KVL KCL Ohm's Law Circuit Practice Problem - (Electrical Engineering Fundamental and Basics Review)

hey gang so I thought I do of gated circuit and I've done in the past where we get to use all three of those principles kerkoff's voltage law Kirchhoff's current law and Ohm's law so here is our circuit that we're going to try to analyze and the question I'd like you to try to solve is the following find all repeat all voltages and currents ah can see the currents anymore but says currents currents and the circuit so we have four different elements and I gave you that guy 30 volts it's a power supply and the voltage is already labeled for you so we need to find you on which the voltage across there B 2 the voltage across that resistor and V 3 and the voltage across that resistor but I also want you to find currents so let's go ahead and identify the currents I want you to find so we have a current going through here that goes through that resistor okay let's call that I one tiny and then with the current going down this resistor let's call that I to right it's coming down through here and then there's a current going out of that node or it could be going in but I'm going to go ahead and point it this way three okay so those are three currents you have a current here and then you have the two currents here right this one in that one so three currents to worry about in our system and again the polarities as indicated for the voltages are kind of arbitrary and if we solve for these voltages and turns out we've a negative voltage all that means is that we need reverse how we drew the polarity here same thing with the currents I drew the current going into this node that way and then out of it that way and out of it this way but then if we solve for eyes if they turn out to be negative then we just have to reverse the way we drew them so where do you start give all these variables right 6:3 voltages recurrence so how do you start to figure out what to do the brute force approach to this which I'll go through is to write down all the relationships you can between all the variables let's start off with something simple for instance we know use the black pen this gives we know we can apply Kirchhoff's current law in here you look at that node well the way I drew it there's only one current going into that node by one so the current going in has to equal the currents going out and based off the way I drew it currents going out or i2 and i3 again don't worry if we drew them wrong we determine if we label the the direction the currents wrong we'll figure it out by having their negative currents if we are wrong so that's one relationship where what else do we know here well let's apply Kirchhoff's voltage law do you see how many loops we have to deal with here two loops and I'm going to define the loops going in that direction so I salute one and then this guy look - just so you can follow along okay so let's add up all the voltages around loop one and remember they have to equal to zero so loop one if we go around this way we have eight volts and then going on this way we have up my equals v1 because that's unknown v1 and then going down this way we have v2 and then going up this way the polarities different now remember positive - positive - and then - - positive so that means you have to subtract that 30 volts and we know that equals to zero now well the second loop let's take a look at loop as we go around this way a positive - so v3 positive no not positive as we label that - the positives so we're gonna have to - v2 equals to zero okay well do we have enough to to solve for all of our right now we only have three questions and we have like six variables here so what else do we know what other whatever their equations can we write or relationships can be right about well luckily hopefully you know Ohm's law and based off those law we can develop a relationship or write a relationship between current voltage and resistance here so v1 equals the current going through here which we defined as i1 times that resistance which is 8 and then we know v2 which is that voltage are down there equals I which we defined as i2 times this resistance which is 3 ohms and then we have one more relationship we can write about that maybe you cannot see there you go v3 equals I and we defined this I that's going through this resistor as a three way up there earlier I three that's the current going through here so I three times this resistance of six right so I three times six equals that voltage drop across there ah well that's better okay because now we have one two three four five six equations and six variables that we can mathematically solve now where do you start how do you start solving stuff well we want to eventually get to one equation with one variable so we have to slowly work our way through this which means I might need to have extra paper and you but there are some relief some relationships you can start with for instance this guy hey v3 minus v2 equals 0 well that kind of means v3 equals v2 well that's good to know so we have that handy well what else here well if v3 equals v2 we can use Ohm's law in here maybe to get into current so v3 is I 3 times 6 equals v2 which is I 2 times 3 and we get saw one of these guys we can solve for let's solve for I 2 ok so I 2 equals and the 6 divided by 3 so 2 3 it was kind of weird outage the parentheses but that's ok you know what I mean right I 2 equals 2 I 3 well that's another relationship we can use well why don't we just start from there so we've we've solved for I 2 in terms of by 3 so how about we try to come up with an equation one equation where all we have is I 3 so what we do to get everything in terms of I 3 I love it up here I 1 is in terms of I 3 but we know I 2 equals two x i3 so how about I 1 equals 2 I 3 because i2 equals two x i3 plus i3 huh very cool now we have I 1 equals 3 times 3 well that's useful to know so we know I want equals 3 times 3 we know I 2 equals two x i3 and that's actually enough to get us to get us someplace where do you want to go from here but let's use this equation let's use that equation this might actually be a long about way of doing it but there's no really there's no one right approach there's lots of ways to solve all of these equations so let's take a look at that dude so view on we can replace I'll put that down here down there v1 we can replace with I 1 times 8 v2 we can replace with I 2 times 3 minus 30 equals 0 and we know we can replace both of these guys with I 3s yeah I 1 equals 3 times i 3 so that's 3 times 3 times 8 plus I 2 which we know right here is 2 times I 3 so 2 times 3 times 3 minus 30 equals 0 ok one equation one variable we're golden oops you can't see that ah there you go all right 3 times I 3 because that is equivalent to I 1 based off that equation 3 times a 3 and then I 2 times 3 well I 2 we know is 2 times I 3 to 2 times R 3 so now we have one equation with just one variable let's rewrite this so 24 hour 3 plus 2 times 3 6 3 minus 30 equals 0 ok combine these guys that equals 30 I 333 and then we can move the 30 the other side here follow the arrows so 33 equals 30 well that means we take 30 divide both sides by 3 equals as I move paper for you equals 30 divided by 30 which is 1 BAM 1amp all that for one okay so now we know i3 is one F now everything's gonna be simple yeah with my thumbs thumbs up simple okay so if I three equals one amp then we know I one has to be let's move it down for you a little bit so where I just left off before I was rudely interrupted by the fact that I'm running out of space on the paper is we saw four I three we found out I finical is one amp and now we can use that to solve for other variables so for instance up here using our KCl equation we notice that I want equals three x i3 so let's move to my paper so I want equals three x i3 and we know I 3 is just 1 amp way down there so let's go ahead and put one amp here so that means three times one so I want equals three amps okay so that's good we have I 1 we know I 3 equals 1/2 so how about I - well we have relate to bright to knowing right here we said that the i2 equals to x i3 okay so 2 x i3 huh - x i3 and i3 is 1 so that just gives us 2 amps so now we have all three currents well that was simple we have I 1 equals i3 we have i3 equals 1 amp then we have I said that funny we have I 1 equals 3 amps we have I 3 equals 1 amp then we have I 2 equals 2 amps fantastic and now we can use Ohm's law to solve for our voltages or can you see right there oh so v1 well we know v1 equals I 1 times 8 so i1 we know is 3 amps times age right 3 amps I just saw fort that means the first voltage is 24 that's kind of high I - I mean I - me too using Ohm's law right there is i2 which we just solved for as being 2 amps so 2 amps times 3 that is 6 for some units in here 6 bolts huh so v2 guess you didn't see that equals I which is 2 amps right there times the resistance which is three homes equals 6 volts and then the last one is for V 3 V 3 equals I 3 times 6 so V 3 equals I 3 which is 1 amp times 6 pounds because that's what we had in the original circuit for the resistor right there this equals 6 volts so now we have all the voltages and and a messy way that ready we have all the currents find three cools 3 amps a high-quality closed three amps are three equals one amp that idea goes to us voila it just took a little longer than I was hoping with more paper but hopefully that helps a little bit

Channel: EE Review Videos

Views: 884,525

Keywords: Kirchhoff's Circuit Laws (Field Of Study), Ohm (Unit Of Resistance), Electrical Network (Invention), Ohm's Law, Physics (Field Of Study), Electronic, circuit analysis, electrical current, voltage in a circuit

Id: YIsGDYVM6RA

Length: 14min 52sec (892 seconds)

Published: Sat Jun 06 2015