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Solving Circuits with Multiple Sources

Patrick Ford
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Hey, guys. So in this data, we're gonna put together everything we know about Kirk AUVs rules to solve a full problem. Let's check it out. All right, So we're gonna combine Kurt Calls or Kirchoff Junction rule, which is a simpler one. It's just that the current in equals current out. We're gonna also use the loop rule, which is the more complicated one where we have to write the loop equations and we're gonna use them together to solve more complicated circus circuits with multiple resistors. Here are the three steps we're gonna follow now. We've used lots of these parts individually. Now we're just putting everything together. First step, we're gonna label directions. We're gonna first label junctions. Remember, Junction is a split on the wire. So this point here is a juncture. I'm gonna call that a And this point down here is a junction gonna call that points be okay. So label the junction easy. We're gonna label loops and remember that loops are arbitrary. This is just the sequence in which you're going to walk through the resist through the, uh, through these, uh, circuit and you can pick the direction. So here's a loop, and we're just gonna go this way. We're just gonna go this way. I'm gonna call this loop one, and we're gonna go this way. I'm gonna call this loop to okay, So this step is done. This step is done. Um, now we want to label the direction of currents. And remember, currents will be the assumed direction. You can try to figure out which one might make most sense, or you could just randomly guessing. So here, notice that this battery has the positive here. Which means current typically would flow out of here if this was a single battery. Um, circuit, but you can't really know that for sure. But either way, I'm gonna just use that to sort of dictate that. I'm gonna assume that the current is gonna 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 gonna have three different currents. Okay, let me clean this up a little bit, so I'm gonna call that I won. Um, there's another battery here, and this is a positive terminal here, So you could think that. Okay, the currents probably going this way. So I'm going to draw it right here. I to this better branch has no batter's just a resistor. Remember, The Junction rule says that current n 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 any somewhat consistent way. All right, so I three we're going to assume that it's going that way. So we got our directions of currents 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, negative, positive and negative. But the resistor will be where the currents 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 on then. This resist right here. If you sort of backtrack, I too you can see that I, too, goes like this and then it goes like this, right? So I choose entering right here. So this is the positive of the resistor, and this is the negative of the resistor. You wanna be very careful setting this up because you could set this up wrong. You're going to get the wrong answer. Cool. So don't screw this part up, all right? Eh? So we're done here. We put our little pluses and minuses everywhere. Now we're gonna rights and equations. First, we're gonna write Junction equations. We're gonna write one for each junction. I have two junctions, two junctions. Therefore, I'm gonna have to Equations we're gonna do is first because they're easier. And then later we're gonna write loop equations for each loop. I have two loops. Therefore, I'm going to have to equations. Okay, but let's focus on the junction first. So for Junction A, I'm simply going to write that currents in equals currents out. Now look at it right here. I one goes in and I two goes in. So in is I one plus I to and then out. Is I three coming out. Okay, so that's it for Junction. Be same thing. I'm gonna write in equals out. So look at be here. I three is going. Um let me make this green. I three is going here because it just keeps going through I three we had already drawn. I two is coming out of B, and if you backtrack, I want if you sort of go backwards here, right, I one is obviously gonna look like this and like this so that it's just flowing. Now, if you look at B, I three is going in and the other two guys were going out, So it's gonna be I three equals out, which is I one plus I to and we're done with that equation. You might notice that these two equations are actually identical. They're both saying I three equals the other two. It's the same. Which means you effectively don't really have to equations. You just have one equations. I'm gonna cross that out so we don't get 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 gonna write these two loop equations, so let's start with loop one. Over here. Loop one, and then we're gonna right, Is that the some of the voltages equals zero. Okay, equals a bunch of stuff, which then equals zero. And this is what we have to fill in. Now, Loop one. Is this guy right here? Okay, I'm gonna pick a starting point, like in the little corner here, Actually, 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 this battery, you have to cross that you're crossing from negative to positive. So your landing on the positive So this is gonna be positive. Nine volts. And I'm just gonna write a nine, because everything here is a voltage survey things in votes. So positive nine. And then you keep walking, keep walking. No one here and then the loop is just this little square here. So we're actually gonna turn here, and then we're gonna jump over, cross over this guy. We're gonna go from positive to a negative. So this is going to be a negative voltage now. Voltage of a resistor is given by OEMs law. Voltage equals I r. Okay, so this is gonna be the current through this resistor, which is I three times the resistance, which is 15. Okay. And that's it. There's nothing else. You keep going around the loop, you're done. So you just get back to the original point. So now we're done. I'm gonna move I three that way. So I get nine equals I three times 15 or I three equals nine, divided by 15 which is a 150.6. And actually, just to save time, gonna save space. I'm gonna put it over here. 9/15 is 0. amps. Cool. So we got that first one down and this is I three. 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 currents through each one of the three branches. One man down. So let's get the second one. We're gonna write loop to here, and hopefully we can get a current out of this as well. Some of our voltages equals a bunch of stuff which equals to zero. Let's look at Luke. Two were looping this way I'm going to start over here and we're gonna go like that. Okay? I don't want you to confuse that with the current. This is the loop direction. So the first element that we're gonna hit up while they were walking here is we're going to jump over this, this resistor, and we're going from a negative to a from a negative to a positive. So this is gonna be positive. I are the voltage of resistance. Are are the current here? Is I to and the resistance actually just plug in the resistance? It's I two times 10. Now, notice that we're walking were looping opposite to current. That's fine. Okay, forget about that. Just go to 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 here right away, and we're jumping from a positive to a negative. We're landing on the negative. So this is negative or minus five, Volz. Okay, because we're going for a positive to a negative and we keep going. We keep going, we keep going, Then we want to complete the loop. We wanna go all the way back to this point right here. So we have to We're gonna go this way because that's the loop we chose. And we're gonna go from negative to positive. We're gonna go from negative to positive. So this is gonna be positive. The voltage of of the resistor, both of your resisters. I r The eye here is I three. And the R is 15 by the way. We already know I three it is 30.6. Okay, so we're gonna be able to plug that in there. So 10 I to minus five plus 50.6 times 15 equals zero. This is simply nine. I'm gonna write 10. I two equals Gonna put the I in that. The nine to the other side. Negative. Nine. I'm gonna move the five to the other side. It becomes a positive. So this is gonna be negative. Four. Negative four. Okay, so this is gonna be a negative four. So I two is going to be negative. Four, um, divided running out of space there. Negative four divided by 10 which is negative. 100.4 amps. Of course. Alright, So that's that. I got the second currents. I got that. I too is negative for 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 eye to that we assumed is actually wrong. Okay, So assumed. Let's write this here Very important. Assumed incorrect direction off I to. So I two is actually the other way. And I'm going to write this over here actual I to is in this direction right here. Okay, now, that doesn't mean anything. Um, that that doesn't mean you stop doing the problem where 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. Um, 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 but deep inside, you know that the current is the other way. Alright, Cool. Um so we wanna find we have I two. We have I three we wanna find. I won, but we ran out of loop equations. We already made two loops. Eso What we're gonna do now is we can actually use this equation right here. We can use this equation right here to find I won because we already have the other two. So I one plus I to equals I three. So I one is simply I three minus i to I three is six amps minus I two. I two is negative. Four. 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, so there's gonna be a positive point. Six plus four is one. So the current through I one is one amp. Okay, Now we're basically done. I mean, we are done. What I wanna do is real quick. Just show you something with these currents. So I one right here Now we know was one amp and I know that I to the actual direction of eye to is this way. And I know