Current through resistor in parallel: Worked example | DC Circuits | AP Physics 1 | Khan Academy

So we have an interesting circuit here. The goal of this video is to figure out what is the current that flows through the 6 ohm resistor. Pause this video and see if you can work through it.

The way that I am going to tackle it is first simplify the circuit. Take these two resistors in parallel and think about what the equivalent resistance would be. We have seen that before:

1 over the equivalent resistance is going to be equal to 1 over 6.0 ohms plus 1 over 12.0 ohms.

1 over 6 is the same thing as 2 over 12. So, 2 over 12 plus 1 over 12 is 3 over 12, and 3 over 12 you could view that as the same thing as 1 over 4.0 ohms.

Thus, 1 over the equivalent resistance is equal to 1 over 4 ohms. Well, that means that the equivalent resistance is 4 ohms.

Now we can simplify our circuit where we replace these two resistors in parallel with one resistor of the equivalent resistance, and that is going to be equal to a 4 ohm resistor.

Now, the next thing we could do is figure out what the current is through this part of the circuit, which would be the same thing as the current right over there. We could call that I sub 1, and we can just use Ohm's law for that.

We would have I sub 1 equal to our voltage drop, which is 24 volts, divided by the equivalent resistance of these two resistors in series. When you have resistors in series, you just add them up to figure out the equivalent resistance.

So this would be divided by 2.0 ohms plus 4 ohms plus 4.0 ohms. 24 divided by 2 plus 4, or 24 divided by 6, is 4. Since we're dealing with two significant digits, it'll be 4.0.

We're talking about current, so this is 4.0 amps or 4.0 amperes.

Now, how do we use that information to calculate this current right over here? We can call that I sub 2. One way to think about it is what is going to be your voltage drop from this point to this point?

If you know the voltage drop from that point to that point, and if you know that the voltage drop from this point all the way down here is, then we can figure out what the voltage drop from here to here is going to be.

So let's do that. The voltage drop across this first resistor, remember your change in voltage, is just equal to your current times your resistance. So this is going to be your current, which is 4 amperes, times your resistance, which is 2 ohms.

2 ohms times 2 ohms is going to be equal to 8.0 volts. If the voltage difference between that point and that point is 24 volts, which we know from this voltage source, but if we drop eight volts as we go to this point,

well, then the difference between this point and this point, or this point and this point right over here, has got to be a 16 volt drop. If our delta V across the 6 ohm resistor is equal to 16 volts, well then we can use Ohm's law again to figure out I sub 2.

I sub 2 is going to be equal to our drop in voltage, so 16 volts divided by this resistance, 6 ohms. So what is this going to be equal to?

16 divided by 6 is 2 and 4 over 6, or 2 and two-thirds, or 2.66666. If we round to two significant digits, you're going to have 2.7 amps.

So we just figured out what we wanted to figure out. This right over here is 2.7 amps or 2.7 amperes.

But we can keep analyzing it for fun. I encourage you to figure out what that current is now. The current I sub 3, and use the exact same technique.

One thing that you should feel very comfortable with is that this current that is flowing through the first resistor, that 4 ampere current, gets split between I sub 2 and I sub 3.

So I sub 2 and I sub 3 should add up to the original 4 amps. Just thinking about it that way, if you do the same type of analysis we just did, you should get 1.3 amps for I sub 3 because 2.7 plus 1.3 is going to be equal to 4 amps.