Main content

### Course: Class 12 Physics (India) > Unit 3

Lesson 9: Series and parallel resistors- Series resistors
- Resistors in parallel
- Parallel resistors (part 3)
- Resistors in series and parallel review
- Parallel resistors
- Simplifying resistor networks
- Example: Analyzing a more complex resistor circuit
- Analyzing a resistor circuit with two batteries
- Simplifying resistor networks
- Finding currents and voltages (pure circuits)
- Finding currents and voltages (mixed circuits)

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Simplifying resistor networks

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

## Want to join the conversation?

- 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)
- 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.(25 votes)

- @7:35you 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)
- "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.(11 votes)

- What do you mean, at1:43, by "pulling out of voltage source"?(1 vote)
- 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.(8 votes)

- how can we decide the equivalence resistor position?!(3 votes)
- 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)

- At 06.31 you get from 1/2 to 2 but saying (1)%(1/2) it get the right result but somehow I dont really get it. We have the formula 1/Rp = 1/R1 + 1/R2... so in the end we get (1/Rp)%(1/Rn) Anyway dont know if that is correct. Im studying to become an electrician so math is not my strongest subject(1 vote)
- There are a lot of reciprocals in the general equation for parallel resistors. The tricky one is the one on the left side. I'll do the math slowly...

1/Rp = 1/R1 + 1/R2 + 1/R3

1/Rp = 1/12 + 1/4 + 1/6 = 1/12 + 3/12 + 2/12

1/Rp = (1 + 3 + 2)/12 = 6/12

1/Rp = 1/2

so now you have one more reciprocal to get Rp...

Rp = 2

Maybe it helps to write it like this...

1/Rp = 1/2 = 0.5

Rp = 1/0.5 = 2(4 votes)

- 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)
- can you please show me how can I find the current and voltage at each resistors?(1 vote)
- 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 to6:54in 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)

- 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)
- From the author:You derive this formula by starting with the reciprocal version of three parallel resistors,

Rp = 1( 1/R1 + 1/R2 + 1/R3)

Then multiply the expression by R1R2R3/R1R2R3 to get the one you like to use.

I've not seen this taught, but if you can keep it in your head then by all means use it.

The reason it's not taught is probably because it is a special case for 3 in parallel (2 in parallel is far more common, so its special case is always taught). The general reciprocal-filled formula is fully general purpose so it's the one thing you have to memorize to master all parallel circuits.

Your formula reminds me of the conversions for Delta-Wye resistor networks:

https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-resistor-circuits/a/ee-delta-wye-resistor-networks.(2 votes)

- Hi Willy,

In what real-world scenerio would a circuit need to be simplified?(1 vote)- Simplification of resistor networks is a central concept in Thevenin's Theorem. Check out this section of Spinning Numbers: https://spinningnumbers.org/t/topic-dc-analysis-special.html

There's a real-world example in the "how-to" article. https://spinningnumbers.org/a/thevenin-howto.html#example-2(2 votes)

- 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)
- As long as you use the correct equation for finding the equivalent resistance of a series or parallel configuration, you will get the correct answer. Whether the values are even or odd does not matter.(2 votes)

## 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.