see if it matches it, what I get. All right. If you're given this a shot, this is what you end up with. I didn't want to, you know, spend time again, writing the whole thing. But again, you can pause the video and just confirm that this is what you got. And if you basically reduce that, you will eventually end up with this equation. And of course, you know, if we had gone the other way around, you have gotten the same equation, you were gotten all positives, but it's the same thing. Right. The equation would have remained the same. And of course, if you have taken a different loop, let's say you have chosen a different loop, let that this big loop, then you would have gotten different equation, but eventually the final answer turns out would have to turn out to be the same. All right. Anyways, for our loop, this would be the equation. And again, if you've got some place wrong, no worries. Just, you know, just go back and check, you know, what, where, where the error was. But now that we have the two equations, our physics is pretty much done. And now all we have to do is algebra, which I believe that you would be able to take care of this. So I'll just quickly brush through the algebra part. So I'm just going to minimize all the physics part. This is now the algebra. So if I just pull that equation over here, I just try to match the coefficients. So I'm going to match the coefficients of I one with this one. So I just multiply this whole equation by three. And that gives me a new equation. All the coefficients multiplied by three. And then I basically subtract this equation to finally give me I two equal to negative two amperes. I'm just brushing through this. I'm pretty sure you can do this all by yourself. And again, if you plug this back in, you will get the value of I one and you'll get I one as negative six amperes. So this is just algebra. And there we have it. We have solved our problem. Let me just get rid of this. Again, if you need some time on this, pause the video. All right, but that's pretty much it. So we have solved our problem. And now finally, just to make sense of these numbers, you might be wondering, what does this negative sign mean? Well, the negative sign is basically saying, hey, we chose the wrong direction when we assumed current to be flowing this way. I one really flows like this, but it doesn't matter if we got it. It means that the current over here is six amperes flowing down. I two is also negative, meaning, hey, we got this wrong as well. So this has to be two amperes downwards. And the current over here is just I one plus I two. So the current over here will be negative eight and not writing that down. And that again means that the current in this branch has to be going in the opposite direction eight amperes this way. And so the final thing is that, you know, you now maybe asked further questions. For example, they could ask you, what is the potential difference between let's say point A and point B? How do you calculate that? Well, you can use the same idea. You can use the same Khrushchev's law, but instead of walking through the entire loop, you start from point A and you end at point B. So I'm not going to do that. I'm pretty sure you can work this out yourself, but you go walk. You start over here, VA, you walk, walk, walk, walk like this. And you'll get VA plus or minus something. And then you stop here. So that will equal VB. And then you do the algebra and find out what VA minus VB would be. And that would be the potential difference between these two points. And because you can choose any path you want, you can walk like this and reach here, you can walk like this and reach here. In all cases, VA minus VB has to be the same. Great idea to confirm that.