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KENDRA PUGH: Hi.

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Today, I'd like to talk
to you about circuits.

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Last time, we finished up the
LTIs, and signals, and

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systems, where we learned how to
both model existing systems

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and predict their long-term
behavior.

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But we haven't forayed into how
to actually create systems

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in the physical world.

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We've created some amount of
systems in software and made

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some brains for our robots.

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But if we want to make something
in the physical

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world, then we probably have to
come up with ways to model

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physical systems or use
physical components.

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That starts our new
model on circuits.

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Circuits are going to be our
first foray into designing

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systems in the physical world,
also designing systems using

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physical components.

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It's worth mentioning now that
the information that you learn

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about circuits is good for more

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things than even circuits.

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You can use basic circuit
diagrams and properties of

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circuits to model all sorts of
kinds of systems, especially

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ones in the human body--

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circulatory system, neurological
system, different

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kinds of fluid flow,
that kind of thing.

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In the next few videos, we'll
go over how to represent

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circuits, and also cover some of
the basic methods by which

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people solve circuits.

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We'll also introduce an element
called an op-amp, and

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use that element in order to
enable us to do things like

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modularity and abstraction
from our circuits.

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First, let's talk about
representation.

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In the general sense, when you
come across a circuit diagram,

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you're going to see--

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at the very broad level--

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a bunch of elements and a bunch
of connections between

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the elements.

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Those things will form
loops and nodes.

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If you don't actually specify
the elements, then your

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circuit diagram actually
looks a whole lot

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like a block diagram.

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And in fact, block diagrams and
circuit diagrams are very

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closely related in part because
block diagrams are

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used to model feedback systems,
which frequently are

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implemented using circuits.

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In this course, we're going to
be focusing on independent

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sources and resistors as the
two major kinds of elements

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that we'll use in
our circuits.

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We'll also use things like
potentiometers, which are

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resistors that you can
adjust, and op-amps.

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And we'll look at op-amps
specifically in a later video.

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But I have one drawn up here
just so you recognize it when

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you see it written.

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Note that it looks a whole lot
like the block diagram symbol

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for a gain.

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And that's intentional, and
we'll cover that later.

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But in the meantime, the other
sources that we're going to be

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using are independent current,
and voltage sources.

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We're going to use resistors
to adjust the amount of

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voltage and current that we're
actually dealing with and then

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sample either the current or
the voltage at a particular

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point in our circuit
to get the desired

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values that we're after.

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On a circuit diagram, when
you're interested in the

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voltage drop across a particular
element, you'll

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indicate it by putting a
plus and minus sign.

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This also indicates
the directionality

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of the voltage drop.

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Likewise, when you're interested
in the current

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flowing through a particular
element, you'll usually see an

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indication of it by labeling the
current i, and then maybe

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i with some sort of subscript,
and an arrow indicating the

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direction of current flow
through that element so that

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you avoid making sign errors
with the person that might be

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reading or writing
your diagram.

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A quick note here.

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This is the reason that
electrical engineers use j to

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symbolize values in
the complex plane.

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It's because i is used
in particular

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for values of current.

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Let's review Kirchhoff's voltage
laws and Kirchhoff's

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current laws.

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You've probably covered this
in 8.02, electricity and

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magnetism, or possibly in
an AP physics class.

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But we're going to go over
it really fast right now.

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Kirchhoff's voltage law is that
the voltage drop around a

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loop is equal to 0.

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Or if you take the voltage drop
across a particular loop

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in your circuit, the sum
of those voltage drop

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is going to be 0.

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Let's demonstrate
on this diagram.

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Or, I'll demonstrate
on this diagram.

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Say the voltage drop
across this element

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is equal to V, right?

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Doesn't matter what it is.

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We're going to stick
with that.

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The voltage drop across these
elements, if I were to move

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around this loop, is
going to sum to 0.

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Note that if I'm tracing out
my voltage drop across this

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loop, I'm actually moving
through this voltage source in

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the direction opposite of
its indicated potential.

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So when I move through this
voltage source, I'm going to

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account for its value as
negative V. As I work my way

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around the rest of the circuit,
the voltage drop

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across these elements is
going to sum to V.

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This is true for all loops
in my circuit.

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So any loop that includes V, the
elements I encounter as a

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consequence of moving around
that loop are going to have

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voltage drop equal and opposite
to the value I get by

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moving through V in
this direction.

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This loop counts, too, but it
doesn't include V. All this

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loop tells me is that the
voltage drop across this

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element is equivalent
to the voltage

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drop across this element.

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Or, the voltage drop in this
direction across that element

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is equal to the voltage
drop in this

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direction across this element.

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That's Kirchhoff's
voltage law.

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Kirchhoff's current law is that
the current flow into a

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particular node is equal to 0.

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Or, if you take all of the
current flows in and out of a

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particular node and sum them,
they should sum to 0.

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I've actually got the
same set up here.

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I'm not going to use
a current divider.

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I'm interested in the current
flowing over this element.

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It's actually the same as the
current flowing over this

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element because resistance
doesn't change current,

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resistors flowing through
a resistor should

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not change the current.

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So this is still the same i.

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Here's my node.

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The current flowing in this
direction and in this

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direction, if I took the linear
combination of these

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two currents, they would be
equal in value to the current

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flowing into this node.

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When I'm looking at the current
flowing through a

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particular node, I
pick a direction.

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It's usually arbitrary.

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I pick a direction.

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It's arbitrary which
direction I pick.

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Typically, you pick currents
flowing into the

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node as being positive.

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I sum up all the currents, and
I set that equal to 0.

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So in this case--.

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Or--

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pretty simple.

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Let's practice on this
particular circuit.

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One thing to note is that when
you're solving circuits in the

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general sense, both when you
want TA help and when you're

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solving for a mid-term and want
partial credit, you want

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to label all of your nodes, all
of your elements, and all

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of the currents that you're
interested in solving.

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See, I've got my voltage drop
across this resistor, this

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resistor, and this resistor
labeled, as well as these

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currents, which I'll also
be solving for.

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The first thing that I would
do when approaching this

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problem is attempt to reduce
this circuit to something that

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is a little bit simpler.

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The first thing that I'm going
to do is try to figure out how

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to change these two resistors
in parallel into a single

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resistor and still have
an equivalent circuit.

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That'll allow me to
solve for I1.

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There will be 0 nodes
in my system.

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I'll just have one
single loop.

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And the current through the
system will just be V/R.

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So if I'm just looking at these
two resistors, I have

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resistors in parallel.

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In the general sense, the way
to solve for resistors in

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parallel is to take
the inverse of the

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sum of their inverses.

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When you only have two
resistors, you can typically

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cheat by saying that this
is equal to their

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product over their sum.

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I'm going to redraw my current
understanding of the circuit.

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The other stuff that I've saved
myself is that because

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these resistors are
in parallel,

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they're a current divider.

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They take the current in and
divide it two ways determined

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by the ratio between
these two values.

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The thing I'm actually
interested in expressing is

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that V2 and V3 are
the same value.

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When you have a current divider,
the voltage drop

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across all elements in the
current divider are the same.

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So the value of V here is going
to be both V2 and V3.

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2R plus 6/5 R. I'm going to
go with 16/5 R for now.

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I've solved for I.

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At this point, I have a voltage
divider, which means

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that the current flowing through
this part of the

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system is going to
be the same.

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But the voltage drop across
this element versus this

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element is going to be
proportional to the ratio

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between these two values.

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V1 is going to be the amount
of the total resistance in

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this simple circuit that this
resistor contributes over the

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entire resistance
in the system.

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Or, 10/5 R over 16/5 R, which is
10/16 R, or 5/8 R. Or, it's

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going to be 5/8 V.

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Same thing happens with V2.

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Note that these two values
should sum to V in order to

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maintain Kirchoff's
voltage law.

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We've also found V3.

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So the two things that we have
to find are I2 and I3.

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Here, I've just done
Kirchoff's current

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law for this node.

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Because I'm working with a
current divider, I can break

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up the total current flowing
into that node into the number

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of parts equal to the sum
of these values and then

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distribute them.

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And then, that's
[? inappropriate ?]

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given that less resistance
means more current.

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What do I mean by that?

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Well, I mean that here my
current is equal to 5/16 V

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over R. I2 is going to be equal
to this value over the

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sum of these two values
times I1.

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Likewise--.

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And just to simplify--.

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That concludes my tutorial
on circuits.

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Next time, we'll talk about
other ways we can solve this

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circuit, and then we'll end
up talking about op-amps.