Welcome back. Let's do some more linear circuits.
Last time we talked about this idea of resistance and looking at individual
elements. Now we're going to start putting them
together, and we're going to be discussing Kirchhoff's Laws.
So we're going introduce Kirchhoff's Current Law and Kirchhoff's Voltage Law,
and then we'll use it to solve a very simple circuit.
So as I mentioned before in our previous lesson we introduced Ohm's law.
We introduced this idea of resistance and then we calculated it.
And then we found this way of relating the voltages and currents within devices.
So now we're going to look at between devices in a system perspective.
After we've covered this, were going to then be able to start talking about
resistors as actual devices that are intended to exhibit Ohm's Law.
So the objectives of this lesson are to first of all have you be able what
Kirchoff's current law and Voltage law are, and describe this voltage
relationship parallel elements. And current relationship of series
elements as well as to use Kirhhoff's Law to find unknown values in a simple
circut. Kirchoff's Voltage Law basically states
it is, that you make your way around any closed path in a circuit summing up your
voltages. Now the sum has to be zero.
And this is because if you start in one place [INAUDIBLE] And you take a trip
around and then you come back to where you started.
You can't have gained or lost any, kind of, electrical elevation.
So as an example, if I have this pathway here, I can sum up all of my various
voltages. VA, VG, VH and VD.
Because this makes a closed circuit and then when I get back to the beginning.
The total sum of the voltage has to be zero.
And you can go in any direction you want, this one happens to be clockwise, but I
can also do it counterclockwise. Or in this case just flip the signs
rather. And so now instead of having the arrows
go from the pluses to the minuses; now they go from the minuses to the pluses.
So if I sum up the a, the c, the f, the h, and the e, they still all have to
equal 0. And this would be equivalent to, if I
instead changed the direction. The E, the H, the F, the C, and the A.
You see that these are 2 equivalent ways of going around and looking at the
circle. So if I look at parallel circuts.
And they make a nice little loop here. If I do my Kirchhoff's law, I see that VA
minus VB has to be equal to 0. And this is because general trend is
you're following the arrow. You can either choose to use the first
sign that you see or the second sign that you see.
So here I used the first sign. So I used plus vA and minus vB.
When I do that, I could actually put the vB on the other side, and I discover that
anytime you have two parallel elements, Which is to say two elements that connect
the same two nodes that they have to have the same voltage across them due to
Kirchhoff's current, or Kirchhoff's voltage law.
For Kirchhoff's current law we're going to be looking at the currents.
And Kirchhoff's current law states that the sum of all the currents going into a
node must equal the sum of the currents going out of the node.
And remember we've already kind of discussed nodes as being the set of
interconnected wires. So if I color the blue node here and I
give some ref instructions I get the equation that ia plus ib must equal id
plus ie because ia and ib are going in id and e are going out.
I'll look at another node this orange node.
I see that IB plus IF, the two going out must equal IC, the one going in.
Let's look at physical descriptions to why Kirchhoff's Current Law is so
important. What if I had this system, where I have a
one mili amp. Current source and I have two nodes
separated by this one centimeter of distance and I have current that's now
flowing through the system and Kirkov's current law does not hold but that means
that I'm going to be building up charge here as I'm pushing You know pulling
electrons, and pushing electrons through. So this is going to get a charge at a
rate of ten to the negative third times T cololms, so T being in seconds.
But on the other note, these, this charge has to come from somewhere.
It's coming from Q 2. So we get negative ten to the third times
T cololms. And then I can use coulombs law,
remembering that KE is equal to this 8.987 times 10 to the ninth newton meters
squared per coulombs squared, and plugging in these values, I discover that
the strength of the attractive force between These two points is equal to
89.87 times t squared meganewtons. And so let's see what's going on here.
In the first millisecond I have 20 pounds of pressure, or attractive force, pulling
these two together. After one second it's twenty million
pounds, and then two seconds, eighty million pounds, or kilogram.
They're all listed over here. So there's a tremendous amount of force
that's going to be pulling these things together.
And it's this electromagnetic force that motivates Kirchoff's law and why it's so
important. If I place two devices in series in this
configuration, it's possible for us to find another relation, similar to like we
did with the parallel circuits. And when we say series, that means that
there's two devices connected to each other, but there's nothing else
connecting in that middle area. Because they're in series, the currents I
a must equal i b because of Kirchhoff's Current Law.
The current flowing in to that center node, here, has to be equal to the
current flowing out. Now we're going to solve a simple
problem. We have a few nodes and from the things
we have learned so far, we know that P is equal to i times v.
We also have Kirchhoff's Current Law and Kirchhoff's Voltage Law.
So let's find the unknown values. First of all, at A, we have one amp of
current flowing through it, and we have three watts.
The arrow goes from the plus to the minus, which means that, because p equals
i times v, this must be 3 volts. Now I can use Korkov's Voltage Law, and
make this big loop around the outside here, so I get minus 3 volts plus vB plus
5 is equal to 0. As long as the vB is minus 2 volts.
Using Kirchoff's current law we see that, since we have 1 amp of current flowing
this way, and A and B are in series but the arrows now point in opposite
directions. This current must be minus one m.
D and C are two devices that are in parallel with each other and so
Kirchoff's voltage law states that they have to have the same voltage.
That's 5 volts. We also know the power in c is five watts
the arrow goes from the plus to the minus so p equals i times b.
That means that the current flowing through here must be one ampere.
Because we have minus one ampere coming here and one ampere going out here that
would be the same as having. One ampere flowing this way.
And Kurcoff's current law states that the sum of that, that, and that have to be
zero, which means that this current is minus two amps.
So finally the two unknown powers pb Is now equal two [UNKNOWN] plus to the
minus, so minus two plus minus one...two watts...and Pd we have they are going
from the plus to the minus so that equals I times V, 5 times -2 is -10 watts And if
I add all of these up, 3 plus 5, plus 2, plus -10, is 0, which means we have
conservation of power, and that gives us some confidence that our answer is
correct. Finally remembering that if they're
positive powers, we say that they're consuming power, and negative that
they're Generating. This is a source, and the others are
consuming power. And so we've been able to solve all of
the voltages, currents, and powers, for this simple example.
So to summarize, we've covered Kirkov's current law, voltage law.
We applied Kirkov's voltage law, to show that parallel elements have the same
voltage. And use Kirchhoff's Current Law to
identify that series elements have the same currents.
We gave an interesting justification for Kirchhoff's Current Law.
And then solved a simple circuit by using Kirchhoff's Laws and, as well as, the
Power Law that we learned before. In our next lesson we'll be introducing
resistors, that are devices that are specifically intended.
To exhibit Ohm's Law and inhibit current flow.
And then we'll apply Ohm's Law and Kirchhoff's laws together to some
resistor circuits. As always, if you've had any questions on
this material, go to the forums. I look forward to seeing you there and
seeing you in our next lesson. Until then.