KENDRA PUGH: Hi.
Today, I'd like to talk to you about circuits.
Last time, we finished up the LTIs, and signals, and
systems, where we learned how to both model existing systems
and predict their long-term behavior.
But we haven't forayed into how to actually create systems
in the physical world.
We've created some amount of systems in software and made
some brains for our robots.
But if we want to make something in the physical
world, then we probably have to come up with ways to model
physical systems or use physical components.
That starts our new model on circuits.
Circuits are going to be our first foray into designing
systems in the physical world, also designing systems using
physical components.
It's worth mentioning now that the information that you learn
about circuits is good for more
things than even circuits.
You can use basic circuit diagrams and properties of
circuits to model all sorts of kinds of systems, especially
ones in the human body--
circulatory system, neurological system, different
kinds of fluid flow, that kind of thing.
In the next few videos, we'll go over how to represent
circuits, and also cover some of the basic methods by which
people solve circuits.
We'll also introduce an element called an op-amp, and
use that element in order to enable us to do things like
modularity and abstraction from our circuits.
First, let's talk about representation.
In the general sense, when you come across a circuit diagram,
you're going to see--
at the very broad level--
a bunch of elements and a bunch of connections between
the elements.
Those things will form loops and nodes.
If you don't actually specify the elements, then your
circuit diagram actually looks a whole lot
like a block diagram.
And in fact, block diagrams and circuit diagrams are very
closely related in part because block diagrams are
used to model feedback systems, which frequently are
implemented using circuits.
In this course, we're going to be focusing on independent
sources and resistors as the two major kinds of elements
that we'll use in our circuits.
We'll also use things like potentiometers, which are
resistors that you can adjust, and op-amps.
And we'll look at op-amps specifically in a later video.
But I have one drawn up here just so you recognize it when
you see it written.
Note that it looks a whole lot like the block diagram symbol
for a gain.
And that's intentional, and we'll cover that later.
But in the meantime, the other sources that we're going to be
using are independent current, and voltage sources.
We're going to use resistors to adjust the amount of
voltage and current that we're actually dealing with and then
sample either the current or the voltage at a particular
point in our circuit to get the desired
values that we're after.
On a circuit diagram, when you're interested in the
voltage drop across a particular element, you'll
indicate it by putting a plus and minus sign.
This also indicates the directionality
of the voltage drop.
Likewise, when you're interested in the current
flowing through a particular element, you'll usually see an
indication of it by labeling the current i, and then maybe
i with some sort of subscript, and an arrow indicating the
direction of current flow through that element so that
you avoid making sign errors with the person that might be
reading or writing your diagram.
A quick note here.
This is the reason that electrical engineers use j to
symbolize values in the complex plane.
It's because i is used in particular
for values of current.
Let's review Kirchhoff's voltage laws and Kirchhoff's
current laws.
You've probably covered this in 8.02, electricity and
magnetism, or possibly in an AP physics class.
But we're going to go over it really fast right now.
Kirchhoff's voltage law is that the voltage drop around a
loop is equal to 0.
Or if you take the voltage drop across a particular loop
in your circuit, the sum of those voltage drop
is going to be 0.
Let's demonstrate on this diagram.
Or, I'll demonstrate on this diagram.
Say the voltage drop across this element
is equal to V, right?
Doesn't matter what it is.
We're going to stick with that.
The voltage drop across these elements, if I were to move
around this loop, is going to sum to 0.
Note that if I'm tracing out my voltage drop across this
loop, I'm actually moving through this voltage source in
the direction opposite of its indicated potential.
So when I move through this voltage source, I'm going to
account for its value as negative V. As I work my way
around the rest of the circuit, the voltage drop
across these elements is going to sum to V.
This is true for all loops in my circuit.
So any loop that includes V, the elements I encounter as a
consequence of moving around that loop are going to have
voltage drop equal and opposite to the value I get by
moving through V in this direction.
This loop counts, too, but it doesn't include V. All this
loop tells me is that the voltage drop across this
element is equivalent to the voltage
drop across this element.
Or, the voltage drop in this direction across that element
is equal to the voltage drop in this
direction across this element.
That's Kirchhoff's voltage law.
Kirchhoff's current law is that the current flow into a
particular node is equal to 0.
Or, if you take all of the current flows in and out of a
particular node and sum them, they should sum to 0.
I've actually got the same set up here.
I'm not going to use a current divider.
I'm interested in the current flowing over this element.
It's actually the same as the current flowing over this
element because resistance doesn't change current,
resistors flowing through a resistor should
not change the current.
So this is still the same i.
Here's my node.
The current flowing in this direction and in this
direction, if I took the linear combination of these
two currents, they would be equal in value to the current
flowing into this node.
When I'm looking at the current flowing through a
particular node, I pick a direction.
It's usually arbitrary.
I pick a direction.
It's arbitrary which direction I pick.
Typically, you pick currents flowing into the
node as being positive.
I sum up all the currents, and I set that equal to 0.
So in this case--.
Or--
pretty simple.
Let's practice on this particular circuit.
One thing to note is that when you're solving circuits in the
general sense, both when you want TA help and when you're
solving for a mid-term and want partial credit, you want
to label all of your nodes, all of your elements, and all
of the currents that you're interested in solving.
See, I've got my voltage drop across this resistor, this
resistor, and this resistor labeled, as well as these
currents, which I'll also be solving for.
The first thing that I would do when approaching this
problem is attempt to reduce this circuit to something that
is a little bit simpler.
The first thing that I'm going to do is try to figure out how
to change these two resistors in parallel into a single
resistor and still have an equivalent circuit.
That'll allow me to solve for I1.
There will be 0 nodes in my system.
I'll just have one single loop.
And the current through the system will just be V/R.
So if I'm just looking at these two resistors, I have
resistors in parallel.
In the general sense, the way to solve for resistors in
parallel is to take the inverse of the
sum of their inverses.
When you only have two resistors, you can typically
cheat by saying that this is equal to their
product over their sum.
I'm going to redraw my current understanding of the circuit.
The other stuff that I've saved myself is that because
these resistors are in parallel,
they're a current divider.
They take the current in and divide it two ways determined
by the ratio between these two values.
The thing I'm actually interested in expressing is
that V2 and V3 are the same value.
When you have a current divider, the voltage drop
across all elements in the current divider are the same.
So the value of V here is going to be both V2 and V3.
2R plus 6/5 R. I'm going to go with 16/5 R for now.
I've solved for I.
At this point, I have a voltage divider, which means
that the current flowing through this part of the
system is going to be the same.
But the voltage drop across this element versus this
element is going to be proportional to the ratio
between these two values.
V1 is going to be the amount of the total resistance in
this simple circuit that this resistor contributes over the
entire resistance in the system.
Or, 10/5 R over 16/5 R, which is 10/16 R, or 5/8 R. Or, it's
going to be 5/8 V.
Same thing happens with V2.
Note that these two values should sum to V in order to
maintain Kirchoff's voltage law.
We've also found V3.
So the two things that we have to find are I2 and I3.
Here, I've just done Kirchoff's current
law for this node.
Because I'm working with a current divider, I can break
up the total current flowing into that node into the number
of parts equal to the sum of these values and then
distribute them.
And then, that's [? inappropriate ?]
given that less resistance means more current.
What do I mean by that?
Well, I mean that here my current is equal to 5/16 V
over R. I2 is going to be equal to this value over the
sum of these two values times I1.
Likewise--.
And just to simplify--.
That concludes my tutorial on circuits.
Next time, we'll talk about other ways we can solve this
circuit, and then we'll end up talking about op-amps.