Kirchhoff's Loop Rule - Online Tutor, Practice Problems & Exam Prep

Kirchhoff's loop rule states that the sum of all voltages in a closed loop of a circuit equals zero. This principle is crucial for analyzing circuits with multiple voltage sources. When writing loop equations, voltages are added or subtracted based on the direction of current and the loop. For series voltage sources, voltages add if aligned in the same direction and subtract if opposing. To verify calculations, ensure all branches have consistent voltage magnitudes and polarities, applying Ohm's law (V = I × R) for resistors to find voltage drops.

Intro to Kirchhoff's Loop Rule

Hey, guys. So now we're going to start talking about Kirchhoff's loop rule, which is going to be massively important when you start solving more complicated circuits. Now we're going to build this up a little bit at a time over a few videos, and I'm going to take it a little bit slow because I want you to have a strong foundation so that when you get to problem-solving, it can be a breeze. This can be a little hairy, so we're going to build it up a little bit at a time. Let's check it out. Alright. So far, we've dealt only with resist, with circuits that had a single source, like a single battery. But now we're going to start to get into circuits with multiple batteries. So, we're going to need new tools. And what we mean by new tools is new rules and new equations.

The new rule we're going to use is Kirchhoff's loop rule and it states that the sum of all voltages in a loop is simply 0. It's conceptually very simple, but it's pretty tricky to write equations. One way to summarize this entire thing here is just to say that the sum of all voltages in a loop is 0. This rule is also called Kirchhoff's or Kirchhoff's voltage law. Voltage law because you're going to be going in a loop adding voltages, or voltage law because you're adding voltages. Whatever you want to call it, I want to point out that this rule actually works for any circuit. So, we could have used this to solve some circuits earlier, but it's going to be especially useful when you have multiple sources. It's too hard for the easy stuff, but it's necessary for the hard stuff.

So, what we're going to do is for each loop in a circuit, we're going to write one loop equation. For example, if you look here, these wires and all these elements form one loop, so we're going to get one loop equation. What we're going to focus on now is how to write this loop equation. What the equations will do is that we'll either add or subtract voltages of the batteries and the resistors. The voltage of the battery here. Now, the voltage of the resistor is simply times r. times r. is the current, and r is the resistance. And this comes from Ohm's law, v equals r.

V = I R .

That's the voltage of the resistor is going to be . So, for example, I have a current here. So it's going everywhere here. And what we're going to do is we're going to write the sum of all voltages which, by the way, is going to equal 0. I'm going to write all the voltages here. I'm going to list all the voltages. So, I'm going to pick a point here. Let's say we're going to start here, and we're going to make a loop all the way around this circuit here. And I can go in any direction. I can go this way or I can go the other way. I'm going to go this way because it's the way that the current's going. So, we're just going to go with the current. And so the first element that you run through here is . So what you're going to do is you're going to write . And then when you get to , you're going to just write plus. And then you're going to keep going here. Imagine you're just moving along. You're going to cross this. So you're going to write plus the current, which is everywhere, . And then you're going to get over here, and you're going to write plus. Now, notice here that I said that the equation is going to add or subtract voltages. And here I just added all of them, which is actually wrong. These voltages are going to be either adding or subtracting, and I don't know yet. We don't know yet which one. And that's what we're going to talk about next, but I just want to give you a basic idea first before we start, that you're going to be adding these things and then setting them equal to 0. Now, are they going to be adding or subtracting is what we're going to talk about now.

They're going to be adding or subtracting depending on two things. On the direction of the current and on the direction of the loop. So direction of the current in this case was clockwise, and the direction of the loop was also clockwise. Just to show you what I mean by direction of current and direction of the loop. So the first thing we're going to do, and we have to do this before we start writing the equation, is we're going to use a direction of current to put positive and negative signs on the ends of each resistor. So let me show you what I mean. So here for a resistor, the positive end is where the current enters the resistor. So look at this diagram down here. The current's going this way. It enters the resistor here. So I'm going to put a little positive on this side of the resistor, and therefore, the negative goes on the other side. The current keeps going keeps going, and the current enters this resistor right here. So we put a positive here and make this negative bigger and a negative here. For batteries, the positive end is always going to be, the positive terminal. So it actually does not depend on the direction of the current. So the positive terminal is a longer one here. So I'm just going to put a positive on this side, so the other side's a negative. And I'm going to put a positive on this side, and this other side's a negative. So the first thing we do is we put these signs everywhere.

Now what we're going to do is we're going to choose the direction of the loop. Remember I mentioned when we started going around the circuit up there that I could have gone clockwise or counterclockwise. We're going to choose a direction, and I'm going to choose to loop this way. I'm going to write loop here. And all that means is the sequence in which you're going to add things. Right? So we're going to start here, and we're going to go this way. So the first element I'm going to cross is , then I'm going to cross, and so and so forth. And it says here, when crossing elements in this direction, this is the direction of the loop, you're going to add the voltage if you crossed that element going from negative to positive. Once you get this, it's going to be super simple, but I want to move slowly here. So let's just see here. If you're going from negative to positive. So if you follow the current here, you're going from positive to negative. That's the opposite here. Right? So if you add when you're going from negative to positive, this means you will subtract subtract if you're going from positive to negative. So this here, the voltage of that will be subtracted. So I'm going to write the sum of voltages is negative . Now let's go to the next element. I'm going to keep going here, and I'm going from positive to negative again. So when I cross this battery, I'm going from positive to negative. So I'm going to subtract that voltage there. Now let's keep going. Now I'm going to go again from positive to negative. So because I'm going into the negative, I'm going to subtract the voltage of, and remember the voltage of every resistor is always. So this is going to be. And finally, I get to this last element here, and I'm going to cross it. And when I cross it, right, imagine your sort of charge is going through. I'm going from negative to positive, finally going to a positive. So this voltage will be a positive voltage when I list it here, . And this entire thing has to be set to 0. If you want to make this a little neater, I can move some things around. Notice that everyone's negative except. So I can move everything to the other side, and it's going to look like this. equals. So that's the loop equation for this loop.

What I want to do now is I want to write a loop equation for the same circuit above, which I drew down here. But now we want to go in the opposite direction of the loop. Let's go in the opposite direction of the loop. So I'm going to put my little starting point here, but now we're going to go this way. What this means is that you're first going to encounter this guy and jump through it. Now remember the first thing you do is you put these signs everywhere. These pluses and minuses. The battery is the easiest one to do because the big one is always plus, this is always minus, always plus, always minus. The resistor, the plus goes where the current enters the resistor. This is one of the most important things to remember. Where the current enters the resistor. So the current hasn't changed direction. So it enters here, so positive to negative, and the current keeps going to here, and then it enters here positive to negative. Now we're ready to start going around the loop. Sum of all voltages in this loop, we're going to start here. So the first element you jump across is the, and it's going from positive to negative. So it's going to be negative. And then you keep going, you jumped from negative to positive. So it's going to be positive voltage of the. Voltage of a resistor is, so. And then we're going to keep going here, and we're going to jump from negative to positive. So it's going to be positive voltage of that battery, which is . And lastly, we're going to get here and jump from negative to positive. And because I'm going into a positive, it's going to be positive, the voltage of of which is. Don't forget at the end to set this entire thing to 0. And you end up with this.

Now if I want to clean this up a little bit, I can move the only guy that's negative here is . So I can move to the other side, and I get that equals. Actually, I'm going to write on the left. I'm going to write. Imagine that all these guys go to the other side, and then you get rid of all the negatives. So you get something like this after a little bit of moving around. And the reason I wanted to put it here is because I wanted to show you, this is the final equation, that this equation is exactly identical to this equation. And the point here that you absolutely have to remember is that the direction of the loop does not matter. The direction of the loop does not matter in that it's going to give you the same equation whether you go clockwise or counterclockwise. So just pick one. Cool. So, this is a quick introduction of how you write these loop equations. We're going to do a little bit more to build up the concept so that you are a beast at this. Let's keep going.

Direction of Current in Loop Equations

Hey, guys. So in this video, we're going to keep talking about writing loop equations, except that I'm going to add one complexity, which is the fact that you won't always know the direction of the currents. Let's check it out. Alright. So in the more complex circuits that we're about to start seeing, you're typically not goin... I is just 6 over 3, which is 2 amps, and we're done. That's it for part a. Now we're going to do the same thing here for part b, write that the sum of all voltage is in this loop equals a bunch of stuff equals to 0 and then it's going to allow us to solve for I. What I'd love for you to do is pause the video right now, try to emulate the steps that I had in part a, see if you can get something for part b. I'm going to keep rolling here but hopefully you gave this a shot. It's important that you're following and then you're able to do this yourself. So we're going to start here and we chose to go in the same direction. So I'm actually going against current now. I'm going to sort of walk around the circuit against the direction of current, which is totally okay. Right? So I'm going to go here and I mean, the first thing I gotta I gotta jump over is this, that I'm jumping from a negative to a positive. So this is going to be positive, positive, the voltage of that resistor. And that the voltage of resistor is IR. So it's I times 2. This is a new problem by the way. So I can't just use this 2 over this 2 as current. Right? We gotta find the current in this one. And we're going to keep walking here and jump from a positive into a negative terminal. So this is going to be negative 4 volts. And we're going to keep walking here and jump from a negative to a positive. And because that end in a positive, this is going to be positive voltage of the resistor. Voltage of resistors are or is IR. So it's going to be I times r which is 1. And finally, we get over here and we're going to jump from a negative to a positive. And because I'm landing on a positive, this is positive 10. K? This may even be already a little repetitive for some of you, hopefully, which means you are getting it. That'd be awesome. So here I can add up these i's. I have one I and 2 i's. That's a lot of i's. So this is going to be 3 I and then 10 minus 4 is 6 plus, plus 6 volts equals 0. So if I'm solving for I, I'm going to move it 6 volts the other way. 3 I equals 6 volts. Therefore, I I'm sorry, negative 6 volts. Oops, almost messed up. And that means that I is negative 2 amps. Okay? Negative 2 amps. Now notice that this here was 2 amps positive and this here was 2 amps negative. Do you think that's a coincidence? It isn't. It's not a coincidence at all. In fact, you should have expected that because these are exactly the same circuits, they would get the same current. This negative here, what's up with that guy? All that negative that means is that you've chosen or the direction that was assumed for you, you didn't choose, it was chosen for you, was wrong. So if you had chosen this direction, it just means that you chose the wrong direction. But the cool part is the answer is still right. You just now know that it was actually the current was actually physically moving in the opposite direction. K? So the answer to what is the magnitude of the current, well, the current is 2 amps. That's the magnitude of the current. But the direction of the current, we now know is opposite to what we were working with, where we're going with the current being counterclockwise and we know the direction of the current is actually clockwise. K? Here, I got a positive which means the direction I was working with was actually the correct direction. Cool? So you pick, you were either given a direction or you assume slash guess 1. And if you and if you, you know, if you pick the wrong one, it's okay. You're just going to end up with a negative current at the end. The number will still be right and you would just know, oh, whoops. I guess the wrong one. That's cool. Flip it. You don't have to resolve the problem again. You would just know that the direction of the current is opposite to the assumed one. K? I said that same thing a few times. Hopefully that sticks. Let's keep going.

Solving Circuits with Multiple Sources

Hey, guys. So in this video, we're going to put together everything we know about Kirchhoff's rules to solve a full problem. Let's check it out. Alright. So we're going to combine Kirchhoff's or Kirchhoff's Junction Rule, which is a simpler one. It just set the current in equals current out. We're going to also use the loop rule, which is the more complicated one where we have to write the loop equations. And we're going to use them together to solve more complicated circuits with multiple resistors. Here are the 3 steps we're going to follow. Now we've used lots of these parts individually. Now we're just putting everything together. Cool. First step, we're going to label directions. We're going to first label junctions. Remember junction is a split on the wire. So this point here is a junction, I'm going to call that A. And this point down here is a junction, I'm going to call that point B. Okay? So label the junction easy. We're going to label loops and remember that loops are arbitrary. This is just the sequence in which you're going to walk through the resist through the circuit and you can pick the directions. So here's a loop and we're just going to go this way. We're just going to go this way, I'm going to call this loop 1. And we're going to go this way, I'm going to call this loop 2. Okay? So this step is done. This step is done. Now we want to label the direction of currents. And remember, currents will be the assumed direction. You can try to figure out which one might make the most sense or you can just randomly guess them. So here, notice that this battery has the positive here, which means current typically would flow out of here if this was a single battery circuit. But you can't really know that for sure. But either way, I'm going to just use that to sort of, dictate that I'm going to assume that the current is going to go this way. I'm going to call this current I₁. Now notice that there are 3 branches here. Right? This is a branch. This is a branch and this is a branch. So because there are 3 branches, I'm going to have 3 different currents. Okay? Let me clean this up a little bit. So I'm going to call that I₁. There's another battery here and this is the positive terminal here. So you could think that, okay, the current's probably going this way. So I'm going to draw it right here, I₂. This middle branch has no batteries just a resistor. Remember the junction rule says that current in equals current out. So if I have 2 currents going into the A, the third current has to be coming out of the A so that we're at least consistent. We're guessing all these directions, but at least let's do it in a somewhat consistent way. Alright? So I₃, we're going to assume that it's going that way. So we got our directions of currents assumed. Once we know that, we can label positive and negative on voltage sources and resistors. So the batteries will be positive on the long side, positive and negative, positive and negative. But the resistor will be where the currents will enter. Okay. So let's look at every resistor. This resistor right here, the current is entering from this side, so this is the positive, this is the negative. And then this resistor right here, if you sort of backtrack I₂, you can see that I₂ goes like this and then it goes like this. Right? So I₂ is entering right here. So this is the positive of the resistor and this is the negative of the resistor. You want to be very careful setting this up because if you set this up wrong, you're going to get the wrong answer. Cool? So don't screw this part up. Alright. So we're done here. We put our little pluses and minuses everywhere. Now we're going to write some equations. First, we're going to write junction equations. We're going to write one for each junction. I have 2 junctions therefore I'm going to have 2 equations. And we're going to do these first because they're easier. And then later we're going to write loop equations for each loop. I have 2 loops, therefore I am going to have 2 equations. Okay. But let's focus on the junction first. So for junction A, I'm simply going to write that currents in equals currents out. Now look at A right here, I₁ goes in and I₂ goes in, so in is I₁ + I₂. And then out is I₃ coming out. Okay? So that's it. For junction B, the same thing, I'm going to write in equals out. So look at B here, I₃ is going here because it just keeps going through. I₃. We had already drawn I₂ is coming out of B, and if you backtrack I₁, if you sort of go backwards here, right, I₁ is obviously going to look like this and ]]>like this.]]>

Combining Voltage Sources in Series

Hey, guys. So in this short video, we're going to talk about how we can merge voltages in a circuit to make that circuit simpler. Let's check it out. Alright. So you can combine voltage sources if the sources are connected in series. This only works if they're connected in series, you can do this to simplify a circuit. So, for example, I have a 10 volt battery here and a 5 volt battery here. They are in series because there's a direct path between all these 4 guys here. The wire doesn't split so they're all in series. So I can combine these 2. So instead of having 2 batteries, I can just have 1. And generally, what you do, you just add up the voltage. Something like v₁ + v₂. Now if the voltage sources are in opposite directions, their voltages are actually not going to add, they're going to subtract. Okay?

And we're going to do these 2 quick examples and I think this is going to make a ton of sense, totally obvious. So look at this battery here. The positive terminal is this way. So it's pushing currents this way. Right? The current is leaving, the charge is leaving the battery through this side. This battery is here. So the positive sign is here. Let's put positive and positive. So this guy is also pushing this side. So you might imagine think of this as forces almost. Right? This thing is pushing clockwise and then this thing is also pushing clockwise. They're helping each other. They're both pushing the same direction. So I can redraw this and just say that these 2 act as a single battery, of voltage 15. It doesn't matter that there's a resistor in between them. Got it?

Similarly, by the way, if you remember, we can also merge these 2 resistors. They're in series, so they're also just going to add up. So an equivalent circuit here would be one with a 15 volt battery. I'm actually just going to draw in this direction here. That 10, I'm instead going to put a 15 volt here. And the resistances we'll add 2 plus 3 is 5. It would be this, 5 ohms. Okay? And by the way, this example here is asking us to find the magnitude and the direction of the current. Because they're both currents are going the same direction, the direction of the current will simply be let's write here. Direction is going to be clockwise. Now what about the magnitude of that currents? The current goes this way. Well, to find the magnitude of a current in a circle like this, you can just write v = I r. So I is going to be v / r. The voltage is 15. The resistance is 5. So it's just 3 amps. Very simple.

Now here, it's a little bit different because this guy over here is pushing currents or at least it's trying to push currents this way. Something that would be really cool is if you could pause this video and try to do it on your own. I think a lot of you are going to get this right because it's very straightforward. But I'm going to keep going here. So this guy is pushing this way and this guy is pushing this way. They're clashing. Well, who do you think is pushing current harder? Right? So the 10 volt is putting more of a pressure, or providing more influence for the current to move. So that one is going to win. Which means that despite the fact that they're pushing against each other, the currents will overall move in this direction. Okay? So the 5 is actually fighting against the 10 volts and making it weaker. Therefore, we can just subtract the 2. We're going to say 10 minus 5 is just 5, which means the effective, the equivalent voltage here is going to be 5 in this direction. 5 volts this way, again, going up against a 5 ohm resistor. The direction is still clockwise, but the current will be different. Current is going to be v = I r. We're going to solve for I and this is going to be v / r. The voltage is 5. The R is 5. So the answer is 1 amp. Cool? That's it for this one. Super simple. Let's keep going.

Find Two Voltages (3 sources)

Hey, guys. So let's check out this example. We have a circuit with three batteries and three resistors, and we want to find two of the voltages. One voltage is given to us, and then we're asked to find V₁, V₂. We're also given two of the three currents. There are three branches, and two of them are already given to us. Now, remember the general steps to solve these problems are that first we're going to label a lot of stuff and then, we're going to write as many equations as we need to find everything we're looking for. Okay?

So what are we labeling? We're labeling junctions, we're labeling loops, we're labeling currents, and we're also putting pluses and minuses on the resistors and the batteries, which is the setup to be able to write these equations. Alright. Junctions are easy; it's just points where the wire splits. So let's call this a and b. You might notice that this 4 a right here, if I keep extending it, it's going to go this way, 4 a. And then this is going to go this way, 4 a. Right? And right away, because of the junction rule, you might already be able to see here that if I have 5 coming in and 4 coming in, there has to be nine coming out this way because current in equals current out. So nine in mean there's going to be nine out, and that's this current. So now I actually already know all three currents, which is good.

Alright. So let's label our junctions out of the way. We've actually already calculated one of the currents which is going to be helpful. What about loops? So the currents on this loop are going this way: 5 amps, 5 amps. Right? I got all the current labels. So, I like to loop remember, the direction is arbitrary. But I like to write loops that follow the current just because everything's going the same way; it's just easier to understand. Right? So we're going to do a loop this way, which, if you notice, will follow the 5 amps all the way around and then it's actually going to follow the 9 amps as well. Not necessary, it's just a little nicer. But don't sweat it too much. You can just make a loop in any direction. Okay? Same thing here. I'm going to start this loop right here at point b, and I'm also going to spin this way, which will neatly go along the direction of 4 and the 9. It doesn't have to. I just like doing that. Okay? Cool. So those are the loops, and the currents are actually already all labeled. So we got this, this and this. And now we're going to put pluses and minuses everywhere. The plus on a battery will always be on the long stick right here. So plus, plus, minus, minus. And the plus on the resistor will be where the current enters. So the current's entering from this side, so that's a plus. So the other is a minus. This is a plus because of the 9 amps right here; this is a minus. And then this is a plus because of the 4 amps entering that way, and this is a minus. So we're done setting this up. It's really important to get this right so you can write the equation properly.

Now let's write equations. Remember, there are junction equations and there are loop equations. The junction equations will help you figure out the currents, and the loop equations will help you figure out anything, but typically voltages or currents. Right? It'll help you figure out either one of those. We already know all the currents, so we don't have to write a junction equation at all. In fact, when we said that 5 + 4 gets to 9, we're essentially writing that in our heads. Okay? So let's just write loop equations. There are two loops, so we can write two equations. So for loop 1, I can write that the sum of all voltages equals a bunch of stuff which equals 0. And all we got to do is fill in this bunch of stuff. Loop 1, if you go around loop 1 this way, there are four elements that you're going to hit up. Right? Two resistors and two batteries. So you should expect they're going to be 4 turns in this thing here.

So the first thing we're going to do is we're going to go through the 18 volts. So we're walking over here. Let's make this a different color. We're walking here, and then you're going to jump from a positive to a negative. So remember that means you're going to add 18 volts. Over here, you're jumping from negative to positive. So that's positive 18. And then the positive to negative. So that's going to be negative in the voltage of the resistor. Remember, the voltage of a resistor is given by Ohm's law, V = IR. So, I'm going to make two little spaces here. This is for my I, this is for my R. The I here is 5, and the R is 8. Okay? I'm going to make a little bit more space here just in case.

Now we're going to keep walking down this path here. My mom turns over here. We're going to jump from negative to positive, and this is going to be positive. I don't know that voltage. It's what I'm looking for. Positive V₁. And we're going to keep walking, jump from positive to negative. So negative, this is a resistor; negative IR. The current is 9 and the resistance is 4. So you get this. Cool? So let's clean this up. 18 minus 40 plus V₁ minus 36 equals 0. If you notice, good news, there's only one unknown. So you can solve for this. And if you move some stuff around, I got it here. You're going to get, the first voltage is going to be 58 volts. Cool. We already got V₁. Now we just got to get V₂. And we're also done with the equation for the first loop. So now it's going to write an equation for the second loop. And hopefully, the V₂ will come out of there. So the sum of all voltages in the second loop is going to be a bunch of stuff equals 0. The second loop is also going to run into two batteries. Do that. Oops, wrong color. We're running around with green. I forgot to put a plus and a minus here, but you know the plus is the big one. We're going to jump from negative to positive. So this is going to be positive V₂. By the way, that's the variable I'm looking for right there. Then I keep going; jump from positive to negative. So it's going to be negative. This is a resistor, so I'm going to put negative IR. I is 4, R is 6. And then you're going to keep walking here. You're going to get here. You're going to jump from here to here, from negative to positive. So it's going to be positive V₁. By the way, V₁ is 58. So we could plug that in as well. And we're going to keep walking, and then we're going to cross over, and we're going to get negative IR, where I is 9, and R is 4. Okay? So if we write this out, you get V₂ minus 24 plus 58 minus 36. And if you move everything around carefully, you get that V₂ is just 2 volts. Okay? So that's it for this one. We got V₁. We got V₂. Along the way, we also got all the currents. So we actually know everything we need to know about this resistor, about this circuit rather. Alright? That's it for this one. Let's keep going.

For the circuit below, calculate 

(a) the voltage V₁ shown, and 

(b) the current through the 6-Ohm resistor.

• A. i_6Ω = 4.67 A, V₁ = 1.8 V
• B. i_6Ω = 4.67 A, V₁ = 24 V
• C. i_6Ω = 0.67 A, V₁ = 26 V
• D. i_6Ω = 0.67 A, V₁ = 30 V

For the circuit below, calculate the voltage across the 100-Ohm resistor.

• A. 10 V
• B. 17 V
• C. 24 V
• D. 56 V

How to Check Your Work (Kirchhoff's Rules)

Hey, guys. So now that we've seen how to solve these complex circuits using Kirchhoff's rules, I want to show you a simple method you can use to double check your work. And the idea here is that solving these circuits is a long process. There's a lot of steps to it. So there's a lot of room for small mistakes. And it'd be great to be able to quickly double check. So let's talk about them in this video. Alright. So once you know everything, you know all the voltages, all the currents, and all the resistances, you can double check your work with a simple rule. And you actually already know this rule, which is that all branches must have the same magnitude of voltage. Same magnitude of voltage. Okay. So you might remember we talk about this many times. If you have something like this and the voltage of the source is 9 volts, then the voltage of this resistor is 9 volts as well. And that's because they're sort of opposite to each other on the two branches, right? You can think of these as, this is one branch and this is another branch. So the voltages have to be the same on opposite sides. Now the direction must always be the same as well. And direction is actually not the right word. The right word here is polarity, but it really just means direction. So what does that mean? So if you have a 9-volt battery this way and then a 9-volt battery this way, it's the same voltage, but this one's positive on the left and this one's positive on the right. So they have different directions. The directions have to match, and the total magnitudes have to match. So let's do this real quick. We're going to check if all the numbers match up. Now just to be clear, you're not going to get a test question that says, here's a complete circuit. Is this right or not? Right. I haven't seen those in tests, but I'm giving you this just to sort of build up the skill that might be useful if you have enough time at the end of a test. Right? So or if you're doing your homework. Alright. So we're going to check that all the voltages on all the branches add up. And a branch, remember, is all of this. But really, you just have to worry about the top part because there's only you only have circuit elements up here. There's nothing on the sides. Same thing here. So does that branch have the same voltage here as this branch and does it have the same voltage as this branch? Cool? So let's look at all the voltages. This guy is 18. What about this guy here? This is a resistor. The voltage of a resistor, remember, comes from Ohm's law, v equals I r. So I can just multiply the I with the r. Same thing here. I can multiply the I against, with the r. So I times r, they're going to multiply. This is going to give you the voltage r times I. Okay. So let's multiply these numbers. This is 40. This here is 36 volts and this here is 24 volts. Okay? So now what we're going to do is we're going to put the polarities and I'm going to write it over here. So this 18 has a positive and a negative. Right? Remember the positive and negative on the resistor depends on the direction of the currents and if the current's coming this way, this is a positive and a negative. So if you go here to the side, I'm going to write that you have 18 V positive and negative. And then here you have 40 V positive and negative. And you can think of it as 18 volts this way and 40 volts this way. What do you think is that net voltage or the equivalent voltage of those 2? If you have if they're going opposite directions, they're going to subtract. The 40 the 40 wins over the 18. So it's 40 minus 18, 40 minus 18, which is 22. So this guy is the winner, which means this entire branch has the equivalent voltage of 22 with the positive pointing to the right. Okay? Now all the other branches have to have the same thing. Let's look here. This is positive and negative. So I got a 58 volts positive-negative, and this is 36. The current look at these 2 currents here. Look at these 2 currents. The third one, they're both going into here. So the other one has to be this way, which means this is a positive and this is a negative. So here I have a positive 36 volts negative. The 58 this way is going to overpower the 36, and then you do 58 minus 36, that's 22 as well. Awesome. So I have a 22 to the right. Cool. So so far it's matching up, and then here at the end, I have 24 here and 2 here. The 24 wins, so it's 24 minus 2, they're also going opposite directions. It's 24 minus 2 so you end up with 22 here and here. Okay? I drew it this way but yeah, but I could also have drawn 22 this way, 22 this way, 22 this way. Maybe that's a little bit easier just to just to sort of for it to make sense. All these things have a polarity of 22 to the right, okay? Or the positive side is on the right side over here of the whole branch. That's it. That's all you got to do. So let's look at this one, and it might be a good idea for you to pause the video and do exactly what I did before and double check it. This might seem a little long because the first time I'm explaining, but once you get the hang of it, you can do it really quickly. Cool. So I'm going to keep rolling here, and I'm going to go as fast as I can to show you that this can be actually pretty fast. So if you do the calculator 6 times this to get the voltage, it's 4 volts. And if you multiply these two numbers to get the voltage, this is going to be 14 volts. Okay. This current is leaving. This current is leaving. So this current must be entering, which means this is positive, negative, positive, negative, positive, negative.

Here’s what students ask on this topic:

What is Kirchhoff's Loop Rule and how is it applied in circuit analysis?

Kirchhoff's Loop Rule states that the sum of all voltages around a closed loop in a circuit is zero. This principle is essential for analyzing circuits with multiple voltage sources. To apply it, you write loop equations for each loop in the circuit. Start at a point in the loop and move in a chosen direction (clockwise or counterclockwise), adding or subtracting voltages based on the direction of the current and the loop. For resistors, use Ohm's Law ($V=IR$) to find voltage drops. The sum of these voltages should equal zero.

How do you determine the direction of current and loop in Kirchhoff's Loop Rule?

The direction of the current is usually given or assumed based on the circuit's configuration. For resistors, the positive end is where the current enters. For batteries, the positive terminal is fixed. When choosing the loop direction, you can go clockwise or counterclockwise. The key is consistency: if you cross an element from negative to positive, you add its voltage; if from positive to negative, you subtract it. The direction of the loop does not affect the final equation, as it will yield the same result regardless of the chosen direction.

How do you combine voltage sources in series using Kirchhoff's Loop Rule?

Voltage sources in series can be combined to simplify a circuit. If the sources are aligned in the same direction, their voltages add up. If they are in opposite directions, their voltages subtract. For example, a 10V and a 5V battery in series with the same polarity will combine to a single 15V source. If they are opposing, the effective voltage will be 10V - 5V = 5V. This simplification helps in writing loop equations and analyzing the circuit more easily.

How can you verify your circuit analysis using Kirchhoff's Loop Rule?

To verify your circuit analysis, ensure that all branches have consistent voltage magnitudes and polarities. Check that the sum of voltages around each loop equals zero. Use Ohm's Law ($V=IR$) to calculate voltage drops across resistors. Compare the calculated voltages with the known values to ensure they match. This double-checking process helps identify any mistakes in your calculations and ensures the accuracy of your analysis.

What are common mistakes to avoid when applying Kirchhoff's Loop Rule?

Common mistakes include incorrect assignment of voltage polarities, inconsistent loop direction, and errors in applying Ohm's Law. Always ensure that the positive end of a resistor is where the current enters and that the positive terminal of a battery is correctly identified. Be consistent with the loop direction throughout the analysis. Double-check your calculations for voltage drops using $V=IR$. These steps help avoid errors and ensure accurate results.