L(2-2) Kirchoffs Laws KVL,KCL

Kirchhoff's Laws are used to find the unknown voltages and currents in a circuit. See cookbook and other reference material here: (www.ece.utah.edu/~ece1250)

Hello, this is Dr. Cynthia Furse,
from the University of Utah.

Today, we're going to talk
about Kirchhoff's Laws.

Kirchhoff's Current Law says that the sum
of the currents entering a node is = 0.

Or equivalently, the sum of
currents leaving a node is 0, or

the sum of currents leaving is equal
to the sum of currents entering a node.

So we can write the equation
four different ways.

When we're talking about a node, we're
talking about something like this right

here, that's in the center
of all four of the wires.

Now, when we say the sum of
the currents entering the node,

we can see that i1 and i4 are entering
the node while i2 and i3 are leaving.

So we're going to say i1 + i4
those are the currents entering.

And i2 which is leaving, we're gonna
put it there with the minus sign and

i3 we're going to put with a minus sign.

When we write the equation where we have
the sum of the currents leaving the node

is equal to 0, what we're doing is
just switching the minus signs.

Minus and plus signs here, as you can see.

When we talk about the sum of the currents
leaving is equal to the sum of

the currents entering.

Then you could say,
here's the sum of the currents entering,

here's the sum of the currents leaving,
and they are equal.

These four different ways of writing
the node equation are all equivalent.

This is Kirchoff's Current Law, or KCL.

Kirchhoff's Voltage Law or
KVL is the other law that we will use.

This says that the sum of the voltage
around a closed path is = 0.

The way you do this is you
run a loop like this, and

you just add up all of the voltage
as you're going around the loop.

So for example,
as we're going through here,

we say -VDC +, see there's the + V1 + V2,

and that's equal to 0.

So -VDC + V1 + V2 = 0.

Alternatively, we could say that the sum
of all the voltage drops is equal to

the sum of all the voltage rises.

So the voltage rise is +VDC and
the voltage drop is V1 + V2.

So there's just a convention
that typically,

we use a systematic clockwise
motion around the loop.

We could go counter clockwise, but

convention is just that
we normally go clockwise.

We also assign a positive sign to
the voltage across an element,

if the positive sign of that
voltage is encountered first.

So that's why I was kinda circling
this thing as I went to your + V1.

Now, let's talk about a simple circuit
where we can apply KCL and KVL.

So here's our simple circuit,
let's first apply the loop equation.

Right here's our loop and
let's apply the equation.

So we can see minus,
see we count the minus sign first,

-VDC + V1 + V2 = 0,
that's the equation right here.

That is the KVL equation or
the loop equation.

Now, let's examine the circuit,
what do we know?

Well, VDC is defined that's 120 volts,

R1 is defined that's 10 ohms,
R2 is defined that's 50 ohms.

What's unknown?

V1 and V2 are both unknown.

So if you remember from math,
if you have two things that are unknown,

then you're going to need two equations.

But we only have one equation,
so what else can we do?

Well Ohm's Law will give us two more
equations, so why don't we try that?

So we can see here for example that if we
have a current I1 that's running through

R1, we know that V1 = (I1) times (R1),
we also know that V2 is (I2) times (R2).

Well, look, we got two more equations.

But we also have two more unknowns,
I1 and I2.

So now, we have four unknowns and
three equations.

So what we always need
is one more equation and

this one's going to come from a node.

So here's our node equation right here,
that's going to be our KCL.

And remember, that's the sum of the
currents coming in to the node is equal to

the sum of the currents
going out of the node.

So I1, which is all the currents
coming into that node is equal to I2,

where the currents are going out.

So we now have four unknowns, V1, V2,
I1, and I2, but we have four equations.

So now, we can solve it.

You're going to say, heavens.

It looks like we could have done this
equation a whole lot simpler and

absolutely, you could have.

But what I'm showing you is the systematic
way of handling the KCL and the KVL.

Okay, so now,
let's actually solve this equation.

So here's our first
equation where we have VDC.

Well, that's know.

We could substitute that in there and
V1 and V2.

We know R1, we know R2,
let's substitute those in there.

And we know that I1 = I2 so
let's just call it a value I.

So just substituting in, what I'm
gonna do is substitute I in there,

and I in here and of course,
all of our constants.

This equation now becomes this right here,
and I can solve if for I.

I = 120 volts, let's just bring
in the 120 over to that side,

divided by the sum of 10 and
50, so I = 2 amps.

Then if we wanted to know V1, we would
just bring our 2 amps back up in here.

Let's substitute it into the equations for
V1 and V2,

and you can see that we have 20 volts
across 1 and 100 volt across the other.

Let's just do a quick check,
let's just check our loop equation.

We have 120 volts, so that's -120

volts + 20 + 100 is that equal to 0?

Yes, it is, so
we've been able to check our equation.

Now, what I showed you was
an extremely simple circuit, and

yes, there's an easier
way to solve the circuit.

But let's use it to define a cookbook that
will let us do more complicated problems.

So if I were to set a cookbook for
this I would say,

the first thing I would want to do
is write down all known values.

What was known?

The 120 volts, the 10 ohms,
and the 50 ohms.

Then I would keep track of all the unknown
values as I went throughout my

circuit because I know that I need as
many equations as I have unknowns.

Now, we would write all of the KVL or
loop equations.

An important thing is that each new loop
must pick up at least one new element,

like a new resistor or
a new voltage element.

Also current sources,

which we'll talk about in a minute,
can't be included in any loops.

Fourth, I'd write Ohm's Law.

In fact, I usually prefer to
write Ohm's Law as I go just for

convenience because I know I'm
going to have to apply it anyway.

Ohm's Law is just applying the current and

the resistor together to get the voltage
any time we have a resistor.

Then I would apply as many KCL or

node equations as are needed
to fill in my unknowns.

Each KCL equation must also pick
up at least one new current,

then I'd solve for all the unknowns.

Okay, let's take this cookbook now and
see how we can apply it.

So the first thing we do, and
let's apply it to a problem,

that's just a little bit more
complicated than one we just did.

I'm going to add a current
source I3 here which is 6 amps.

So let's apply this, and as you can see,
we're going to have another element.

So let's just apply each of the steps.

First of all,
let's write all of the known values.

Well, VDC is known, R1, R2 and
I3 are all known values.

Now, let's keep track of our unknowns.

I've already drawn two unknowns here,
I1 and I2.

So we have two unknowns, and
we need to have at least two equations.

So let's write, The loop equation's next.

Well, this same loop equation is the one
that we just wrote and it hasn't

changed at all, even though we've added
another element over here on the right.

And so for the loop equation,
it's -VDC + (I1)(R1),

so here is -VDC + (I1)(R1) + (I2)(R2) = 0.

This is what I mean by
applying Ohm's Law as I go.

I'm just doing + (I1)(R1)
instead of saying + V1, and

later having to go back in and substitute.

Well, we could say each loop needed
to pick up at least one new element.

But which elements did this loop pick up?

It picked up the VDC, the R1 and the R2.

Now, current sources cannot be
counted as new elements, and

in fact, I can't include them in my loops.

So there's no other loop that
I could do in this picture,

because any other loop would have to
go through the the current source.

This is a current source right here, and
I cannot include that in a KVL loop.

Now, let's apply Ohm's Law,
as I went, that's right there.

And then let's apply as many node
equations as needed to fill in

the unknowns.

Well how many unknowns do I have?

Two, how many equations do I have?

One, so I need at least one node equation.

Here's my node right there, and
let's apply the node equation.

Let's say that every current
that's coming in, I1 + I3.

So here's I3, it's coming into that node.

I1 + I3 = I2.

Okay, that's my second equation.

So now, I can solve for
all of my unknowns.

Let's just do the solution,
just one time, for fun.

So here was my VDC, I1, R1, I2,

R2 = 0 and I1 + 6 amps = I2.

So what I'm going to do is substitute
in order to remove the variables.

What I'll do here is I'll
substitute in for I2,

so this equation now
becomes this right here.

Let's solve for
the remaining variable, that's I1.

That gives us the equation that you
see here, and now let's go back and

actually do the math that
tells us that I1 is -3 amps.

And then,
if I go back to my original equation,

in order to find the other variables.

Let's say I want to find I2, I2 = 3 amps.

Now, let's talk about what
it means to have -3 amps and

what it means to have +3 amps.

So right here, when I did my analysis,
I said that this value I1 was a variable.

And when I solved for it,
I discovered that it is -3 amps.

What does that mean?

Well, it means I've guessed the direction
of the current incorrectly.

It means that the current is actually
going from right to left instead of from

left to right, so that I have three
amps going from left to right.

Now, if I had actually put my
arrow this way in the first

place when I did my solution, I would have
found out that I1 was equal to 3 amps.

And I would have had
the correct direction.

It doesn't matter which way
you guess your current.

If you guess it wrong, it's gonna
come up with a negative number for

the current, and that's just fine.

So go ahead and leave it like this, and
just tell me that you have -3 amps for I1.

Okay, let's go one step further.

Let's add one more loop.

So let's apply our cookbook
to this system right here.

First of all,
we need to write all known values.

V1 will be known.

I1 will be known.

R1, R2, and R3.

Then we're going to keep track
of all of our unknown values.

I already have an I1, an I2, and an I3,
that I can see are going to be unknown.

Then we're going to write
all the loop equations.

Let's see how many loops we could do.

Well, here's our first loop.

That's a loop we've been doing all along.

Here is a second loop.

Is there room for another loop?

Can I do a third loop here?

No, I can't, because I have this
current source right there.

Can I do a third loop around this way?

No, I can't,
because that's including a current loop.

Could I do a third loop here?

Well, that's a legitimate loop.

But I can't do that, because it
doesn't pick up any new elements.

My first loop picked up R1 and R2.

My second loop picked up R3.

There isn't anything else to be picked up.

So I am only going to have the two
blue loops that are shown here.

Then I would apply as many KCL or

node equations as are needed
to fill in the unknowns.

So I'm going to have two loop equations,
the blue lines that are shown.

And I'm going to have one node equation
here, where things are orange, and

that's going to fill in my unknowns.

And then we would solve for the unknowns.

Okay, now,
let's take a look at the special node.

So the node remember was right there,
but there was a line or

a short circuit right here.

Any time you have a short circuit, it just
means that you have a really big node.

You don't need to include
two separate nodes.

This is just one really big node.

Okay, now, let's imagine even
adding another node, sorry,

even adding another loop.

So let's talk about how
we might do this problem.

Okay, so
write down all the known values, V1,

R1, R2, R3, R4, and I1.

Those are all known.

Keep track of the unknowns.

They would be I1, I2, and
I3, and let's call this IDC.

I know it says I1, let's call it IDC.

All right, so those are the unknowns.

So we know that we're going to need,
and there's also going to be an I4.

So we're going to have four equations and
four unknowns.

Now, let's see how many loops we can do.

First of all,
we've got our first loop right here.

That pick up V1, R1, and R2.

That's one loop equation.

Here's a second loop equation.

Picking up R1 and R3.

There's at least one new element, the R3.

Can we do a loop here for I4?

No, we can't, because that's going
to include the current source, so

we cannot do that as a loop.

All right, so we have two loop equations,
here's one node equation.

And in fact, this node is actually going
to encompass all of this area where you

can see it short circuited together.

I need at least one more equation and

somehow I need to pick
up R4 in that equation.

So let's see what we would do.

Okay, here is a very good choice for
handling that.

We would do a loop all the way around
the outside like this picking up R4.

So that would give me
three loop equations,

and one node equation for
the four unknowns.

So my problem would end
up looking like this.

3 loops, 1 node would give me 4 equations,
sorry, 4 unknowns.

3 loop equations, 1 node equation
would be 4 unknowns and 4 equations.

Okay, now let's take a look and just kind
of expand our understanding of the nodes.

Could I have used this as
a node equation instead?

Well, the answer is, no, I can't,
because I don't know how much

current is going to be coming
out of this voltage source.

So just like the fact that I cannot
include loops when I have a current

source, I cannot include nodes that
are attaching voltage sources.

So I'm gonna add that information
to my cookbook as well.

Okay, so let's do this problem.

Maybe it looks a little complicated,
but it really isn't.

So let's do each of our loops.

My first loop is the VDC
loop right here which

is going to say minus,
this is the minus plus,

-VDC + I1R1 + I2R2 = 0.

This is my loop 1.

Now let's do loop 2.

And remember, I've got the plus minus
going from my current like this.

So +I3R3, this is I3 + I3R3,

you hear that's all short

circuited minus I1R1 = 0.

Again, let's call this IDC, okay?

And let's do this outside
loop right there.

So I'm gonna say -VDC

+ I3R3 + I4R4 = 0.

Now, let's do the node.

So right here is the big node
that I'm gonna be interested in.

So the node equation is going to say,

all of the currents that come in has to
equal all of the currents that go out.

Or all the currents
coming in are positive,

all the currents going out
are negative and that's equal to 0.

So I3 is coming in, that's positive.

I4 is going out, that's negative.

I1 is coming in, that's positive.

I2 is going out, that's negative.

I also have to include plus IDC,
and that whole thing is equal to 0.

Okay, now let's keep track
of how many unknowns I have.

What are my unknowns?

My unknowns are I1, I2, I3, and I4.

How many equations do I have?

I have th3 ee loop equations and
1 node equation.

So I have four equations and
four unknowns.

Okay let's talk about the math for
this for just a minute.

We can solve this using algebra, and

I know that you've had
some practice doing that.

But I want to show you one specific
method of being able to solve the KCL and

KVL math.

It's gonna be very handy
when we apply it in MATLAB.

Any time we have four equations like this,
and

four unknowns, we can write them as
a matrix, as I'm gonna do right here.

So let's take our first equation.

Here's our first equation right there.

So the unknowns are gonna be I1,
I2, I3, and I4.

Anything that does not have an unknown
moves over to the right-hand side.

So we have +VDC on the right-hand side.

Anything that does have an unknown,
I've put the multiplier, like R1 times I1.

See here's R1 times I1 and R2 times I2.

That's right here, R2 times I2.

Okay, and then there are no I3's and
there are no I4's in that equation.

So they are 0 and 0.

Let's go to our next equation.

Our next equation right here says, there
is nothing that doesn't have an unknown.

So I have 0 over here
on the right hand side,

then I have -R1 times I1.

I have nothing times R2 and

I have R3 times I3 and nothing times R4.

Okay, let's go down to the third equation.

The third equation has VDC,
which has no unknowns.

So let's bring it over thee there's +VDC.

Do I have anything times I1?

No, I2?

No, I3?

Yes, R3?

Yes, I4?

Yes, R4?

So there's the matrix, and
then this very last one,

what is over here that doesn't
have a unknown in front of it?

Take that over, that's gonna become -IDC,

what do we have times I1,
a +1, I2-1, I3 + 1, I4-1.

So here's my matrix.

Here's my vector of unknowns.

Here's my vector of constants.

Then so
this is writing the matrix equation for

this particular circuit element.

Next, you can use Gaussian elimination or
MATLAB matrix solution.

So I'm going to leave this here.

We're not actually going to
do the algebra to finish it.

Because in general, I'm going to assume,
you'd use MATLAB to solve this problem.

So in short,
we've talked about two important laws.

One is Kirchhoff's Current Law that
says the sum of all the currents

entering the node = 0.

The second law is
the Kirchhoff's Voltage Law or KVL,

that says the sum of the voltage is
around a closed path or loop is = 0.

Finally, we wrote down a cookbook for
the KCl and kvl.

The cookbook is this.

First, write down all the known values.

Keep track of how many
unknown values you have,

write all the loop equations
that you can write.

Remember that each loop needs to
pick up at least one new element and

that current sources can't be in loops.

Then apply Ohm's law which I usually
choose to do as I go for convenience.

And that is equal to the voltage is equal
to the current times the resistance.

And then apply as many nodes as you
need in order to fill in the unknowns.

Each KCL equation must pick
up at least one new current.

Shorts as they are here can be used
to combine nodes into a single node,

and nodes can't touch voltage sources.

Then you solve for
the unknowns using matrix math.

So the picture that's on
the front that's in Red Canyon.

That's near the bottom of
Thunder Mountain, near Bryce Canyon, Utah.