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.
