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# Kirchhoff's Loop Rule

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Sections
Intro to Current
Resistors and Ohm's Law
Power in Circuits
Microscopic View of Current
Combining Resistors in Series & Parallel
Kirchhoff's Junction Rule
Solving Resistor Circuits
Kirchhoff's Loop Rule

Concept #1: Intro to Kirchhoff's Loop Rule

Transcript

Now, we're going to do is we're going to choose the direction of the loop. Remember, I mentioned, when we start going around the circuit up there that I could have gone clockwise or counterclockwise, we're going to choose the direction and I'm going to choose to loop this way, I'm going to write the 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 R1 then when I cross V2 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, okay? 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, 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 oposite 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, okay? So, this R1 here, the voltage of that R1 will be subtracted. So, I'm going to write this, sum of all voltages is negative i, R1, okay? 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, okay? V2. 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 R2 and remember the voltage of every resistor is always i, R so this is going to be i, R2 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 go into a positive. So, this voltage will be a positive voltage when I listed here, V1 and this entire thing has to be set to 0 okay, if you want to make this a little neater I can move some things around. Notice that everyone's negative except V1. So, I can move everything to the other side and it's going to look like this V1 equals V2 plus i1, r, I'm sorry I meant i, R1 plus i, R2. So, that's the loop equation for this loop, okay? So, what I want to do now, is I want to write a loop equation for the same circuit above which I draw down here, but now we want to go in the opposite direction of the loop, let's go in the opposite direction the loop. So, I'm going to put my little starting point here but now we're going to go this way and what this means is that you're first going to encounter this guy and jump through it, okay? 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, the plus goes where the current enters the resistor, this is one of the most important things to remember, okay? For the current enters a resistor. So, if current hasn't change direction so it enters here so positive to negative and the current keeps going till here and 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 V1 and it's going from positive to negative. So, it's going to be negative V1 and then you keep going you jump R2 from negative to positive. So, it's going to be positive of voltage of the R2 voltage of a resistor is i, R. So, i R2 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 V2 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 I, of R1 which is i, R1, don't forget at the end to set this entire thing to 0, okay? And you end up with this. Now, if I want to clean this up a little bit I can move, the only guy it's negative here is V1. So, I can move you one to the other side and I get that V1 equals V2, actually I'm gonna write V1 on the left, okay? I'm gonna write V1 of the left, V1 equals, imagine that all these guys go to the other side and you get rid of all the negatives. So, you get something like this after a little bit of moving around and we can order to put it here is because I wanted to show you, this the final equation, that this equation is exactly identical to this equation and the point here that you absolutely have to remember is that direction of the loop does not matter, okay? The direction of the loop does not matter and that it's going to give you the same equation whether you go clockwise or counterclockwise. So, just pick one okay, 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.

Example #1: Direction of Current in Loop Equations

Transcript

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 current, let's check it out. Alright, so there are more complex circuits that we're about to start seeing, you're typically not going to know the direction of the current. So, what you're going to have to do is you're going to have to assume the direction of the current, okay? This is the word are you typically hear but what this effectively means is that you're guessing them in other words, if you don't know, if you have no way of knowing you just pick one and then you solve the problem that way, okay? And don't worry about picking the wrong one it's going to be okay. I'll talk about that at the end, okay? So, before I talked about having there being these two steps to writing these equations but there's actually a new step here that you have to take before you can even get steps one and two and it is that you have to assume, assume the direction of all currents unless it's given to you, right? So, if you don't know you assume, if you're given a direction then you use that direction, okay? So, then you're going to label positive and negative signs and you're going to cross elements in the chosen direction of the loop.

First, let's talk about number one real quick and we've done this before but just as a refresher for every one of these elements you want to put pluses and minuses to the left and right or on the opposite ends of the elements. So, plus or minus here, plus or minus here and we're going to do this based on this rule here. So, the battery is going to be positive on its longer terminal. So, you put a little positive on the longer terminal and remember. So, I'm going to posit here, which means you put a negative on the other side. So, positive here, negative on the other side, okay? Same thing here by the way, positive here, negative on this side positive here, negative on this side, okay? Resistor is going to be positive where the current enters the resistor, where the current enters a resistor. So, this depends on the direction the current, if you don't know the direction of the current then you're going to use the assumed direction of the current, okay? Alright, cool. So, in this question here, before we get into number two, we're being asked to write loop equations. So, we're going to be following these steps over here to do that, okay? And here, it's asking to write a loop equation, if the current is clockwise. So, here current is clockwise means it goes this way so the current is entering this resistor here. So, I'm going to put a positive and then this one must be a negative the current keeps going and it enters this resistor here. So, this is the positive end and this is the negative end and then here we want the current to be counterclockwise which means it is going this way and it's entering this resistor here. So, that's positive, the other ones negative and then it's entering this resistor here. So, this is positive and this is negative, okay? So, step one is to label the positive and negative signs, we got that done and step two is to cross elements in the chosen direction of loop, in other words you're going to go around the loop and then you're going to sort of jump over each one of these elements and as you do that you're going to add the voltage, if you're crossing from negative to positive and you're going to subtract the voltage if you're crossing from positive to negative, okay? Remember that. So, now let's just do that, let's pick, let's pick a direction of the loop here, we're going to loop this way for both of them and we're going to start right here, okay? Right here, and right here, that's my starting point. So, if I'm going in this direction. Remember, the direction of the loop it's just a sequence in which you're going to sort of walk around this circuit it's not the direction of the car necessarily, okay? So, it's just a sequence that you're going to add these things. So, I'm going to write the sum of all voltages in this loop is a bunch of stuff equals 0, okay? And, what we have to do is write all these different things, they're going to be four numbers here or four elements here adding or subtracting, so the first one we get to is this true resistor then I'm going from positive to negative, I'm going from positive to negative. So, I'm going to subtract that voltage.

Remember, the voltage of a resistor comes from Ohm's law V equals i, R. So, here I'm going to write, we're going to the negative. So, I'm going to write negative i, R and the resistance here is 2 ohms. So, I can do this, I cannot plug anything for i because we're looking for the current, okay? We're looking for the current through the whole thing. Alright, now we're going to keep walking here, I'm going to jump from a positive to a negative therefore that voltage is also going to be subtracted negative for V, I'm going to keep going and we're going to jump from a positive to a negative, this voltage is also going to be subtracted negative, this is a resistor. So, it's going to be i, R, i is i and R is 1 and then finally I'm going to get over here and jump over this guy from it's going to go from negative to positive because I land on a positive, this is a positive 10 volts, okay? Everything is equal to 0 and this is going to allow us to solve. So, I have 4 volts and 10 volts or 10 minus 4 volts, if I combine these two, this is going to be 6 volts and then I have negative 2, actually let's move these guys to the other side. So, if you move negative 1i is going to be positive 1i, and then this is 2i it's going to become positive 2i so this is obviously just 3i and at this point, I'm going to be able to solve for i, 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 the sum of all voltages in this loop equals a bunch of stuff equals to 0 and then it's going to be, it's going to allow us to solve for i, what I've loved 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 card which is totally okay, right? So, I'm going to go here and the first thing I gotta jump over is this and I'm jumping from a negative to a positive so this is going to be positive, the voltage of that resistor and the voltage of that resistors i, R. So, it's I times 2, this is a new problem by the way so I can't just use this to over these two as current, right? We've got to find the current in this one, I'm going to keep walking here and jump from positive into a negative terminal. So, this would be negative 4 volts and we're going to keep walking here and jump from a negative to a positive and because end in the positive thi is going to be positive voltage of the resistor, voltage of resistors r or is i, R. So, it's going to 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, okay? This may even be already a little repetitive for some of you hopefully, which means you are getting, that'd be awesome. So, here I can add up these i's, I have one i and two i's, it's a lot of i's. So, this will be 3i and then 10 minus 4 is 6 plus 6 volts equals 0. So, if I'm solving for i, I'm going to move with 6 volts the other way 3i equals 6 volts therefore i, I'm sorry, negative 6 volts, almost messed up and that means that i is negative 2 amps, okay? Negative 2 amps Now, notice that this here was a 2 amps positive and this here was a 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 the exactly the same circuits that you 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 have 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, okay? 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 currents 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, okay? Here, I got a positive which means the directional is working with was actually a correct direction, cool? So, you picked, you were either given the direction or you assumed/guess one and if you, you know, if you pick the wrong one, it's okay? You'll just end up with a negative current at the end, the number will still be right and you would just know, oh 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, okay? I said that same thing a few times, hopefully that sticks, let's keep going,

Example #2: Solving Circuits with Multiple Sources

Transcript

Hey guys. So in this video we're going to put together everything we know about kirchoff'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 the simpler one, it says that 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 three 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. B, okay? So, label the junction, easy, we're going to label loops and remember that loops are arbitrary, this is just a sequence in which you're going to walk through the circuit and you can pick the direction. 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 loop1 and we're going to go this way, we'll call this loop2, okay? So, this step is done this step is done. Now, we want to label the direction of current and remember current will be the assumed direction you can try to figure out which one might make most sense or you can just randomly guess them. So, here notice that this battery has a 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 currents is going to go this way, I'm gonna call this current one. Now, notice that there are three branches here, right? This is a branch, this is a branch and this is a branch. So, because there are three branches I'm going to have three 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 a positive terminal here. So, you could think that, okay, the current is probably going this way. So, I'm going to draw it right here, i2, this bit of branch has no batteries just a resistor. Remember, the junction rule says that current in equals current out. So, if I have two 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 current 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 current 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 it goes like this right? So, i2 entering right here. So, this is the positive of the resistor and this is the negative of the resistor, you will 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 two junctions, two junctions therefore I'm going to have two equations, I'm going to do this first because they're easier and then later, we're going to write loop equations for each loop, I have two loops therefore I'm going to have two equations, okay? But, let's focus on the junction first. So, for junction A I'm simply going to write that current in equals current out. Now, look at A right here, i1 goes in and i2 goes in. So, in as i1 plus i2 and then out is i3 coming out, okay? So, that's it for Junction B same thing, I'm going to write in equals out. So, look at D here, i3 is going on, let make this green, 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 I want you to sort of go backwards here, right? i1 is obviously going to look like this and like this so that it's just flowing. Now, if you look at B, i3 is going in and the other two guys are going out. Soit's going to be i3 equals out which is i1 plus i2 and we're done at that equation, you might notice that these two equations are actually identical, they're both saying i3 equals the other two, it's the same which means you effectively don't really have two equations you just have one equation. So, I'm going to cross that out so we don't make confused, it doesn't mean it's wrong it just means that it's excessive, we don't need both of them, okay? Cool, so we got one equation out. Now, we're going to write these two loop equations. So, let's start with loop1 over here, loop1 and then we're going to write is that the sum of all voltages equals 0, okay? Equals a bunch of stuff which then equals 0 and this is what we have to fill in. Now, loop1 is this guy right here, okay? I'm going to pick a starting point like in the little corner here, actually I'll start over here so we hit that battery first. So, we're going to start here and go in that red direction which looks like this, okay? So, you're walking over here and then you get to this battery, you have to cross that, you're crossing from negative to positive. So, you're landing on the positive so this is going to be positive 9 volts and I'm just going to write a 9 because everything here is a voltage so everything os in volts so positive 9 and then you keep walking, keep walking, no one here and then being loop is just this little square here. So, we're actually going to turn here and then we're going to jump over, cross over this guy, we're going to go from positive to a negative so this is going to be a negative voltage. Now, voltage of a resistor is given by Ohm's law, voltage equals i, R okay? So, this is going to be the current through this resistor which is i3 times the resistance which is 15, okay? And that's it, there's nothing else to keep going around the loop, you're done, okay? So, just get back to the original point. So, now we're done, I'm going to move i3 that way so I get 9 equals i3 times 15 for i3 equals 9 divided by 15 which is 0.6, I'm actually just save time I'm going to save space, I'm going to put it over here, 9 over 15 is 0.6 amps, cool? So, we got that first one down and this is i3 and, in writing this equation we already solved for one of the currents and if you look at the problem, the problem is asking for the current through each one of the three branches, one main down. So, let's get the second one.

We're going to write the loop2 here and hopefully we can get a current out of this as well, sum of all voltages equals a bunch of stuff which equals to 0, let's look at loop2, we're looping this way, I'm going to start over here and we're going to go like that, okay? I don't want you to confuse that with the current so this is the loop direction. So the first element that we're going to hit up while we're walking here is we're going to jump over this resistor and we're going from a negative to a, from a negative to a positive, so this is going to be positive i, R, the voltage of resistors are R, the current here is i2 and the resistance, actually I can just plug in a resistance, it's i2 times 10. Now, notice that we're walking, we're looping opposite to current, that's fine, okay? Forget about that just go, just look at the positive and negative. So, we're jumping into a positive. So, it's positive and then here we get this guy right away and we're jumping from a positive to a negative, we're landing on a negative so this is negative or minus 5 volts, okay? Because, we're going from positive to a negative and we keep going, we keep going, we keep going and then we complete the loop, we will go all the way back to this point right here. So, we have to, we're going to go this way because that's the loop we chose and we're going to go from negative to positive, we're going to go from negative to positive, so this is going to be positive the voltages of the resistor, voltage of resistors i, R, the i here is i3 and the R is 15, by the way, we already know i3 it is 0.6, okay? So, we're going to be able to plug that in there. So, 10 i2 minus 5 plus 0.6 times 16 equals 0, this is simply 9. So, I'm going to write 10 i2 equals, I'm going to put the i and then the 9 to the other side negative 9, I'm going to move the 5 to the other side, it becomes a positive so this is going to be negative 4, negative 4, okay? So, this is going to be a negative 4. So, i2 is going to be negative 4 divided, running out of space there, negative 4 divided by 10 which is negative 0.4 amps of course. Alright, so that's that, I got the second current, I got that i2 is negative 4 amps. Now, what does this negative mean, okay? I got a positive here, which is, cool? But, what does that negative mean? Well, what that means is that the direction of i2 that we assumed is actually wrong, okay? So, assumed, let's write this here, very important, assumed incorrect direction of i2. So, i2 is actually the other way and I'm going to write this over here, actual i2 is in this direction right here, okay? Now, that doesn't mean anything, that doesn't mean you stop doing the problem or you do it again or you flip the, you flip the arrow and start over. Now, you just know that the direction was wrong but the answer is still correct, the magnitude of the current is 0.4, in fact, we're going to keep using the wrong number. Alright, as we keep going here but deep inside you know that the current is the other way. Alright, cool. So, we want to find, we have i2, we have i3, we want to find i1 but we ran out of loop equations, we already made two loops. So, what we're going to do now, is we can actually use this equation right here, we can use this equation right here to find i1 because we already have the other two. So, i1 plus i2 equals i3. So, i1 is simply i3 minus i2, i3 is 0.6 amps minus i2, i2 is negative 0.4. Notice that you plug as a negative even though you know that that means that it's in the other way just keep rolling with it, okay? Negative minus, so I have two negatives here si this is going to be a positive 0.6 plus 0.4 is 1 so the current through i1 is 1amp, okay? Now, we're basically done I mean, we are done, what I want to do is real quick just show you something with these currents. So, i1 right here. Now, we know was 1 amp and I know that i2 the actual direction of i2 is this way and I know.

Practice: For the circuit below, find the current through each of the 3 branches. Concept #2: Combining Voltage Sources in Series

Transcript

Hey guys. So, 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, in a 5 volt battery here, they are in series because there's a direct path between all these 4 guys here, the Y doesn't split so they're all in series. So, I can combine these 2 so instead of having two batteries I can just have one and generally when you do you just add up the voltage is something like v1 plus 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 two 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 current this way, right? The current is leaving, the charge is leaving the battery through this side, this battery is here. So, the the positive sign is here, that's for positive and positive. So, this guy is also pushing this side. So, you might imagine, think of this as force as almost, right? This thing is pushing clockwise and then this thing is also pushing clockwise, they're helping each other they're both pushing this interaction. So, I can redraw this and just say that these two 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 can remember we can also merge these two 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 instead, I'm going to put a 15 volt here and the resistances will add 2 plus 3 is 5, it would be this 5 ohms, okay? And by this example here is asking us to find the magnitude and the direction of the current because the both currents are going to 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 current? The current goes this way, well, to find a magnitude of a current in a circle like this you can just write V equals i, R. So, I is going to be V over 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 of this guy over here is pushing current or at least it's trying to push current this way, something that'd be really cool is if you could pause this video and try to do in 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's pushing this way, this guy's 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 effect 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 of the 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 equals i, R, we're going to solve for i and this is going to be V over R, the voltage is a 5, the R is a 5 so the answer is 1 ampere, cool? That's it for this one, super simple, let's keep going.

Example #3: Find Two Voltages (3 sources)

Transcript

Practice: For the circuit below, calculate

(a) the voltage V1 shown, and

(b) the current through the 6-Ohm resistor. Practice: For the circuit below, calculate the voltage across the 100-Ohm resistor. Concept #3: How to Check Your Work (Kirchhoff's Rules)

Transcript