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Welcome back to Linear Circuits.

This is Doctor Ferri.

This lesson is on the Current Divider Law.

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

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

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

a current source across them.

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The Current Divider Law is given right here.

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

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

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

we had to find voltage drop across a particular resistor.

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

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

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

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

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

Let's go ahead and derive it now.

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

and I can replace them with their equivalents.

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Where this right here is, R1, R2,

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

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

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

That means that this voltage drop here.

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

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

R1 over R1 + R2, times I sub s.

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

from the branch that we're looking at.

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

And that makes sense because if R1 becomes really large,

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

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Now the solution here is to use the formula,

I2 is equal to R1, which is 300,

over the sum of the two, which is 6000,

times I sub s, which is 0.1, so

in this case, it's 0.05 amps.

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

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

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

But what happens if R2 is say, 1000?

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

Then I2 would be equal to 3000, again,

it's the opposite resistor over this sum,

which is now 4000, times 0.1,

the source, and that gives 0.075 amps.

So the current, actually, went up, so

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

more currents are going to want to flow that direction.

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

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

because I've got this resistor here.

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

tells me that the current is going this way.

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

part between this branch.

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

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

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

which gives me the opposite resistor for this branch.

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

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

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

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

But I've got three branches right here.

Now, the Current Divider Law only allows two branches.

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

reduce it down to one equivalent resistance, and

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

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

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

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

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

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

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

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

All right, thank you.

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