Simplifying resistor networks

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 our 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 to equivalent resistors are 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 1 over R parallel this is for three or more resistors 1 over r1 plus 1 over r2 plus one over r3 etc 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 to 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 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 it'd 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 and parallel patterns and there one and is this this is not in series nope okay oh there's a possibility there'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 its going to be relatively straightforward application of these two formulas we've decided what our unknown is when we decided where to start and that strategy is the key to this 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 8 ohms equals 10 ohms so I can remove these resistors here and replace 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 is equal to 100 over 20 or 5 ohms and I could have done this a little quicker if I 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 5 ohm resistor you can see where this is going we just basically are going to collapse this circuit one step at a time 5 ohms plus one ohm is this series combination here and that is equal to 6 ohms so I can replace this now with 6 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 twelve 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 twelve plus two over twelve and that equals six over 12 which is one-half and one over RP that's one over RP so RP equals two over 1 or 2 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 2 ohms 2 ohms final step is 1 2 ohms in series 1 plus 2 equals 3 and finally we get to the last step which is all right so we're done what that means is from the viewpoint from the 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 turn 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