Simplifying resistor networks

- [Voiceover] We've learned about series and parallel resistors, we've learned how to simplify series and parallel resistors into an equivalent resistor, and just to review, for the series resistor, R series, an equivalent R series is equal to the sum of resistors in series. R1 plus R2. And we learned that if we have resistors in parallel, meaning they share the same nodes, if they're in parallel, we can get a parallel equivalent resistors R parallel, and if there was two resistors, if there's two resistors in parallel, the formula was R1 times R2 over R1 plus R2, plus R2. And if there's three resistors or more in parallel it's a little more complicated. This formula look like this. One over R parallel, this is for three or more resistors. One over R1 plus one over R2, plus one over R3, etcetera, for as many resistors as you have. These two parallel formulas do the same thing. This one's a little more convenient to use when you just have two resistors. So now, this problem in this video, we're going to look at a complicated resistor circuit. There's a lot of resistors here. The question we're going to ask is, if we plug in this voltage source here into our resistor network there's the two little plug points, the question is what kind of current is going to be pulled out of this voltage source? That's what I want to know. And if I stare at this, I go, well how am I going to figure that out? The important part of what we're doing here is to stop for a minute and just see what this circuit looks like, just be with it for a second. Let your eyes wander over it and find resistor patterns that you recognize. And you just wait and do that. All right, so we're looking for series in parallel patterns and there's one, and this is not in series, no. Okay, oh, there's a possibility, here's a possibility of parallel resistors. This, is there a series one over here. I don't see a series head to tail connection here and there's a series connection here that we can use. Okay, so there's a couple of possibilities here for simplification. And one of the questions is, okay, where do we start? We want to start this problem at the end farthest away from the thing we care about, what we care about is this right here, this is what we're looking for. And so, a good way to start this problem is to start at the very far end of the circuit and work our way backwards. And this is opposite the way you read sentences and things like that, but in circuits it's sometimes what we want to do. So what I'm going to do is start over here on this side and we're going to disassemble this circuit and simplify it as we go. So we've actually at this point, we've actually done the hard work of this, and the rest of it's going to be relatively straightforward application of these two formulas. We decided what our unknown is, and we decided where to start and that strategy is the key to making this a simple straightforward process. All right, so let's go after it. We're going to, we're basically now going to do the work of simplifying this network. So, what I see here is two resistors in series, two ohms plus eight ohms equals 10 ohms. So I can remove these resistors here and we place them with a 10 ohm resistor, that's step one. What do we do next? Now we look at it again. Right here we see we have two parallel resistors. So we're actually going to use this form. We're going to use this form of the parallel resistor formula, and that says that 10 times 10 over 10 plus 10, this is equal to 100 over 20 or five ohms and I could've done this a little quicker if I know that if two resistors in parallel have the same value, 10 and 10. I know that the parallel combination of those is exactly half. So now we have an equivalent resistor for those two guys. We can take them out and we replace that with a five ohm resistor. You can see where this is going, we're just basically going to collapse this circuit one step at a time. Five ohms plus one ohm is this series combination here and that is equal to six ohms, so I can replace this now with six ohms, we're just consuming our circuit and erasing and rewriting as we go. And I said six ohms here, and what do we have now? Now this is a little more challenging. I have three resistors in parallel and I'm going to use now this formula here for resistors in parallel, so one over the parallel combination is one over 12 plus one over four plus one over six and the common denominator is 12 so that is something over 12, which will be one plus three over 12 plus two over 12, and that equals six over 12 which is one half and one over Rp, that's one over Rp, so Rp equals two over one or two ohms. And now I can replace all three of these. I get to erase all three. We're almost done. And that was, I said two ohms, two ohms. The final step is one and two ohms in series. One plus two equals three and finally we get to the last step which is All right, so we're done. What that means is, from the viewpoint, as far as that voltage source can tell, it's putting out enough current to drive three ohms. And that whole rest of that complicated network, looks like three ohms to that resistor, to that voltage source. And we have taken a complicated circuit and turned it into a really simple one. So that's the steps of simplification. The key was figuring out what you want and then going to the far end of the circuit and working your way backwards and simplifying as you go.