0:00

[MUSIC]

Â Welcome back to Linear Circuits.

Â This is Doctor Ferri.

Â This lesson is on the Current Divider Law.

Â The objective is for you to be able to use the Current Divider Law.

Â And we can look at it from a perspective of finding the current through this

Â branch, when we've got two resistors in parallel with one another and

Â a current source across them.

Â 0:40

The Current Divider Law is given right here.

Â Now it's very similar to the Voltage Divider Law,

Â in terms of trying to find a current through one of our branches.

Â Before, when we've done the Voltage Divider Law,

Â we had to find voltage drop across a particular resistor.

Â Here, we're trying to find the current through a particular resistor.

Â If we've got this formula right here, where we've got the opposite,

Â so we're trying to find the current right through this resistor, and what we take is

Â the opposite resistor, over the sum of them, times the current source.

Â So it's the opposite over the sum times the current source.

Â Let's go ahead and derive it now.

Â Let's recognize that we've got two resistors in parallel with one another,

Â and I can replace them with their equivalents.

Â 1:33

Where this right here is, R1, R2,

Â over R1 + R2, because I've got two of them.

Â So let's call this I, well, that's I sub s right there, and

Â that's going in and splitting into both of these, so this is still I sub s.

Â That means that this voltage drop here.

Â 2:24

So, from here, I have that, I sub 2 is equal to V out, over R2.

Â So, if I just take V out and plug it in there, what I will get is this formula,

Â R1 over R1 + R2, times I sub s.

Â 2:43

Now, one thing to note about this is that we take the opposite resistor

Â from the branch that we're looking at.

Â And that means that if R1 becomes really large, this current becomes large.

Â And that makes sense because if R1 becomes really large,

Â the most of the current's going to want to flow through R2 and it becomes larger.

Â 3:12

Now the solution here is to use the formula,

Â I2 is equal to R1, which is 300,

Â over the sum of the two, which is 6000,

Â times I sub s, which is 0.1, so

Â in this case, it's 0.05 amps.

Â And, I do want to point out, whenever the resistors are the same,

Â then the current is going to be divided equally between the two of them.

Â So if this is 0.1, then this 0.05 makes sense.

Â But what happens if R2 is say, 1000?

Â So I've reduced the resistance for R2 and left everything else the same.

Â Then I2 would be equal to 3000, again,

Â it's the opposite resistor over this sum,

Â which is now 4000, times 0.1,

Â the source, and that gives 0.075 amps.

Â So the current, actually, went up, so

Â the proportion that went to R2 got larger because the resistance went down, so

Â more currents are going to want to flow that direction.

Â Now I want you to recognize when to use the Current Divider Law.

Â In this particular case, it doesn't look like what we've been looking at,

Â because I've got this resistor here.

Â Well, that resistor doesn't really matter because its current source

Â tells me that the current is going this way.

Â And part of that current's going to split between this branch, and

Â part between this branch.

Â Now these two resistors are in parallel with one another, though they don't look

Â that way to be, physically, they don't look that way, but electrically, they are.

Â So the current is 0.6 going this way, and I can still use the Current Divider Law,

Â which gives me the opposite resistor for this branch.

Â The opposite resistor over the sum, times the current that goes into that node.

Â 5:09

Here's another case, where in this case, I've got three resistors in parallel.

Â Actually, these two are combined, right there, to be one resistor.

Â If I look at them in series, I can combine them.

Â But I've got three branches right here.

Â Now, the Current Divider Law only allows two branches.

Â So that means I'm going to have to take this part of the circuit and

Â reduce it down to one equivalent resistance, and

Â if I do that, I would find that it's equal to 50 omes, and

Â in this case, if I've got two amps going in, then it's going to be split equally

Â between these two branches, and each one will get one amp.

Â To summarize, we've got the key concept of the Current Divider Law.

Â 5:52

We're looking at a circuit in this configuration with two resistors in

Â parallel with one another, this is the formula here, where it's the opposite

Â resistor over the sum of the two, times the source, or times, actually,

Â the current that's going into the node that connects the two parallel resistors.

Â All right, thank you.

Â [MUSIC]

Â