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Last video, we saw what
happens when we have

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resistors in series.

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Now let's see what happens
when we have

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resistors in parallel.

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All right, let me pick
a new color.

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New color will be magenta.

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There's my battery: positive,
negative.

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There's my ideal conducting
wire.

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Here's my ideal conducting wire,
but then-- and this is

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new-- it branches off, and
I have two resistances.

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I have one here.

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And that's another resistance.

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And let's say that this has a
resistance R1, this has a

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resistance R2.

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And, of course, the
non-intuitive convention is

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that the current flows from the
positive to the negative

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terminal, but we know that the
electrons are actually flowing

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in the other direction.

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And I want to keep saying that,
because I think it's so

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important to understand what's
actually happening as opposed

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to the convention.

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Well, anyway, in the previous
video, we said, well, when we

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have devices or components in
series, that the current

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through the entire circuit
is constant.

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But let's think about
what happens here.

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So we have these electrons--
actually, let's think about it

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from the electron flow.

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The electrons are flowing,
flowing, flowing at a given

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rate, and then here they
have a choice.

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Some of them can take this top
path, some of them can take

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this bottom path.

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So if you think about it, the
flow of electrons in this

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branch plus the flow of
electrons in that branch have

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to add up to the flow
of the electrons in

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this branch, right?

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And then they're going to meet
back up, and then the flow of

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electrons here-- so if we think
of it this way, and now

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I'm going to go back to the
convention, this is I1.

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So you have these electrons
flowing at a given rate.

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This is the current
right here.

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They're going to branch off,
and maybe half of them go--

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we'll see if the resistances
are equal, if both of these

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branches have an equal amount
of capacity in terms of how

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fast the electrons
can flow through.

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If they're equal, or since we're
going to current in this

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direction, let's talk about
positrons or positive charges.

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If the positive charges--
although I just want to keep

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saying that it is
not the positive

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charges that are moving.

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It's the electrons.

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But if you say that the lack of
electrons can flow equally

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easily between both paths,
that's if the resistances were

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the same, we could imagine that
the current, the flow,

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would split itself up,
and then over here

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would meet back up.

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And then we would say that the
current here would also be I1.

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But let's figure out where
the current's going.

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Let's call this current I2 and
let's call this current I3.

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So I think it is reasonable, and
you can imagine with water

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pipes or anything, that the
current going into the branch

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is equal to the current
exiting the branch.

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Or you could even think of it
that the current entering--

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when the currents I2 and I1
merge, that they combine and

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they become Current 1, right?

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I mean, think about it.

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In a given second, if this is
5 coulombs per second-- I'm

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just making up numbers-- and
this is 6 coulombs per second,

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in a given second right here,
you're going to have 5

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coulombs coming from this branch
and 6 coulombs coming

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from this branch, so you're
going to have 11 coulombs per

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second coming out once they've
merged, so this would be 11

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coulombs per second.

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So I think hopefully that makes
sense to you that this

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current is equal to the
combination of this current

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and that current.

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Now, what do we also know?

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We also know the voltage along
this entire ideal wire is

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constant, so then voltage-- let
me draw that in another

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color, in blue.

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So, for example, the voltage
anywhere along this blue that

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I'm filling in is going to be
the same, because this wire is

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an ideal conductor, and you can
almost view this blue part

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as an extension of the positive

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terminal of the battery.

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And very similarly-- I'll do it
in yellow-- we could draw

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this wire as an extension
of the negative

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terminal of the battery.

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This is an extension
of the negative

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terminal of the battery.

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So the voltage difference
between here and here-- so

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let's call that the total
voltage, or let's just call it

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the voltage, right?

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The voltage difference between
that point and that point is

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the exact same thing as the
voltage difference between

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this point and this point, which
is the exact same thing

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as the voltage difference
between this

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point and this point.

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So what can we say?

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What is the total current
in the system?

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If we just viewed this as a
black box, that this is some

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type of total resistance, well,
the total current in the

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system would be the total
voltage, the voltage divided

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by-- let's call this our total
resistance, right?

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Let's say we couldn't see this
and we just said, oh, that's

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just some total resistance,
and that is equal to the

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current going through R1.

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This is I1.

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This is a 1 right here.

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This is current I1.

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What's current I1?

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Well, it's going to be the
voltage across this resistor

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divided by the resistance,
right?

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That's what Ohm's law tells
us: V is equal to IR, or

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another way we could say it is V
over R is equal to I, right?

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So I1 is equal to the voltage
across this resistor, but we

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just said that voltage
is the same thing as

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this voltage, right?

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The voltage here is the same
thing as the voltage here.

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The voltage here is the same
thing as the voltage here.

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So the voltage across that
resistor is still V, and so

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the current flowing across that
resistor is V over R1.

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And the same logic:
what is I2?

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I2 is this current.

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What is the voltage across
this device?

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Well, that's just
V again, right?

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It's the same thing as the
voltage across this device, so

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it's V over R2 by Ohm's law.

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Well, all these V's are the
same, so we can divide both

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sides of that equation by V,
and we get 1 over the total

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resistance is equal to 1
over R1 plus 1 over R2.

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And you could make that argument
if we had an R3 here.

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Let's say that we had another
device, and that is R3.

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You could use the exact same
argument, and you would have a

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plus 1 over R3.

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And if you had Rn or 10 of them,
you'd just keep that 1

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over R4, R5, et cetera.

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So let's see if we can use
this information we have

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learned to actually solve a
problem, and I actually find

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that it's always easier to
solve a problem than to

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explain the theory
behind a problem.

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You'll see that with most of
these circuit problems, it's

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actually very basic
mathematics.

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So let's say I have a 16
volt battery plus,

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minus, it's 16 volts.

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And just to hit the point home
that you always don't have to

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draw circuits the same, although
it is nice if you're

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actually drawing complicated
circuits, I could

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draw it like this.

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I could draw the circuit like
this, and let's say that

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there's a resistor here.

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And then let's say there's a
wire and then there's another

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resistor here, and that this
decides to do some random

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loopy thing here and that they
connect here, and that they

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come back here.

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This strange thing that I have
drawn, which you will never

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see in any textbook, because
most people are more

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reasonable than me, is the exact
same-- you can almost

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view it topologically as the
exact same circuit as what I

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drew in the previous diagram,
although now I will assign

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numbers to it.

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Let's say that this resistance
is 20 ohms and let's say that

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this resistance is 5 ohms. What
I want to know is, what

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is the current through
the system?

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First, we'll have to figure
out what the equivalent

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resistance is, and then we could
just use Ohm's law to

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figure out the current
in the system.

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So we want to know what the
current is, and we know that

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the convention is that current
flows from the positive

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terminal to the negative
terminal.

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So how do we figure out the
equivalent resistance?

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Well, we know that we just
hopefully proved to you that

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the total resistance is equal to
1 over this resistor plus 1

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over this resistor.

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So 1 over-- I won't
keep writing it.

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What's 1 over 20?

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Well, actually, let's just
make it a fraction.

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So it's 1/20 plus-- 1/5
is what over 20?

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That's 4/20, right?

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So 1 over our total resistance
is equal to 5/20, which is

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equal to what?

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1/4.

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So if 1/R is equal to 1/4,
R must be equal to 4.

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R is equal to 4 ohms.

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So we could redraw this
crazy circuit as this.

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I'll try to draw it
small down here.

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We could redraw this where this
resistance is 4 ohms and

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this is 16 volts.

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We could say that this whole
thing combined is really just

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a resistor that is 4 ohms. Well,
if we have a 16-volt

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potential difference, current
is flowing that way, even

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though that's not what the
electrons are doing.

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And that's what our
resistance is, 4

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ohms. What is the current?

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V equals IR, Ohms law.

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The voltage is 16 volts.

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It equals the current times 4
ohms. So current is equal to

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16 divided by 4, is
equal to 4 amps.

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So let's do something
interesting.

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00:10:17,150 --> 00:10:20,640
Let's figure out what the
current is flowing through.

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00:10:20,640 --> 00:10:21,470
What's this?

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00:10:21,470 --> 00:10:27,330
What's the current I1 and
what's this current I2?

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00:10:27,330 --> 00:10:29,430
Well, we know that the potential
difference from here

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00:10:29,430 --> 00:10:31,330
to here is also 16
volts, right?

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00:10:31,330 --> 00:10:35,530
Because this whole thing is
essentially at the same

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00:10:35,530 --> 00:10:37,420
potential and this whole thing
is essentially at the same

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00:10:37,420 --> 00:10:40,990
potential, so you have 16
volts across there.

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00:10:40,990 --> 00:10:46,610
16 volts divided by 20 ohms,
so let's call this I1.

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00:10:46,610 --> 00:10:54,830
So I1 is equal to 16 volts
divided by 20 ohms, which is

217
00:10:54,830 --> 00:10:55,300
equal to what?

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00:10:55,300 --> 00:10:56,660
4/5.

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00:10:56,660 --> 00:11:02,820
So it equals 4/5 of an ampere,
or 0.8 amperes.

220
00:11:02,820 --> 00:11:05,440
And similarly, what is the
amount of current flowing

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00:11:05,440 --> 00:11:06,810
through here?

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00:11:06,810 --> 00:11:07,590
I2?

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00:11:07,590 --> 00:11:09,110
I'm going to do this in
a different color.

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It's getting confusing.

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00:11:11,546 --> 00:11:13,030
I'll do it in the
vibrant yellow.

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So the current flowing through
here, once again, the

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potential difference from here--
that's not different

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enough-- the potential
difference from here to here

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is also 16 volts, right?

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00:11:22,190 --> 00:11:26,240
So the current is going to be
I2, is going to be equal to

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00:11:26,240 --> 00:11:35,390
16/5, which is equal
to 3 1/5 amps.

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00:11:35,390 --> 00:11:37,730
So most of the current is
actually flowing through this,

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00:11:37,730 --> 00:11:40,530
and that makes sense because the
resistance is less, right?

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00:11:40,530 --> 00:11:42,230
So that should hopefully give
you a little bit of intuition

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of what's going on.

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00:11:42,980 --> 00:11:45,910
And less current is flowing
through here, so I2 through

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00:11:45,910 --> 00:11:53,630
the 20-ohm resistor is 0.8 amps
is I1, and I2 through the

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00:11:53,630 --> 00:11:58,250
5-ohm resistor is equal
to 3.2 amps.

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00:11:58,250 --> 00:12:00,740
And it makes sense that when
you add these two currents

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00:12:00,740 --> 00:12:04,560
together, the 3.2 amperes
flowing through here and the

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00:12:04,560 --> 00:12:07,000
0.8 amperes flowing through
here, that when they merge,

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they merge and then you have 4
amperes flowing through there.

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00:12:10,840 --> 00:12:14,280
Anyway, hopefully, I have given
you some intuition on

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00:12:14,280 --> 00:12:17,850
what happens when we put
resistors in parallel.

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I will see you in
the next video.

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