Simplifying resistor networks | Circuit analysis | Electrical engineering | Khan Academy

We've learned about series and parallel resistors. We've learned how to simplify series and parallel resistors into an equivalent resistor.

Just to review, for the series resistor, our series equivalent ( R_{series} ) is equal to the sum of resistors in series: ( R_1 + R_2 ). 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. If there were two resistors in parallel, the formula was ( \frac{R_1 \times R_2}{R_1 + R_2} ).

If there are three resistors or more in parallel, it’s a little more complicated. This formula looked like this:

[
\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \text{ 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 two resistors.

So now, in 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 into our resistor network—there are 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.

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. Just wait and do that.

All right, so we're looking for series and parallel patterns. And there's one! And there's this... This is not in series. Nope. Okay, oh, there's a possibility—there's a possibility of parallel resistors. Is this 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 are a couple of possibilities here for simplification. 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.

So a good way to start this problem is to start at the very far end of the circuit and work our way backwards. This is opposite the way you read sentences and things like that, but in circuits, it’s sometimes what we want to do.

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, done the hard work of this, and the rest of it's going to be a relatively straightforward application of these two formulas.

We've decided what our unknown is and we've 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 basically now going to do the work of simplifying this network. What I see here is two resistors in series: 2 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. That says that:

[
\frac{10 \times 10}{10 + 10} = \frac{100}{20} = 5 \text{ ohms.}
]

I could have done this a little quicker if I had known 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 replace that with a 5 ohm resistor. You can see where this is going. We are basically going to collapse this circuit one step at a time: ( 5 \text{ ohms} + 1 \text{ ohm} ) is the 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. I said 6 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:

[
\frac{1}{R_{parallel}} = \frac{1}{12} + \frac{1}{4} + \frac{1}{6}.
]

The common denominator is 12. So that is something over 12 which will be:

[
1 + \frac{3}{12} + \frac{2}{12} = \frac{6}{12} = \frac{1}{2}.
]

Now, ( \frac{1}{R_{p}} = \frac{1}{2} ), so ( R_{p} = 2 \text{ ohms}. )

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.

The final step is one and two ohms in series: ( 1 + 2 = 3 ). 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 3 ohms. That whole rest of that complicated network looks like 3 ohms to that voltage source.

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.