Kirchhoff's Laws in Circuit Analysis - KVL and KCL Examples - Kirchhoff's Voltage Law & Current Law

okay so here we have the next problem on the board we have a circuit given like this we have a hundred and fifty volt source we have a 20 ohm resistor we have five ohm resistor we're given that the voltage drop across this resistor is a hundred volts that's given to us so you take a meter out and you measure and that's what you have we have another resistor here but we don't know what it is the value is just R the question is find the value of R very simple what is the value of this resistor okay how many ohms is that right so here in all honesty in my opinion is where we start to get into some circuits that are not as easy to solve all of the circuits up until now if you just know some basic ideas about Ohm's law you can figure out what's going on you can figure out the currents and the voltages you can kind of get away with not really understanding kirchoff's laws here is the first circuit where it's very difficult to do that because we have a resistor here we have two resistors and basically in parallel which we haven't really talked too much about but it's it's difficult to see exactly what the currents going to do we do know the voltage drop across one of these but this is an unknown resistance so this is sort of a little bit difficult to get to the answer to without using Kirchhoff's current laws so what I recommend is anytime you see a circuit that's a little bit beyond the sort of basic complexity you what you really need to do if you haven't if you're not already given everything in your drawing is redraw the circuit with some labeling right so redraw the circuit again or you could clutter up your original drawing if you want to but that's kind of not a good idea a lot of times and without a test you may not have time so you can just use the original drawing and draw your own labels for some other things for instance we know that this is a source so current is going to come out here and it's going to go this way we know that because we've had a discussion on this when the current gets to this node it's going to go down this way and it's going to go over this way when the current comes back around it's going to rejoin with this one and it's going to go back to the source so we kind of know what the current sort of going to do even if we're not totally right we have an idea so what we need to do is redraw this guy with some labels so that we can then use Kirchhoff's current law and Kirchhoff's voltage law if we try to use Kirchhoff's current law here at this node what are we going to do we don't know this is what we're calling the current here what we're calling the current here what we're calling the current here so a lot of students will look at it'll be like I have no idea what to do well it's kind of like in physics on sometimes in physics when you're given a problem you have to draw your own diagram and then you have to make your own labels for your own variables and then you have to write an equation that have those variables and then eventually you kind of see how you can solve for whatever it is you need to solve for same thing here we're going to need to do some labeling so that we can write our kerkhof current law equations and voltage law equations so that we can have a chance of actually of actually solving it now in order to save a little bit of time I'm going to do some of that right here instead of redrawing the guy I'm going to just draw it here in this particular circuit and the main reason is because this is not a very complicated circuit but certainly if you had a large circuit with lots of labels going on everywhere you want to make sure that you're keeping everything straight from what the problem gave you see the problem gave us this voltage here - what you're adding yourself so we know there's going to be a current coming in here so let's label this kind i sim one you can label it whatever you want you can label the I sub s for I sub source or I source or you can do whatever you want I call it I sub one just makes it easy it's going to come here some of its going to branch down so I'm going to call this I sub two and I'm going to call this I sub three so I've got all the currents labeled everywhere and if you remember I told you most of the time you're going to be writing even though it's a kerkhof voltage law you're going to be writing most of the time in terms of the currents everywhere because we know V is equal to IR so every time we get to a voltage drop we can always write it in terms of IR so usually you want to have the currents labeled everywhere that you think you're going to need to have them in your circuit because usually that's the direction you're going to take most of these problems so we label our currents and since this is kerkhof voltage law section of the course let's go ahead and write a kerkhof voltage law there's two places we can do that we can write a kerkhof voltage law in this this part of the circuit right here and then we can write another one here and then if you really wanted to you could write a larger one we talk about that but you know there's no knees in the reason to go crazy let's just see what we can do with what we have initially so we have two different loops that we can write so let's go ahead and do that so let's say KVL at left loops KVL at left loop and let's write that equation so we're going to start here we're going to go through the source and around so we go up the current and notice we're going in the direction of current flow currents going this way right so once we go through the source we've gone negative to positive so that's a negative stent so it'll be negative 150 right then when we go through our we have a voltage drop across R and you meet and I have it labeled it here but you need to know that when the current flows through resistance like you this it's plus minus so we're traveling through a voltage drop plus to minus so this is going to be a plus sign and you can instead of writing the voltage here you'll write it as I sub one times R because V is equal to IR you're going to be doing that constantly when you write your voltage drops everywhere it's IR we know is labeled we know R is here they're both unknown but that's okay just leave it there you're fine and then you're going to come around to here now normally you would write it as I are here the I sub two times twenty that would be IR but in this particular case this problem gave us the voltage drop so it's going to be plus 100 is equal to zero the reason it's plus is because again we're going in the direction of the voltage drop plus to minus so we have a plus sign here make sure you understand that equation it's negative here because we're going against the convention it's positive here because we have we would have is not drawn here but you know if you go in the direction here it's going to be plus minus this is plus minus so we get plus plus everywhere we have an IR drop here and we're given this drop so there we have a valid we have a valid kerkhof voltage law equation so you look at this and say well can I do anything with this can I solve anything here can I find an answer here I've got a number here a number here I don't know what I want is and I don't know what the value of the resistor is so I have two variables so I can't really use that as a single equation to find the answer so I need to continue writing some sort of equations it's kind of like in physics I mean you draw a picture you've learned what what you're trying to learn in that section so if you're its buoyancy or if it's whatever then you're writing equations with buoyancy so you write some equations you see what variables you have and then eventually you'll say oh I can put these together I can solve for these variables it's the same thing here we don't know exactly what equations we're going to need to write ahead of time you just need to start writing them down and see where you end up so we're done with this loop let's write one over here so let's start down here just below the resistor let's go this way over and then down like this now notice in this case since we're going up we're going from minus to plus because this voltage drop is given to us minus two plus that is a negative negative sense so this is going to be at the right I can spell right we okay so since we're going up like this it's going to be a negative 100 we go this way we go down and then over here the voltage drop here again we're going from the direction of the current flow alright the direction of the current flow so what we're going to have here this is going to be Plus this is going to be minus so we're going in the direction of that so it's going to be plus I sub three times five and there's nothing else because once we do that we get back to where we started there's only two terms here because of that so negative 100 because we're going in the opposite sense positive because we go from positive to negative voltage here and it's IR in this case we use the current that we label that's why it's so important to label stuff in your circuit because we're writing our equations in terms of what we label much like in physics you make your labels write your equations then you have to solve it here we have to make our labels so we look at this and we say well can we do anything with this equation well I have a number here 100 I have five here and I have only one variable here so in this particular case I actually can do something just with this second equation here so let me solve for I sub three so this is the KVL so let me kind of kind of draw a little bracket here kind of a little aside to solve for it just so you don't confuse ourselves so what we're going to have let's move the 100 over so we'll have i-35 is equal to 100 and then we'll divide by 5 so I 3 is equal to 20 after e 20 amps so I'm going to kind of I'm not going to box it I'm just going to have a circle in because it's important it doesn't find our answer the only thing we're asked to find in this problem is the value of this resistor right and this certainly is not the value of a resistor but much like any other problem if you can find anything at all in a problem that's useful then make sure and do it okay so we write this equation we can immediately see that we can solve for I sub 3 so we know what the current is in this particular branch going off this way we know that it's 20 amps so we take stock of our situation we don't know exactly where we're going to go yet but we know that we wrote a KVL here we have two variables so we can't really do much with it we wrote a KVL here and we actually solve something so there is another KVL that you can write in this direction and you can probably get some more information out of that but for right now I want to switch gears and write a kerkhof voltage law at this node here because ultimately we're trying to find the value of this resistor you know I've said this multiple times all of these circuit problems become a game of writing equations until you get enough of them on paper to find what you're trying to solve for so here we have two valid equations I'm able to solve for something I've used kerkhof voltage law for both of the main the main loops that we have here so let me go and switch gears to do a kerkhof current law kerkhof current law at top node at the top node which means this node up here remember current flowing in is negative so since I of 1 is flowing in it is negative so we have negative I sub 1 I sub 2 is flowing out so it's positive I sub 3 is flowing out so it's also positive is equal to 0 ok so that's what I have there uh I have a cut-off current law here but notice I do know that this value is actually a number this value is 20 right but I don't know what I sub 1 I sub 2 R I don't know what this guy is and what this guy's but then I look back at my drawing and I say oh but I do know what this current is because I was given the voltage drop across the resistor I was given the value of the resistance if you know the voltage drop across a resistor and you know the resistance of the resistor then you also know the current because you have Ohm's law which always applies so let me switch gears over here and and use that fact so what basically I'm going to do is I'm going to say V across the 20 ohm resistor is equal to I R so the voltage across that 20 ohm resistor was 100 volts the current we're labeling is I sub 2 and the resistance is 20 so we just divide here and what we get is I sub 2 is equal to 5 amps and again I'm not going to box it I'm just going to circle it because it's useful so here we have a situation where because of the way the thing was drawn we knew the voltage drop we knew the resistance I mean if you had noticed that first then you can calculate I 2 right away and just sort of keep it on the side of your paper in this particular case you could just say I didn't notice that at first I wrote some equations down and then I look back and said oh I know that I can write that down now look where we are we have a kerkhof current law at the top node with these variables involved I know what I 2 is because I just found it I know what I 3 is because I already had it so when I come down here I'll just kind of work under here I guess let me rewrite the Kirchhoff's current law equation so let me rewrite it again the one that we just wrote was negative I 1 plus I 2 plus I 3 is equal to 0 so we have negative I 1 plus I 2 is 5 plus I 3 we already found is 20 is equal to 0 so you have 25 you move it over to the other side so you have negative I 1 is equal to negative 25 so you have I 1 is equal to 25 the unit is amps and again I'm going to circle it because this is not the answer but it's an important result so what we've done based on what we have is now we know the current here we know the current here and we know the current here which is awesome right so then we go back and look at the equations that we've written and see if we have enough information to solve it we've already used this equation we've already used this equation we go back to the other one and we say ok this is the number this is a number now I know what I won is in a number right so now I can find the resistance because it's the only other variable there so let me rewrite that equation that guy that we use that we wrote down is negative 150 plus I 1 R plus 100 is equal to 0 now we know what I want is so we have negative 150 plus 25 R plus 100 is equal to 0 so here when we have add these guys we're going to get negative 50 right so we have negative 50 plus 25 R is equal to zero move the 50 over sort of 25 R is equal to 50 and the value of the resistor R is to the unit ohms because resistance is always in ohms so the value of that resistor is 2 ohms

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Keywords: kirchhoff's current law, kirchhoff's voltage law, kirchhoff's laws, circuit, circuit analysis, current, voltage, resistance, branch current, kirchhoff, kcl circuit analysis, kirchhoff's circuit laws, kvl and kcl circuit analysis, kvl and kcl examples, kvl circuit analysis, kirchoffs current law, kirchhoff's current law problems, kirchhoff's voltage law examples, kirchhoff's voltage law tutorial, electrical engineering kvl, electrical engineering kcl, electrical engineering

Id: Kz6a-tDxiFU

Length: 14min 26sec (866 seconds)

Published: Fri Nov 02 2012