27. Resistors & DC Circuits

Kirchhoff's Junction Rule

# Anderson Video - Complex Circuit Example

Professor Anderson

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And let's see if we can attack this together. So um- we have a circuit that looks like the following. We have a battery sitting right here, I'll give you not my numbers, it is 58 volts, there is a resistor right there which is 120 ohms, there is a resistor here which is 82 ohms, and then it comes back up through a resistor right here and that resistor is 64 ohms. And then there's another part of the circuit. Okay, over here there is a resistor of 25 ohms, there is a battery right here of 3 volts, there is a resistor here of 110 ohms, and then the wire connects back around like that. Okay, that's what the circuit looks like, and there are parts a through infinity it looks like. And each part is basically saying what's the current through this resistor, and this resistor, and this resistor. Okay, we need to find the currents through each of the resistors. So how do we- how do we attack this? Well, let's go back to Kirchhoff's rules for circuits. Kirchhoff's rules. What were they? One was the current at a junction has to be continuous so whatever current is going in has to equal whatever current is going out. And then we also had the loop rule which said that the voltages that are rising minus the voltages that are dropping has to equal zero around a closed loop. Okay. So let's consider two things. Let's consider the following. We've got current in this circuit right here but we also have current in this circuit right here. And let's think about the first rule, the junction rule. Current in equals current out. If i have I 2 coming into that junction, and I have I 1 coming out this way, Then I have I 3 coming down. But we know the junction rule says that the current in has to equal the current out. So I 2 going in has to equal I 1 plus I 3 going out Alright, I 1 is going to come all the way back around to here, I 3 is coming down, and then it picks up again to I 2 and so we have the same rule over here, I 1 plus I 3 going in equals I 2 going out Alright, so that tells us what the currents are doing at those junctions. Now how do we analyze the loop rule? The loop rule says the following: Start at some point, go around the closed loop, and whatever voltages are rising, minus the ones that are dropping, that has to equal zero. So if I start at this point right here and I follow that path, what is the first one that I come to? Well the first thing I come to is the battery so I get plus 58 volts. The next thing I come to is a resistor And that resistor is a voltage drop and we said that we have current I 1 flowing through those wires. And so we have to subtract I 1 times 120 ohms. We continue around this loop and we hit this resistor down here, the 82 ohm resistor, that is another voltage drop. And so we have I 1 times 82 ohms. And now we get to the 64 ohm resistor, right, we have to continue around this loop. And now we have to decide, is the voltage drop across this thing? Should I put a positive here or a negative? Well, there is current I 3 in that resistor going down. And if I was going down, then it would be a negative. Right, I would have a voltage drop. But since, in our path, we're coming back up through it, it becomes positive. We're going to increase in voltage. And so we get plus what is the drop there? It is I 3 in that arm times the resistance, 64 ohms. And that is a complete loop there. And all of that equals zero. Okay, one equation. Now we need to do something similar for the loop on the right. And let's start at the same point and we will go around in that direction, and we'll see what we get in this arm. So the path on the right is going to be what? We're going to start at this point, we're going to go down through this resistor, there is current I 3 going through it, and so we need minus I 3 times 64 ohms. We come along here and we are now with current I 2 and there is no resistor here, we don't have to worry about anything there, but now we're going to go through the 110 ohm resistor. And we have minus I 2 times 110 ohms. We then get to the battery, and if you look the short line is on the bottom, the long line is on the top, which means we are increasing in voltage. So we have to add 3 volts from that battery, and then finally we go through the 25 ohm resistor and it is in the direction of the current and so we drop again I 2 times 25 ohms. And all of that equals zero So we have one, two equations, but we, in fact, have three unknowns. The other equation that you need is from the junction rule, which tells you that the currents are going to add up. And now you can solve for whatever you like. Let's see if we can do that. let's see if we can solve for I 1 and I 2. We have I 2 equals I 1 plus I 3. okay And just real quickly, let's remind ourselves what the circuit looked like. It looked like this, we had I 1 in that arm, I 2 in that arm, I 3 coming down the middle. Okay. So let's solve this thing for I 1 and I 2. And we can ignore all the units on all these things. So this first equation, what do we have? we have 58 then we have minus 120 I 1 minus 82 I 1 plus i3 times 64, but I 3 we know is i2 minus i1 and all of that equals 0. okay and so we can put a bunch of terms together here let's do that we've got 58 and then we've got minus i1 times 120 plus 82 plus 64. and then we have plus 64 I 2. All that equals zero and we can add these things up. 120 plus 82 is 202. And then we've got another 64 so we get 266. And then we still have 64 I 2. So here's one equation that we can use, and now let's simplify the second equation. So this one becomes what? We've got a minus 64 times I 3. But again, we know I 3 is I 2 minus I 1. And then we've got minus 110 I 2, we've got a plus 3 and then we've got minus 25 I 2. All that equals 0. And now we can put a bunch of common terms together, it looks like we have a bunch of I 2's. So minus I 2 times 64 plus 110 plus 25. And then we've got plus 64 I 1 and we've got plus 3. All that equals zero. And now we can simplify this, what do we get? 110 plus 64 is 174, and then we're going to add 25 to it so that's 199. So this becomes I 2 times 199 plus 64 I 1 plus 3 equals zero. Two equations, two unknowns. Let's solve for I 1 and I 2. Okay, so let's solve this first equation for uh- how about I 1. So I can rewrite it, I have 266 I 1 equals what? I just move that over to the other side so this becomes 58 plus 64 I 2. And so we get I 1 equals 58 plus 64 I 2, all that over 266. And now we have this equation down here and we just solve for I 1 in the first one. So let's rewrite this equation. And what do we get? We get minus 199 I 2 plus 64 I 1 plus 3 equals zero. And we will plug in what we got for I 1, 58 plus 64 i2 all over 266 plus 3 equals zero. And now we can solve this equation for I 2. We've got minus 199 i2 plus 64 times 58 over 266 plus 64 times 64 divided by 266 I 2 plus 3 equals zero. And let's combine our I 2's, so we have I 2 times 64 squared over 266 minus 199. And then we have this term, plus 64 times 58 over 266 plus 3. Now, somebody punch this stuff into your calculator and tell me what you get. What do you get? 64 squared over 266 minus 199. What do you get when you do that? Negative 183.6. Really? What about this one? Anybody do that one? 16.95, including the three? Okay. Okay, so let's take a look at that. And if that is confirmed, then do this calculation and tell me what you get for I 2. Okay, and what do you get for I 2 right here? .092? Like that? And the units are amps. 0.092 amps, and if we go back to our- the circuit that we had, it looked like this and we had I 1 right here, I 2 right there, and I 3 right there. So let's try the first part A and see if we're right, because it says it wants the current in the 25 ohm resistor, which is this arm right here, in I 2. And so let's see if we're right. 0.092 amps, submit it, and guess what, we're right. Actually says correct. So, believe it or not, after all those numbers we did get the correct answer on that one. Okay, but we need I 1 because we need to know these other resistors, so let's calculate I 1. I 1 is 58 plus 64 times I 2, which we just said is 0.092. And then we're going to divide all that by 266. And what do you get if you do that? 0.24? 0.24 amps. And that's the second part B which is the voltage- or, the current through the 120 ohm resistor. So let's try that. 0.24 That's a comma we don't want that, we're not in Europe. 0.24 amps, and voila, that one's also correct. And now the last part is they want to know what's the current through the middle resistor here I 3. And we have a relationship which is I 2 equals I 1 plus I 3. Right, the junction rule for that, and so I 3 is I 2 minus I 1. So if I take that and I subtract that, what do I get? Negative what? 0.15 Okay. And, I don't know if they actually want plus or minus signs here, so I think they just want magnitudes, but I'll be your guinea pig, let's try it out. 0.15 amps. Submit it, correct. So they're not worried about the direction, they just want to know the current through it. And now you have all the other parts, because the other parts are going to be part D is going to be the same as part A, and part E is going to be the same as part B. Okay? Alright, good solid challenging problem. Everybody okay with that one? It got pretty complicated pretty quickly. We could have carried resistors R 1 and R 2's all the way through, but I wanted to make sure that you guys could relate those values to the values that you have in yours.

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