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Simplifying resistor networks

A systematic approach to simplify a complicated resistor network by looking for series and parallel resistor patterns. Created by Willy McAllister.

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  • leafers ultimate style avatar for user Lauren  Curphey
    If you can sometimes collapse a resistor network down to a single resistor why would you start with so many resistors in the first place?
    (14 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      When we do a simplification problem like this the first thing we do is pick a location in the circuit we care about. In this case, I picked the far left end. Then the question is, "What does the rest of the circuit 'look like' from the far left end?" If we wanted to know about a different location, the answer would turn out different. The resistor network I worked with is contrived. It doesn't do anything. The point of the video is to show how you can apply the resistor series and parallel formulas over and over. (It's sort of like I was teaching you how add and subtract multiple numbers, but I didn't tell you what numbers were for.) When we get to more complicated circuits, the method in this video will be useful.
      (21 votes)
  • blobby green style avatar for user Duke Ellington
    @ you said the voltage source is putting out enough current to drive 3ohms. What does it mean to "drive" 3ohms? How does this differ from a resistor "absorbing" (V/3) amps? And even more confusingly, what does it mean for a resistor to "draw" (V/3) amps?
    (6 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      "It (the voltage source) is putting out enough current to drive 3 ohms". I use the word "drive" here to refer to the current the voltage source is required to provide. For the sake of illustration, let's say the voltage source is set to V = 3V. Ohm's Law tells us the amount of current is 3V/3ohm or 1 amp. That 1 amp current is provided by (or "driven by") the voltage source. A common way to say this is "the voltage source has to drive 1 amp into the resistor if we expect the voltage across the resistor to get up to 3V."

      We can say something similar from the "point of view" of the resistor, the resistor "demands" or "pulls" or "draws" 1A from the 3V source. We might say "When connected to a 3V source, our 3 ohm resistor draws 1A of current from the source." We use the word "draw" in the sense of "withdraw".

      These terms are examples of EE jargon creeping into my descriptions. Please let me know if you come across other places where I do this.
      (10 votes)
  • male robot hal style avatar for user KEVIN
    What do you mean, at , by "pulling out of voltage source"?
    (1 vote)
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    • spunky sam orange style avatar for user Willy McAllister
      If you connect something (a resistor, a bunch of resistors, anything that conducts current), it causes a current to flow out of the current source. So "pulling current out of a voltage source" is just a casual turn of phrase I used to give the sense that the resistors connected to the voltage source are "demanding" or "pulling" current. If the resistors were not there, there would be an open circuit, and therefore no current. When the resistors are connected, in a sense they are causing the current to flow.
      (7 votes)
  • piceratops ultimate style avatar for user Trần Nhật Nam
    how can we decide the equivalence resistor position?!
    (3 votes)
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    • purple pi purple style avatar for user APDahlen
      Hi Trần,

      Here are my recommendations:

      1) redraw your circuit so it has a left to right flow as Willy has done in this video

      2) start at the far right and work to combine the series / parallel branches

      3) take your time to redraw the circuit

      Personally I find this operation easy but oh so time consuming with many potential errors.

      Please leave a comment below if I missed the point of your question.

      Regards,

      APD
      (2 votes)
  • aqualine seed style avatar for user Sean Francisco
    How about simplifying 3-D models of a circuit? Our professor gave that as a quiz one time. I got so low there :(
    (1 vote)
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  • blobby green style avatar for user ngochongnguyen3797
    can you please show me how can I find the current and voltage at each resistors?
    (1 vote)
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    • spunky sam orange style avatar for user Willy McAllister
      Good question. To find the current in each resistor and voltage at each node, start with the simplified circuit and work backwards.

      It might help to assign a real value to the voltage source, like V = 3v.

      The whole resistor network simplifies down to 3 ohms. Therefore, the current in the voltage source is 3V / 3ohm = 1 amp.

      Then back up a step to in the video where you have 1 ohm in series with 2 ohms. This circuit lets you find the voltage at the node between the two resistors. That voltage is V (1ohm)/(1ohm + 2ohm) = 3 volts * 1/3 = 1 volt.

      Next, back up to where the 2 ohm resistor is three parallel resistors. You know the voltage across the parallel resistors, and you know the resistor values. Use Ohm's Law to find the currents. Hint: they should add up to 1 amp.

      Keep going backwards, alternating between finding currents and voltages. Notice how you are careful to only compute values for nodes or resistors that are part of the starting circuit. You never need to compute current in a resistor that is a simplification of several others.
      (3 votes)
  • blobby green style avatar for user mcmeyer47
    For Three resistors in parallel I prefer to work with Rp= (R1*R2*R3)/(R1*R2+R1*R3+R2*R3). Beyond three resistors this equivalent idea may get a bit cumbersome. Is this a common way of teaching this or is it considered more of a pain than it is a help?
    (1 vote)
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  • blobby green style avatar for user Maalkum
    Hi Willy,

    In what real-world scenerio would a circuit need to be simplified?
    (1 vote)
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  • winston baby style avatar for user Ethan Blea
    Is there a rule when adding ohms for example I cant add 2 ohm levels like 3 and 4 because the outcome will be odd or does the outcome have to be even.
    (1 vote)
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  • stelly orange style avatar for user BootesVoidPointer
    Can any register network be simplified like this?
    (1 vote)
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Video transcript

- [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.