03 - What is Ohm's Law in Circuit Analysis?

hello and welcome to the section of a circuit analysis tutor in this section we're going to cover one of the most important things that you will learn in all of your studies of electric circuits it truly does serve as the foundation for everything that we're going to cover beyond this point and that is what we call Ohm's law right it's fundamentally a relation between current voltage and resistance now we've talked about that already in section one we talked about the concept of the the current midden being that's the electricity that's actually flowing in the circuit that's what's moving we've talked about the voltage being the pushing force that's pushing this current around in the circuit and we've talked about the resistance being whatever is in the way trying to slow you know slow down the flow of this current right Ohm's law is the math behind all the stuff that we talked about in section one so you already know what elms law should state but we're gonna talk about mathematically here and then we're also gonna look at a few really simple circuits to show you how to use it and believe me when I say simple they're going to be very simple circuits but there are a couple of things that I really want to point out for you that you get very comfortable with early on so that as we build the complexity and branch out and make these complicated looking circuits you know you'll have a really good foundational bedrock so I'm gonna point those things out along the way all right so Ohm's law it's probably one of the simplest relations you'll ever see all right so it's extremely simple and what we see is V is equal to I our V is equal to IR all right you can probably guess what a lot of this is is serving to to represent this guy is the voltage V represents voltage right we talked about that before I we are also talked about that this represents current so this is what's flowing in the circuit and then our you might guess is resistance so that's it ladies and gentlemen it is probably the simplest algebraic equation you can come up with V is equal to IR the voltage in a circuit right is equal - exactly the current flowing through some device in the circuit times the resistance of that object so when you think of Ohm's law you really should think about it in terms of it being applied to any specific element in a circuit so envision a circuit you've got a source that's some battery or something pushing electricity out and then there's some other stuff the other stuff could be a little lots and lots of things could be a fan could be a lightbulb whatever but in our study of circuits we're going to begin by talking about resistors resistive networks so think about some resistors over there so the current is going to be going through those resistances so whatever the current is that's flowing through a resistor multiplied times the resistance value itself in ohms is going to tell you what the voltage drop what the voltage is across that resistor right so I think let's talk about it in a little more detail and and you'll see it with some pictures as well now before we get to that point most books will introduce Ohm's law being V is equal to IR now it's a simple algebra equation you can solve for I if you wanted to calculate the current you could just divide both sides by R I would be equal to V over R so a lot of times in books this is honestly the way I like to remember it myself I is equal to V over R it's exactly the same relation there it's not like this is a separate equation from this it's it's all the same thing this is the same this is the relation when you solve for current it means you divide by the resistance like this if you want to solve for the resistance you just divide by the currents would be the over I all right but this particular form is nice to talk about verbally because you can see very easily a few nice things about Ohm's law think about it what we're saying is if you have an object in your circuit virtually gotta remember Ohm's law applies to everything in the circuit all the resistors later on it'll be it'll be a slightly different story but basically it's going to apply two capacitors and inductors and everything else going on at any point in the circuit if you measure them if you get a meter out and you put the little probes onto the circuit and measure the voltage across some some object right then this voltage divided by the resistance of the object of the resistor in this case is going to tell you how many amps current is flowing through that resistor in this relation whatever one you want to look at it applies to every single object every single resistor and everything else in the circuit it applies everywhere all right so it's universally true but when we look at it written in this form we can see a few nice things if the current through an object in a circuit like a resistor truly is equal to B over R then if we increase the voltage just forget about numbers if we just increase the voltage then this numerator gets bigger then that means the current must get larger right if we increase the voltage think about what the voltage is it's the push if I increase the voltage if I turn a knob up and increase the voltage I'm pushing more so to speak on the electrons what this is saying is I'm going to get more current because this numerator gets bigger and it's exactly what we said in section one now what if I do the opposite what if I hold everything else constant I increase the resistance right what's gonna happen I have a circuit with a resistor in there and I slowly increase the value of that resistance and hold everything else constant this denominator is gonna get larger and larger and larger and that's gonna drive the current down because this gets bigger so that means the current goes down so if I increase the resistance we talked about in section one if I make more resistance my current is going to go down right if I increase the voltage I'm pushing more my current goes up so I guess what I'm trying to say as you can see from the math that Ohm's law as we've written it is everything that we talked about in section one it's just mathematically saying the same thing you push more you get more current increase the resistance you get less current it's basically all it says so it's a little bit trickier when you're applying it to complicated circuits but it's basically that's basically it now what we want to do is apply this this knowledge here to a real circuit and again I told you these are gonna be simple circuits because believe me you need to be able to crawl before you walk so here we have a source and let's say this is a 10 volt source and we're gonna have a very simple circuit like this it's just one loop that's all it is this resistor over here is known to be 5 ohms 5 ohms this is 10 volts all right now once we hook this thing up there is going to be a current that goes this way remember current always comes out of the positive terminal and goes around and around around around around around like this it just keeps going on and on and on until the battery dies or if it's plugged into some kind of power supply then it'll just stay going on forever right the question is what is the value of this current what is the value of this current so the way you should think about this there's a lot of different ways to think about it but basically you need to look at Ohm's law here right the current is equal to the voltage divided by the resistance so you need to apply Ohm's law to something in the circuit so we're gonna apply it to this resistor now this is where I need to really spend some time making sure you understand because you have this voltage source connected by a wire and by the way when we use wires and circuit diagrams we presume that those wires are perfect they're not they don't have any resistance they're not really any they're just absolutely perfect wires that carry electricity perfect perfectly so the only resistance in this entire circuit is this it's not in the wire it's not in anywhere else it's only in this this because it's a perfect circuit now in real life the wire really does have a tiny bit of resistance so we have five ohms here because we have connected this battery up directly to the terminals of this resistor then this voltage that's here plus minus is also present across the resistor is also present across the resistor where the positive is the positive side of the voltage is here and negative here so it's just the same sign convention the convention of what you had before all right so let's go through the problem and I'll come back to explaining it so what is this voltage drop across the resistor well it's the same voltage that we have in our source because it's physically connected to the resistor so if we got a meter out and touch the edge of the other each side of this resistor the meter would read 10 volts because that's what's physically it connected to the source right if we put the voltage leads here it would measure 10 volts if we put them here it would measure 10 volts if we put them here with measure 10 volts because it's all connected to the same place so we know that the voltage drop across resistor is 10 volts we know this to be true because 10 volts is connected here so what we do is then we say the current in this circuit or the current through the resistor which the current here I've drawn it here is the same value of the current all the way around it's the same value of the car because it's one circle so the current going through this resistor which is the same as this current in this part here is equal to the voltage across that resistor divided by the resistance itself the voltage across the resistor is 10 volts because it's exactly equal to the source since it's physically connected there and the resistance is 5 ohms so when you're doing these calculations your honor you always want to deal with volts and ohms those are the base units we talked about so I is going to be 10 divided by 5 which is 2 in the unit of current is amperes or amps so there we go you circle that so what this is saying is that in this circuit there are as 2 amps of current flowing through this resistor and of course since it's simply one loop it's two amps of current going all the way around round and around and around and around all the time so you know very very important then you understand that the other thing to make sure you you realize and just just know as we study all these circuits is that every one at every component that you have we see we call this a passive component passive because there's no power supply going on it's just kind of receiving the current this would be an active component it's supplying electricity to everything else or it's called a source really but everything else in the circuit where the currents just flowing is called a passive component these passive components once you energize a circuit or you let the currents start flowing if you put a wire you know the circuit meter they've got a multimeter from the store and put it across that resistance you're going to see a voltage drop because there has to be a mechanism to push the current through through this guy obviously it's coming from the source but there's going to be a voltage drop across every single resistor in your circuit and that voltage drop is going to be governed by Ohm's law because we know there's a current going through it we know there's a resistance here so there has to be a voltage across that resistor and across every resistor in a circuit that Holmes law tells you all right so I just want to make sure you understand that because it's so very important now let's go and do another one here again they're gonna be all pretty simple let's go and do another one and then we'll come back and and talk a little bit more let's say we had slightly different numbers let's again keep this constant at 10 volts it's gonna be a single loop circuit like this and this is going to be now 10 ohms all right now obviously they're still going to be current and going through here like that and it's the same current in every little part of the circuit because it's one loop right so what we know is we have 10 volts same as before but we've increase the resistance so we expect since we've increased the resistance we expect there to be less current right so what we need to do is figure out using Ohm's law what that is the current is going to be equal to V over R right so you have to apply it to something in the circuit let's apply it to this resistor the current through this resistor is going to equal the voltage drop across this resistor divided by its resistance what is the voltage drop here well it's not labeled in your diagram but since you've connected this source directly to the terminals the voltage drop across if you were to measure it with a meter would be 10 volts divided by its resistance of 10 ohms so when you do all of this the current is equal to 1 ampere 1 amps so if you were to measure you put a meter in there and measure how much currents actually running around the circuit notice that the current now is half as much as the current was in the first case the reason it's 1/2 is as much current is because we have exactly doubled the resistance it's because Ohm's law is a simple algebra equation if you double the resistance you're gonna half the current I mean there's no squares anywhere there's no square roots it's not really complicated but you do need to understand what you're doing now since they have two problems on the board even though they're basically the same circuit I want to make sure you understand and start to visualize as you look at circuits you know this is so simple you know the current is going here but anytime you see current going through a resistor there's going to be a voltage drop positive to negative across that resist in the direction of the current flow because you need to think of it as almost like the voltage here right that's across the resistor is what pushes the current through this resistance for lack of a better word and the way we do the sign convention and engineering is that whenever you're talking about a passive component like a resistor or almost anything else a capacitor or inductor everything we're gonna talk about is called passive component then the voltage drop is going to be called a voltage drop for a reason because it goes in the direction positive to negative of the current flow positive to negative right when the current comes back around if you notice the current goes this way it goes from negative up to positive that's totally opposite of what it's doing here but that's because these are special this is a source it basically blows now two sources are generating they're they're basically supplying the power to the circuit so anytime you see current going from negative to positive it's kind of unnatural it has to be a source supplying that from some external supply of energy right but everything else in the circuit you're going to see a drop in voltage from positive to negative now I want to take a little bit of an aside and tell you sort of like an analogy here to make sure you understand what I'm talking about here voltage is kind of like potential energy from physics right potential energy if I go up to the top of the mountain I have more potential energy more potential to do work right because I'm higher off the ground right that's what we would sort of think about is a high voltage up there right high voltage means more potential to do something more potential to push electricity right if I take a bowling ball and roll it down the mountain right a certain distance then and then you stop it halfway down the mountain right and then you look and see where you are halfway down the mountain you have less potential energy you have lost a little bit of potential energy because you started at the top of the mountain and you use some of that energy to get halfway down because the ball rolled down right and you stopped again before you hit the ground so you're halfway up this mountain you still have some potential energy but not quite as much as you had to begin with you've lost you could sort of say you've dropped your potential energy you went through a potential energy drop right that's all we're saying here this source is supplying volt it's like giving a lot of potential energy to this resistor and that potential energy is pushing this current and as the current goes through the resistor it's kind of like once it makes its way through this resistance it's dropped its potential energy as it comes around to the other side so I guess I'm trying to explain the sign convention in ways that you can kind of understand that's why it's positive up here and negative don't get too wrapped up in the signs here it just means that this side of the resistor had a higher potential energy for instance once it forced the current down to the bottom of the mountain then over here we have less potential energy than we then we started with now if we had like five resistors in a row like one wired up after another each one of those resistors would take away a little bit of potential energy from the source and by the time we got down to the bottom we'd be back to or be back to the bottom of the mountain again so to speak so think of your batteries or your wall suckers as supplying energy to the circuit almost like the top of a mountain as you walk around the circuit you're losing voltage so to speak so you're doing a voltage drop from here to here as you cross this resistor and as you get back around to the source now I'm gonna have a few more examples to kind of illustrate this a little bit more but I wanted to make sure you kind of understand that bottom line if you don't want to remember anything else if current is flowing through a resistor you're going to have a voltage drop positive to negative always always always that's the voltage we're using an Ohm's law all right so let's go and do another one I'm this is gonna be the same thing I'm just gonna change the numbers again because I want to give you a lot of experience so if it's ten volts again all right and make it 20 ohms okay 20 ohms and again I want to know what I is here now notice I'm drawing the current I in different places all over the place I'm trying to kind of drill that in that the current is the same everywhere so what we need to do to find this current is I is equal to the / R of course that's exactly the same thing as V is equal to IR you can use this guy you're basically rearranging the equation either way it's the same thing so if you want to calculate this current we need to know the voltage across this resistor what is the voltage across that resistor well it's physically connected to the entire voltage source so the voltage drop across this resistor is 20 is 10 volts and this resistance is 20 ohms so we do this division it's 1/2 which is 0.5 and the unit of current is always amperes 0.5 amps half an amp all right so I guess I wanted to show you the progression if you have the same source in all the cases you start with 5 ohms you have two amps of current increase the resistance by doubling it to 10 ohms you drop your current to 1 amp alright if you increase your resistance again to 20 ohms you get even less current now we're at half an amp all right so that kind of just illustrates what we talked about in section 1 if you keep jacking up the resistance your currents going to go down down down all right now let's do a little different the circuit will look the same the numbers will be a little bit different let's change this to 1 volts this is now a 1 volt source and now instead of a resistance that we've been talking about before this will be 50 ohms 50 M so this is quite different see what we've done now is we've decreased the voltage which means we've decreased the push right and at the same time we've also increased the resistance so both of those things combined together should dramatically decrease our current is now we have less of a push and we have more resistance in the way more resistance getting in our way so again V is equal to IR we can use this version if we want the voltage we have to talk about something let's talk about this resistor here notice that the current is going to come this way because it's going around and around any time we go through a resistor there's going to be a voltage drop this voltage drop is the same as the source because it's connected like this that voltage is 1 volt so we have to talk about a component when we comply when we apply Ohm's law like this is equal to I we don't know what I is but the resistance we know is 50 ohms so we have volts and ohms so the units are fine so the current is 1 divided by 50 so the current 1/50 is 0.02 amps now it's perfectly fine notice that that we have a very very tiny current now because we've decreased our voltage or source and at the same time we've increased our resistance quite a bit so both of those things together drive our current down just like we talked about now notice that we have a very small current much less than 1 amp so now it might make sense to rewrite this in terms of milliamps and that's exactly why I'm presenting this to you here if you think about it for a second this is going to be equal to 20 milliamps because merely means you shift the decimal point three places so if you take 20 it's the same thing as like 20 times 10 to the minus 3 because of milli milli ohms milli amps is 10 to the minus 3 so if your decimal is here and you move it 3 places to the left it'll be 1 2 3 and you get your zero there which is exactly here another way to think of it is if you're going to go from here to milliamps you need to shift the decimal three places 1 2 3 so it's gonna be 20 milliamps if you get confused on how to go quickly from milliamps it's back to amps or kilo ohms back to ohms or a thick metric conversion stuff for you go get the unit conversion tutor because the unit conversion tutor that I have just just gives you lots of examples we're dealing with the metric system but this is just moving the decimal point by 3 spots all right let's say you have a problem where your source is not a nice number of volts that you might think about maybe it's instead of 50 volts maybe it's 50 milli volts the MV means milli volts one one thousandth of a volt in other words and let's say again that's connected to just a single resistor this resistor let's say we don't even know what it is it's just resistance R we don't even know what it is and let's say that the current flowing through this circuit is denoted by AI and that's 10 milliamps right so here I'm giving you a problem where we know the voltage and you know in the voltage right we know the current but instead of volts and amps we're giving millivolts and milliamps right so what you need to be careful about is whenever you use Ohm's law you always have to use volts and amps I guess once you get some experience so you can start mixing units and you kind of know it should come out the other end but in the beginning I always recommend for you to just convert everything to volts and amps and the easiest way to do that is if you want to know what this resistance here just use Ohm's law V is equal to I R we have to apply it to this resistor now we know that there's going to be a voltage job through this resistor because the current is coming this direction you always need to get in the habit of seeing that when you see the current going through something plus minus that should just you should just think of that if you put a meter here that's what you would measure this voltage drop is what's pushing the current through of course that voltage drop is also coming from the source that's the original genesis of the whole thing but whenever you use this guy the voltage here is 50 millivolts now you don't want to put 50 in here you want to put 50 times 10 to the minus 3 because that's what millivolts is you take the number and you multiply by 10 to the minus 3 equals the current is 10 milliamps you don't want to put 10 you want to put 10 times 10 to the minus 3 and this will give you all the credit on your exams 2 if you make a calculation almost 8 times R so it's the same thing to voltage across the resistor the current through the resistor the value of the resistor is unknown now if you grab the calculator and you take 50 times 10 to the minus 3 and you divide it by 10 times 10 to the minus 3 just moving this over here basically giving you the resistance the resistance is going to basically be equal to 5 ohms and that's what's gonna spit out there 5 ohms so make sure I guess what I'm trying to do by showing you this problem is make sure that when you're presented with milliamps or micro amps and kilo ohms or whatever when you're doing your calculations in Ohm's law don't forget that you need to always work with volts ohms and ants all right so you need to work with those base units if you accidentally put millivolts in here and then put instead of milliamps maybe you accidentally put 10 here instead of 10 times 7 minus 3 and you mixed it with something different over here that didn't match then you're gonna get into trouble so what we've done is implicitly converted back to 2 volts and amps and so on to be able to do the calculation all right so so far we've done a single loop right with a single resistor and most important thing I wanted to drill India is how to use Ohm's law how to deal with the unit's right and also to visualize those voltage drops that go across those components when the current flows through them because it's so important we're gonna be doing methods later on when we do analysis where you're gonna have to use that knowledge over and over and over again so I'm really trying to drill it into you now what I want to do is erase the board and show you a couple of circuits with more than one resistor not going to be too complicated but it'll allow us to visualize a few things that the simpler circuits you know we need to be able to build beyond that so we're gonna do that right now all right now our next circuits gonna be quite a bit different from what we've talked about before but it illustrates something very very important so let's say we have 12 volt source and this source is connected to a resistor here and also another resistor in series with it we call this in series when something is just one after another like that all right now this resistor here we know because we bought these resistors that this is a 1 ohm resistor and this resistor is a 2 ohm resistor so we have one ohm in series with 2 ohms and we also know is because we happen to measure let's say that the current going through the circuit is 4 amps now don't forget what this means this means that 4 amps is going all the way around both through both of these resistors back up to the source and around and around again if we had a branch here if we had another part of the circuit that came off like this then this current would end up splitting here and then coming back and rejoining here and going back to the source so anytime you see circuits with lots of branches right you can envision that current splitting off and then circulating around and coming back together much like a water hose might do in a sprinkler system water split off go back and do its thing come back together and go back you know if you have a circuit going a circulation pump it'd be coming back to the pump right same thing here but this is much simpler because we only have one basically one branch there it's just that in this branch we have two resistances there so question number one right and let's just call this just to make it you understand let's call this r1 and r2 resistance wanted resistance to now let's say what is resistance 1 what is resistance 1 right what is the specifically what is the voltage across resistance 1 so we might have something like this drawn on our paper we might have little holes like this plus minus v1 you might see this drawn like something like this all this means is here is the circuit right the certain electricity is going like this we stick a probe in and we measure the voltage only across resistance 1 what is that voltage right and we call it v1 because it matches up with r1 how do we calculate that voltage well anytime you have to calculate voltage you're always going to use Ohm's law right so we've used V is equal to IR and we're going to apply it to this resistance because that's the one we care about that's the one that this problems dealing with so what is the voltage across this guy we call it v1 let's say we don't know what it is so we just leave it there equals what is the current flowing through this resistor the current here is 4 amps it's the same four amps that keeps going around over and over again so we call it four right the resistance is known to be one own we just put a 1 here so v1 is equal to 1 times 4 which is 4 volts right so in other words this is more volts right there right is equal to v1 so if you put a probe there a voltmeter and you measure it you're going to see 4 volts across v1 notice when we applied Ohm's law we didn't even care about what r2 was doing it didn't matter if we know what the resistance of this r1 is and we know what the current is going through that resistance all we need to know is what the value of the resistor is and what the current is going through it to calculate the voltage what voltage are we talking about into the voltage across the element that we're talking about so you see a lot of times before you take circuit Theory people toss around voltage and current like they're just these broad terms right but when you're analyzing a circuit when you talk about a voltage you're not just talking about a voltage here or here you're always talking about the voltage robbed across some circuit element that's what you mean when you say voltage when we say this is a 12 volt battery we're saying that this is a voltage difference between the terminals of these two battery the ends of the battery let's say or if it's a wall socket it's the voltage across the two the two ends that's how you develop that push it's always between two points so when we talk about the voltage here right it's labeled across r1 that's the voltage drop coming from if we had that well actually put plus minus here hi dropping down just a little bit or as we drop across kind of coming down the mountain so to speak now you might guess the next thing I want to ask is what is the voltage drop across r2 right so it would be v2 is equal to I armed - right so the voltage drop across this guy is going to be equal to I times r2 so if we were going to just to make sure it's absolutely clear 12 volts resistor here resistor here this is r1 which we already labeled as one ohm this is 2 ohms and we know the current is 4 amps that's going through the circuit if we're talking about this guy if you you might see it drawn in a book like this the little circles here indicate that it's not really part of the circuit it's just that you're measuring the voltage across this element right and it's gonna be plus minus because the current is going and dropping across through that resistor so we always have a voltage drop like that across these passive elements v2 which is what this is voltage across resistor 2 is equal to the current times R to the equals IR so v2 is equal to the current flowing through this element is the only thing we care about and that current is 4 amps because it's the same guys I was fooling around everywhere else times 2 because we have the resistance of two homes here so V 2 the voltage across resistor 2 is 8 volts right so what you'd have if you actually measure this guy is you'd have eight volts across this element alright so what I really wanted to do by illustrating this there's lots of ways to analyze a circuit I could have done lots of things I could have showed you how to add these resistors together and then do something slightly different what I'm trying to show you is that the only thing that matters is the current flowing through that particular resistor and the value of that resistance in order to calculate the voltage or likewise if another circuit gave you the value of this voltage drop across this resistor and you know the value of the resistance then you have enough information to calculate the current all you care about with V is equal to IR is the voltage across the element the current through the element and the resistance of the element and that's what it applies to so when you apply Ohm's law around your circuit you just need to be able to zero in on whatever it is you're talking about now before we close the section I want to sum this some of this up because there's actually a nice a lesson in here that I really want you to be aware of so let me change colors a little bit just to make it pop off the board I'm gonna redraw the circuit one last time this would be the last time 12 volts so we have a resistor here and we had a resistor in series like this and basically I'm not gonna label everything again but I am gonna put that this is 1 ohm and that this is 2 oh and that this is 4 M just that was flowing through the circuit and basically what we figured out is that if we put a probe across this resistor like this what we figured out that the voltage drop would be appear 4 volts so this is 4 volts here all right and we also figured out that if we measure the voltage drop across the second one so we can have this like a common little terminal here and then the voltage drop from here to here would be again plus minus and the value would be 8 volts so literally if you went and got a 12 volt battery from the store it hooked a 1 ohm resistor in series with a 2 ohm resistor and then you got a voltmeter and you measured across here you would measure 4 volts and if you got a voltmeter and measure it across here you would see 8 volts in accordance with our calculations but what do you notice 4 + 8 is equal to what 12 12 is the total voltage that we started with this is a very important lesson that I'm really trying to point out to you here that you know a lot of times you don't really see it until you're working of circuit problems what we have here is a situation where you see remember back to the other problems it's good a simpler problem briefly I kept telling you over and over again well we had 10 volts and we had 5 ohms right and this voltage was physically connected across the resistor so the voltage drop across this resistor was the entire source voltage 10 volts right but that's slightly different from what we have here here we have two resistances the entire source voltage of 12 volts is kind of connected across both of these guys so basically the 12 volts is spread across both of these elements and that's why both of these voltages actually add up to 12 volts in fact they must add up to 12 volts if they don't add up to the source voltage then you have some kind of like a crazy invention that you can go make lots of money on it has to it's conservation of energy thinking about the mountain analogy right we climb up this mountain to the top right energy-wise ok and we're at the top so it represents 12 volts let's say now we come down a little bit down the mountain through this guy and we bleed off a little bit of the energy right so we bleed off 4 volts worth down here but there's still lots and lots of mountain left so once we get all the way down to the bottom we bleed off the remaining 8 volts when we get down to the bottom there's nothing else left right we're at the bottom of the mountain there's really no more voltage drop that you could possibly do because you're at the bottom of the mountain once you get down back around to the source to the source here you've already used up your 12 volts it's all been used here so to speak to push the current so there's really your back down to where you started with your back down to the base of the mountain that's just as an analogy it's not perfect but it tries to get the point across that whenever you have these things in series like this you should expect to see a voltage drop here you should expect to see another voltage drop here but the sum of the voltages across everything here should always be 12 volts if we put a hundred and 50 resistors in series and we measured the voltage across all of them they should all add up to 12 volts of course in that case each little tiny voltage that we measure is gonna be real small right so that you'll get different values as you can see this happens here as well the voltage that you get across the elements is going to depend on the elements that we have here you can also kind of see that you have two ohms here in one ohm here so because this is two ohms most of the voltage drop is going to occur over the two ohm resistor a little bit less of the voltage drop is going to happen over the one ohm resistor so it's kind of like proportionally this kind of splits up the voltage so to speak this circuits actually something we'll talk a lot more about later it's called a voltage divider and it's very practical if you have a 12 volt source let's say and one part of your circuit way over here maybe doesn't need 12 volts maybe this part of the circuit over here maybe only needs 4 volts maybe if you put more than 4 volts on this part of the circuit way over here it'll blow it up right maybe it's a delicate chip or something like that so you could and if all you had was a 12 volt battery then you could construct a voltage divider like this - basically you would have 4 volt here you're basically taking the source and forcing it to be split up so that 4 volts lies across this guy 8 volts lies across this guy so I could take my delicate circuit over here and build it and wire it up right here to the two terminals here the 4 volts would then be supplying my little delicate circuit the rest of the energy would be going through over here so a voltage divider something very practical if you're building something and you need to control voltage right at least from a big-picture point of view it's not something you do every day and really a delicate equipment where you're trying to get really precise things but it certainly works in a pinch if you have a few components laying around and you need to convert voltage it's a very easy way to do it alright that is basically what I wanted to talk about in this section it's one topic and one topic only and it's so important that it deserves its own section we call it Ohm's law basically it governs what we've talked about in words that if you have the push which is the voltage and your resistance which is in the way so to speak then if you know those two values you can calculate the current flowing in a circuit Ohm's law applies to the whole circuit but it also applies to the individual elements that the power of the electricity is flowing through in a circuit so basically what we do is we need to look at our elements and say alright I is equal to V over R what's the voltage across this in this case it was the source voltage so he put it in there what is the resistance we put it in there and we get the current as we change the value of the resistance or as we change the value of the source voltage the current flowing in the circuit is going to be different each time and that's just exactly what we talked about in the section 1 qualitatively speaking there and then finally we did this problem which is very very telling and very helpful for you later on because what it's basically trying to show you is that look these circuits don't have to be quite so simple if you just know you know if you just know the resistance and the current going through something you can calculate the voltage across an element you can calculate the voltage across another element and very very important that you can add these two voltages up and that's going to always equal the source voltage there are a lot of things in here that I could talk a lot more about but in truth we're going to get to all of them even in more detail as we go in the next few sections so stick with me I promise you that if it's not clear now it'll definitely be clear as you work more problems and what I encourage you to do is work all of these problems again make sure you understand the concepts these circuits are not terribly hard but what you'll find is that if you blow through this and you don't quite understand what I'm talking about and you just keep going then what's gonna happen is once the complexity gets a little bit out of your bounds a little bit you're not going to have the foundation to be able to handle it but just knowing that hey there's gonna be a voltage drop across here it's gonna be plus minus because the currents going this way hey there's gonna be another voltage drop it's gonna be a different voltage drop but the sum of these two voltages has to be equal to the source that kind of stuff you need to internalize because as we do our node voltage methods and our mesh current methods later on for more complicated circuits we're going to be walking around different loops of the circuit and trying to figure out what the current is doing so this is the bedrock for that so make sure you understand it move on to the next section and work with me and I believe that you'll be very good at circuit analysis and feel very comfortable with these methods

Channel: Math and Science

Views: 1,168,260

Keywords: ohm's law, what is ohm's law, circuit analysis, electrical engineering, ohm's law explained, ohm's law for dummies, ohm's law definition, circuit analysis tutorial, circuit analysis lectures, ohm, law, v=ir, i=v/r, resistance, voltage, current, kirchhoff voltage law, kirchhoff current law, ee, engineering, electronics, electrical engineering 101, electrical engineering tutorial, what is ohm's law explain, how to use ohm's law, ohm's law circuits, electric circuit

Id: lf0lMDZVwTI

Length: 39min 40sec (2380 seconds)

Published: Thu May 03 2018