27. Resistors & DC Circuits

Kirchhoff's Loop Rule

# Find Two Voltages (3 sources)

Patrick Ford

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Hey, guys. So let's check out this example. So here we have a circuit with three batteries and three resistors and we wanna find two of the voltages. One voltage is given to us and then were asked to find V one. V two were also given two of the three currents. There are three branches, one branch to branch, three branches. So they're gonna be three cards and two of them already given to us. Now, remember, the general steps to solve these problems are that first we're going toe label a lot of stuff and then to we're going to write as many equations as we need to find everything we're looking for. Okay, so what do we lately we're labeling. Junctions were labeling, loops were labeling currents. And we're also putting at the end pluses and minuses on the resistors and the batteries, which is the set up to be able to write these equations. Alright, So junctions are easy. It's just points where the wire split. So let's call this A and B. You might notice that this for a right here. If I keep extending it, it's gonna go this way for a and then this is gonna go this way for a a right and right away because of loop because of junction rule, you might already be able to see here that if I have a five coming in and a four coming in, there has to be a nine coming out this way because current and equals current out. So nine in means there's going to be nine out, and that's this current. So now I actually already know all three currents, which is good. Alright, so let's label our we got the junctions out of the way. We actually already calculated on the currents, which is gonna be helpful. What about loops? So the currents on this loop is going this way. Five amps, five amps, right? I got all the currents labels, so I like to loop. Remember the direction of Lupus? Arbitrary. But I like to write loops that follow the current just because everything is going the same way. It's just easier in the brain, right? So we're gonna do a loop this way, which, if you notice it will follow the five amps all the way around, and then it's actually gonna follow the nine teams as well, not necessary. It's just a little nicer, but don't sweat it too much. You could just make a loop in any direction. Okay? Same thing here. I'm gonna start this loop right here at point B, and I'm also gonna spend this way which will neatly go along the direction of four. And the nine doesn't have Thio. Um, I just like doing that. Okay, cool. So those are the loops, and the currents are actually already all labeled. So we got this, this and this, and I'm gonna put pluses and minuses everywhere. The plus on a battery will always be on the long stick right here. So plus plus minus minus. And the plus on the resistor will be where the current answer. So the current entering from this side, So that's a plus. So the other is a minus. This is a plus because of the my name's right here. This is a minus. And then this is a plus because of the four amps entering that way, and this is a minus. So we're done setting this up really important that you get you get this right so you can write the equation properly now Let's write equations. Remember, there are junction equations and there are loop equation. So the junction equations will help you figure out the currents and the loop equations will help you figure out anything. But typically, voltages or currents right will help you figure out either one of those. We already know all the currents so we don't have to write a junction equation at all. In fact, when we said that five and four give you a nine were essentially writing that in our heads. Okay, so let's just right loop equations. So there are two loops, so we can write two equations. So for loop One, I can write that the sum of all voltages equals a bunch of stuff which equals zero. And what we gotta do is fill in this bunch of stuff. Loop one. If you go around loop one. This way, there are four elements that you're gonna hit up right to resistors and two batteries. So you should expect that they're gonna be four terms in this thing here. So the first thing we're gonna do is we're gonna go through the 18 volts. So we're walking over here. Let's make this a different color. We're gonna make it green. We're walking here and then you're gonna jump from a positive to a negative. So remember, that means we're gonna add 18 volts over here. You're jumping from a negative from positive to a negative. Sorry. Negative to positive. So that's positive. 18 and then positive to negative. So that's going to be negative in the voltage of the resistor, Remember, the voltage of a resistor is given by owns law. The equals IR. So I'm gonna right, I'm gonna make too little spaces here. This is for my eye. This is for my are the eye here is five and the R is eight. Okay, I'm gonna make a little bit more space here, just in case. Now we're gonna keep walking down this path here. Mom turned over here. We're gonna jump from a negative to a positive and this is going to be positive. I don't know that voltage. That's what I'm looking for. Positive you want, and I'm gonna keep walking jump from positive to negative. So negative. This is a resistor. Negative. I r the current is nine and the resistance is four. So you get this cool so let's clean this up. 18 minus 40. Plus, the one minus 36 equals zero. If you notice good news, there's only 11 variable here. Only 11 unknown. So you can solve for this. And if you move some stuff around, I got it. Here. You're gonna get You're gonna get 58. The first voltage is gonna be 58 volts. Quote. We already got V one. Now we just gotta get V two, and we're also done with equation for the first loop. So now it's gonna write an equation for the second loop, and hopefully the V two will come out of there. So the sum of all voltages in the second loop is gonna be a bunch of stuff equals zero second loop is also gonna run into two batteries in two resistors. So we expect four things to show up in this equation and let's keep going. So we're gonna jump here. Whoops. We're on color. Were running around with green. I forgot to put a plus and a minus here, But, you know, the plus is the big one. Um, we're gonna jump from positive from negative to positive. So this is gonna be positive. V too, by the way, that's the very bottom looking for right there. Um, I'm gonna keep going here, jump from positive to negative. So it's gonna be negative. This is a resistor. So I'm gonna put negative I r. I s four or a six, and then you're gonna keep walking here. You're gonna get here. You're gonna jump from here to here from negative to positive. So it's gonna be positive V one Positive V one, by the way, if you want is 58 so we could plug that in a swell, and we're gonna keep walking, and then we're gonna cross over, and we're gonna get negative. I are where I is the nine and our is the four. Okay, so if we write this out, you get V Tu minus 24 plus 58 minus 36. And if you move everything around carefully, you get that V two is just too Volz. Okay, so that's it for this one. We got the one. We got V two along the way. We also got all the current, so we actually know everything we need to know about this resistor. Uh, about this circuit. Rather. Alright, that's it for this one. Let's get going

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