Parallel resistors (part 1) | Circuit analysis | Electrical engineering | Khan Academy

In this video, we're going to look at another familiar pattern of resistors called parallel resistors. I've shown here two resistors that are in parallel. This resistor is in parallel with this resistor, and the reason is it shares nodes. These two resistors share the same nodes, and that means they have the same voltage, and they are called parallel resistors. So if you share a node, share the same node, then you share the same voltage, and you are in parallel. That's what that word means.

Now, if we go look closer here, we'll see some interesting things. It's hooked up; we have a battery here, some voltage V, and because there's a path—a complete path around here—we're going to have a current. We're going to have a current I flowing in this circuit. Let's label these resistors; let's call this one R1 and this one R2. Those are our parallel resistors. When the current reaches this point here—when a current reaches this node—it's going to split. It's going to split into two different currents: that current and that current. We'll call that one I1 because it goes through R1, and we'll call this one I2, and that goes through R2.

Now, we know any current that goes into a resistor comes out the other side; otherwise, it would collect inside the resistor, and we know that doesn't happen. This one comes here, and they rejoin when they get to this node and flow back to the battery. So the current down here is again I, the same one that's up here.

Now what I want to do is I want to replace these two resistors with an equivalent resistor—one that does the same thing. By the same thing, we mean causes the same current to flow in the main branch. And so that's what's drawn over here. Here's a resistor; we're going to call this V again, and we'll call this R parallel, RP. This resistor causes the same current I to flow here, and now we're going to work out an expression for that. We want to figure out how do we calculate RP in terms of the two parallel resistors here.

Okay, so let's go. What we know is the voltages on the two resistors are the same. We know there's two different currents, assuming that these are two different valued resistors. And now, with just that information, we can apply Ohm's Law, and we use our favorite thing, Ohm's Law, which says that voltage on a resistor equals the current in the resistor times its resistance.

So let's write down Ohm's Law for R1 and R2. Okay, we know the voltage; we'll just call the voltage V. This is for R1: V equals I1 times R1, and for R2, we can write a similar equation, which is V, same V, equals I2 R2. Now there's one more fact that we know, and that is that I1 and I2 add up to I, and these are the three facts that we know about this circuit.

What I'm going to do now is come up with an expression for I1 and I2 based on these expressions and plug them into this equation. Okay, I can rewrite this equation as I1 = V over R1. I can write this one as I2 equal V over R2, and now I'm going to plug these two guys into here. Let's do that. I equals I1, which is V over R1, plus V over R2.

Let me move the screen up a little bit. Okay, now we're going to continue here. I just want to rewrite this a little bit: I equals V * (1 / R1 + 1 / R2). Okay, so here we have an expression. It actually sort of looks like Ohm's Law. It has an I term, a V term, and this R term here. So let me go back up here. Here's our original Ohm's Law. I'm going to write this; I'm going to solve for I in terms of V just to make it look a little more obvious.

I can say I equals V / R, and what I hope you see here is the similarity between this equation and this one down here. So I have this R here, and what's happening is this term is playing the role of that resistance. So I'm going to bring this equation down here and write it right down here times (1 / R). I'm going to call this RP because what I want is for this expression to equal this expression. I'm going to set those equal; same I, same V. These guys are equal.

I can write it all; I'll just write it over here: 1 / RP = 1 / R1 + 1 / R2. This says we have a resistor; we're going to call it RP or R parallel that acts like the parallel combination of R1 and R2. So this is the expression for a parallel resistor. If you want to calculate a replacement for R1 and R2 in parallel, you do this computation, and you get RP.

So let's do one of these for real. Here's an example. Here's an example where I've actually filled in some numbers for us. So I have a 20 Ohm resistor in parallel with a 60 Ohm resistor driven by a 3V battery, and what I want to do is I want to combine these two parallel resistors and find out what is the current right here. Find out what is the current; that's my unknown thing here. I know everything else about this.

So let's use our equation. We said that 1 / RP was equal to 1 / R1 + 1 / R2, and let's just fill in the numbers: 1 / RP = 1 / 20 + 1 / 60. That equals, uh, let's just make 60 the common denominator. So I have to multiply this one by three: 3 / 60 + 1 / 60, and that equals 4 / 60. And so now I'm going to take the reciprocal here: RP equals 60 / 4, or RP equals 15 Ohms.

So what this is telling us is if we have two resistors in parallel, 20 Ohms and 60 Ohms, that is for the purposes of calculating the current here, that's the same as 15 Ohms. It's hooked to 3 volts, just like that. Let's check what the current is. Current is I = V / R = 3 volts / 15 Ohms. That's equal to 0.2 amps, or you can say it's the same as 200 milliamps.

So we actually have now simplified our circuit from two resistors to one resistor, and we are able to compute the current here, which is 0.2 amps. I would invite you to check this by going back and computing this current up here to make sure it's the same. The way you would do that is you would calculate the voltage. The voltage here is 3 volts—3 volts across 20, 3 volts across 60. You'll get I1 and I2, and if you add those together, you'll get the total I, and it should come out the same as this.

I think that's a good exercise for you to do to prove that the expression for a parallel resistor, 1 over R parallel, can be computed from 1 over R1 + 1 over R2.