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Now we're ready to go on and talk about Kirchoff's Current Law.

Kirchhoff's Current Law says that the sum of all the currents entering or leaving a

node in a circuit is zero. And so let's say we have a node in a

circuit. That's a connection point of any number

of wires or components. And let's say that there's a current i

one. And i two flowing into the node, and a

current i three flowing out of the node. Now the first thing you have to do is

pick a consistent sign convention. This is probably the most confusing part

of this whole, application here. So we're going to just say that any

current flowing into a node is positive. So I'd think of the node as like a bucket

and then, pouring water into the bucket. And so, it's going in, the bucket's going

up, so that's positive. And current flowing out of the node is

removing water from the bucket, so that's, we'll call that negative.

And so then with that sign convention Kirchhoff's current law says that at this

node I can write a relationship between these three currents.

I one and I two are flowing in so they're both positive.

I three is flowing out so that's negative.

And the algebraic sum of all of those. Has to be zero.

So that's Kirchhoff's current law. Kirchhoff's current law is really just

the conservation of charge. It says the charge is neither destroyed

nor created in a circuit. It has to go somewhere.

So what goes into a node has to come out. Okay, now we're going to use Kirchoff's

current law and Kirchoff's voltage law to solve another useful circuit which is a

current divider. Now the current divider is nothing more

than the two resistors in a parallel combination.

So the, I connect the battery across the parallel

combination of these two resistors. And now, a current is going to flow out

of the battery. And when it reaches this node A, it has a

choice. Some of the current can flow down through

r one to ground. And some of the current can flow down

through r two to ground. And so what we want to find out is, what

is the total current flowing out of the battery in this circuit, and then what is

i one, the amount of current going through r one and what is i two, the

amount of current flowing through r two. So, to solve this circuit what we're

going to do is apply Kirchhoff's voltage law and Kirchhoff's current law,

everywhere we find an opportunity. So let's first look at loop number one.

Here just this loop from the battery going through the battery and the

resistor R one, and we're going to apply Kirchhoff's voltage law to that loop.

Now that says that the sum of the voltages going around the loop.

The algebraic sum has to be zero. So if I start down here at the ground, I

go up through the battery, I pick up a voltage v, and then I go down through the

resistor. And the current through r one is i one.

And so the voltage drop across this resistor is i1 r1.

And so the sum, algebraic sum, v minus i r has to be 0.

So that tells you that i1 is just v over r 1.

Now Kirchhoff's voltage law in the outermost loop so I can take a loop

around the circuit this way, as well, and that is a similar thing.

It's says I start at ground, so I go up through the battery, and I have reached

the voltage V, and then I go around and down through this resistor, and the

voltage drop across this resistor is i 2 that's the current going through r 2

times r2, and then V minus i2 r2 has to be 0.

So that tells you i 2 is V over R2. Now, notice this.

Both ba, both the resistors have the same voltage across them.

That's the, the characteristic of parallel combination, is that both

elements have the same voltage drop across them.

And so the currents then are, this from Ohm's law V over R1 and V over R2.

Now, the third equation here. Now, notice this.

We're looking for three quantities. I, i1, and i2.

So, in general, when you're looking for three quantities, you need to have three

independent equations to be able to solve them.

That's just a, a, a rule of linear algebra.

So the, we're going to apply Kirchoff's current law at this node, a, to get a

third equation. And that just say that the current

flowing in Is i, and the current flowing out is i one and i two.

So it just says says i minus i1 minus i2 is 0.

So, those are the three equations that we're going to solve to find all three

currents. Now, the first thing we're going to do.

It take equation three here. And then I 1 is V over R 1.

I'll plug that in. And I 2 is V over R 2.

And so that's equation 3, becomes that. And then I can factor the V out of both

of these terms. So I have 1 over R 1 plus 1 over R 2.

Now this equation, I equals V times this 1 over r 1 plus 1 over r 2 has the form

of i equals v over some new r. And this r I'll call the, the parallel

combination of r one and r two. And so 1 over r parallel equals 1 over R1

plus 1 over R2. And, so I can put this over a common

denominator. so this becomes.

I multiply this one by R2, this one by R1.

And then downstairs I have R1 and R2 in both terms.

And, so this becomes R2 plus R1 over the product.

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And the parallel combination then, if I take one over this, I invert this.

Our parallel just becomes the product of the resistors R1 and R2 over their sum.

Now this is in accordant formula to remember.

Okay, now continuing with the current divider the first thing I'm going to

notice is that from these two equations, I can write V is i 1, R 1, and that's

equal to i 2 R 2. And then, I can take and write this, and

when I look aside I'm just looking at my notes so I don't get lost.

So bear with me please. the I'll write this then i1 is equal to

i2. Then it's r2 Over R1.

So now I'm going to take, and I'm going to substitute this in up here, and then

I'll have one equation that just in terms of i and i2.

And I already know i. So, I can find i2.

So, I have, then, i equals i2 times r2 over r1 minus.

Oh, I'm sorry. Well, it's going to be plus i2.

So I put both of these, i equals i1 plus i2 on the other side.

So there's I1 plus I2. So let's see, I can write this then as

I2, so I'll factor out I2 from these two terms.

So I have 1 plus r2 over r1, and that equals i.

And so then i2 is i divided by all of this, one plus r2 over r1.

Now, let's clean this up a little bit. I'll take and I'll multiply the top and

the bottom, by R1. So I'm just writing 1 as R1 over R1, and

then when I multiply that through I get i, R1, and then downstairs here I have an

R1. For the first term, and then the R 1

cancels this R 1 down there. So, i 2 is the total current coming out

of the battery, and the fraction of that current that goes through R 2 Is given by

this. Okay.

Now, Also, if you go back here, to this

equation. We can figure out what i1 is.

So, i1 Is just I2. It's going to be all of this times r2

over r1. And so that's going to be.

If I plug this in, it's going to be I. And then r2 is going to show up up here.

This 1 over r1's going to cancel that one.

So this is going to be r2. All over r 1 plus r 2.

So there you have it. We have everything now.

The the current i is just v over the parallel combination of the two

resistors. And the parallel combination is just r1

r2 over r1 plus r2. So, I find i.

So first, so, a complete solution to this that is.

I first find our parallel. Compute that.

Then I can compute i. That's the total current coming through

the battery. And then from that, that current divides

into i 1 and i 2. And so the amount of current that flows

through i 1 is proportional to r 2 over the sum.

The current that goes through i 2 goes like r 1 over the sum.

Now the point here and you can remember this.

it seems a little strange that, you know, why would the expression for i1 have an

R2 up here? Well, the reason for that is, let's say,

I have a very large resistor here, R2. And a very small resistor for R1.

Then what's going to happen you know intuitively that most of the current is

going to go through the lower resistor. And so if R2 is the bigger number that's

going to give you the larger value of I1. So, so having this large resistor over

here forces the current to go through I1. And so that's why this expression for I1

has an R2 upstairs. if I were to make r2, go to infinity, so

it gets, if this grows without bound, then this term, r2 over r1 plus r2, that

would go to unity. And so all of the current would go

through this resistor. And vice versa.

If, i, if r2 were to go to zero. So this becomes a, a short circuit.

Then, what's going to happen is, the amount of current going through i1 is

going to go to zero. Because all of the current is going to go

through the lower resistance path. The short.

So, there you have it. This is the full solution of the current

divider.