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Mujtaba: Today, we're going
to learn about Ohm's Law.

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Mujtaba: I'm curious how
many of you have heard of

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or know about Ohm's Law?

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Student: I know.

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Student: I do.

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Student: I've yeah, yeah.

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I've heard about it.

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Yeah.

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Mujtaba: So the simplest way it
is represented is V equals IR.

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That is, the voltage drop,
or the potential drop

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across a resistor
in this case is

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just the current that
passes through the resistor

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times its current.

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So if we were to draw a resistor
right here with a value R,

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and this is the current,
then the voltage drop

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across this resistor, this V
is equal to the current times

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that resistor value that the
resistor has so the resistance

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value of the resistor
and the current path

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that passes through it.

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Current is the flow of electrons
and V is the potential.

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So think of it as a
force essentially,

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that allows or enables
these electrons

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to move from an area
of higher potential

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to low lower potential.

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So if you guys have drunk
boba bubble tea before,

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what happens is in
the cup, you guys

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have these bubbles
at the bottom.

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And when you try to at
the end, once you're

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done with your
drink, you're trying

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to suck these bubbles up.

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The when you're trying
to suck these bubbles out

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of this straw, the pressure
that you're putting on to it

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is essentially analogous
to this voltage

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and the flow of these
bubbles in this straw

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is essentially like the
current and the resistance

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is essentially the friction of
these bubbles with this straw.

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And that is the
pressure you will

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need to overcome the
flow of these bubbles

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in this straw against this
friction to to get it.

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This is essentially
analogous to that.

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So what we're doing
is essentially this.

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Oh, sorry about that.

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This concept is
pretty much described

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in almost every high school
and high school physics

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and early college physics.

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But a lot of people have a hard
time using this in real life.

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So one of the ways we can use it
is if we have a simple circuit.

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Mujtaba: So if we
have a simple circuit,

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and we have an LED here.

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Mujtaba: So
essentially, these are

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little light
bulb-looking things that

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emit light when you drive a
current through them so they

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light up.

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So in a circuit like
this, what happens

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is we have a potential drop
or voltage source right here.

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Let's say it is two volts.

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And there's a
resistor right here.

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And we need a current
to flow through this LED

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to light it up.

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So right here I have
a red LED that I'll

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show you guys in a minute.

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But red LEDs have a current
rating of 20 milliamps.

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And in order to make a
circuit where we can light up

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this red LED, we
need to know what

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should be this resistor value
for the current in the circuit

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to be exactly this current
that this LED needs to turn on.

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Mujtaba: Okay, so we will
be finding this resistor

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value to do this, and this
is a real-life application.

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For example, if you want
to turn on this LED,

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so we will need to use Ohm's Law
right here--V equals IR to do

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this.

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And we know the voltage--two
volts--and we know the current

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that is 20 milliamps.

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Mujtaba: So if the current
is 20 milliamps what

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would the resistor value be?

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Student: 100.

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Mujtaba: Exactly.

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So, now we will use
this in a circuit

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and see if our LED turns on.

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So I'll share my camera with
you guys and see if this works.

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Mujtaba: Can you
guys see me now?

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Students: Yes.

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Uh huh.

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Mujtaba: So now.

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Okay, so now what we
have is the circuit.

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Mujtaba: Can you guys
see this circuit?

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Students: Yeah.

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Mmm hmm.

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Mujtaba: We will be not
worrying about this part

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of the circuit for now.

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So the only thing we worry--we
worry about is so there is

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a two volts input right here
in this port and we found that

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a resistor value of 100
ohms will do the job,

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so this is 100 ohms resistor and
we will plug this in from a two

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volt source in this breadboard,
so this is a breadboard.

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If you guys have dealt with
prototyping electronics

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you might have used this.

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It's just a bunch of holes that
are connected to each other.

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So all these five holes in a
row are connected to each other

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and the columns
are not connected

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so you can use them to
essentially make your circuit.

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So we will be connecting
that there and we will

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be using a red LED.

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This is an LED, and
this one is a red one.

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And LEDs have two
legs, essentially.

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The longer one is
the positive one

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because they're directional.

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And if you connect
them the opposite way

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they don't turn on.

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So the longer leg
will go to positive

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which is connected to
the resistor essentially,

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and then the negative
one will go to ground.

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So we will connect
this right here.

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Mujtaba: And we will
connect the other leg to

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Student: I can't see it.

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Mujtaba: Oh sorry.

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Yeah.

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So we will connect the other
leg to ground over here.

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Mujtaba: So yeah, so
ground right here.

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So in this breadboard, I've
set up this rail at the bottom.

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So this negative the blue rail
is ground and the positive rail

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is the positive volts that
I'll connect the battery to.

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And I have a voltage
divider set up right here

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that I'll explain in a minute.

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But this essentially takes
my battery which is nine

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volts--this is a nine volt
battery and divides this

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into essentially the required
voltage that I need and that is

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two volts right here that
will go into the resistor.

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So we will connect this battery
now and see if it works.

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Mujtaba: Can you guys see?

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Mujtaba: There we go.

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So we will have a closer.

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I will connect the battery...

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Mujtaba: battery pin

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Mujtaba: Do you see that?.

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Student: Yes.

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Mujtaba: The red LED turns on.

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So this is a real-life
application of this.

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Mujtaba: So, of Ohm's Law.

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You can use it to turn on LEDs.

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You can use it for your personal
projects or anything like that.

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You can buy LEDs,
they're very cheap

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and you can set up the
circuitry on your own

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and make an LED turn on and
do a lot of other cool things

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with it.

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Mujtaba: Now, LEDs
have--any questions so far?

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Student: No.

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Mujtaba: Okay, so now LEDs
have different current ratings.

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Essentially, if you're
turning on a red one,

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you need a different
current to go through it.

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And if you're turning
on a blue one,

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you would need a different
current to go through it.

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Mujtaba: So a blue one, so
this one needs 20 milliamps

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to go through it to turn on.

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A blue one will
need 30 milliamps

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to go through it to turn on.

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Do you guys think if I switch
this red LED with a blue one,

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will it turn on right now?

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Student: Probably not.

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Mujtaba: Exactly.

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It will not turn on
because because we set up

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this circuit for a 20
milliamp current, which

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is lower than the
30 milliamp current

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and that's why the blue
one will not turn on.

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We can test this and
here I have a blue LED.

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And we can--

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Mujtaba: we can plug
in the blue LED.

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Mujtaba: Okay, now I
can plug in the battery

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and see if it works.

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Mujtaba: Oh, battery
is connected,

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but the blue LED does not
turn on, or it's very dim.

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No, I don't think
it's on at all.

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So, in order to do this--to make
the blue LED turn on--what do

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you guys think the
resistor value should be?

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We know the voltage
is two volts.

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We know the current
to be 30 milliamps.

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What should the
resistor value be?

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Student: Lower than
this one by two thirds.

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Mujtaba: Exactly.

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Something like 67 ohms.

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Mujtaba: So, if
you plug in that,

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we can turn on the blue LED.

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Mujtaba: So now I will
go on to the concept

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of the voltage divider
which essentially

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is the reason we can take
the nine volt battery

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and connect it to it without
burning anything down.

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And that is essentially when
resistors are in series,

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that is they're
connected to each other

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and they divide the voltage
that goes through them.

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Think of it as having two
bubbles in your straw.

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When you're trying to
pull out the bubbles

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at the bottom of
your boba drink,

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if you have two
bubbles in the straw,

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you need to suck on it a
little harder or essentially

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the pressure drop
across the pipe

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needs to be higher for both
of those bubbles to come up.

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And that is essentially
what happens here.

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So two resistors in series,
divide the voltage across them

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depending on the
resistor value they have.

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Mujtaba: So without
going to my screen,

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I can just use paper here.

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Mujtaba: So that is

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Mujtaba: Can you guys see this?

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Students: Yes.

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Mujtaba: Okay.

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So if we have one resistor
in series with other resistor

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and if they're--so, the sum of
them the voltage drop across

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one of them, so say there is a
voltage drop across the whole

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thing, this goes the
opposite direction, okay.

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So the voltage drop across
one of them--this is R1,

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this is R2--will be the resistor
value of one over the sum

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of both of them times
the voltage okay?

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Mujtaba: So, essentially the
voltage drop across one of them

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will be the
resistance of that one

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over the sum of both of them
times the total voltage.

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Mujtaba: So, that
is essentially what

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we are using here to do this.

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So in order to get
two volts out of this,

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what do you guys think the
ratio of these resistors

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is supposed to be?

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Essentially, this one resistor
over the sum of resistances

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times nine should give us two.

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And that way we can
find two divided by nine

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should be the ratio
of these resistors.

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And that is essentially
the resistor value that I

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picked--it's random.

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So you can pick many different
combinations that gives you

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that ratio, which is
I think 0.2--0.22.

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So to get that I chose 1.6
kiloohms resistor and a 5.6

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kiloohm resistor.

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And as you guys can do the
math, 1.6 over the sum of both,

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which is seven
point two will be...

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it's a little odd ratio
but it will be 0.22.

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So that times nine gives us
the two volts that we need.

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And that is essentially what
I'm doing in this circuit right

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here.

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Mujtaba: In here.

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Mujtaba: So you guys can see
the two yellow resistors.

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Those two yellow resistors,
one of them is 1.6.

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The one that is
horizontal is 1.6.

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And the other one is 5.6.

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And 1.6 over the
sum of both of them

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gives us the ratio
and that connected

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to the battery which is the
blue and red rail at the bottom.

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Mujtaba: Right here.

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Mujtaba: This rail is nine volts
and that divides it and make

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to two volts.

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And that's how we find the two
volt from nine volt battery.

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Mujtaba: Okay.

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Any questions

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Student: Hmm I have a question.

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Mujtaba: Yeah.

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Student: Like when
we connect something.

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So, in in the
voltage divider, we

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can add something that has
like finite resistance.

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So, we change effective
resistance of one

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of the resistors
in voltage divider.

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How like how can how do we
need to pick resistances

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of voltage divider by the
resistance to make this more--

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Mujtaba: Which affects lower?

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So you are you talking about
the two volts that at the end

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we are getting you want
to make that lower?

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Student: No.

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00:16:01,656 --> 00:16:07,254
I mean like there are two
volts in the voltage voltage

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divider if you don't
connect anything to it,

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but if you connect something
the effective resistance

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of something we
connect and resist.

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And there's a
resistance too, which

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we connect with the scheme
in parallel, changes.

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And we want to have two
volts, exactly two volts.

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Mujtaba: That is,
that is correct.

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That is correct.

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If you connect it in series, it
changes the effective voltage

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drop or the effective
resistance and therefore

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the effective voltage
drop across one resistor.

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But, so the way it is
connected to this circuit,

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and this diagram or in
any other situation when

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you're connecting
something, you're

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usually connected in parallel.

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So in parallel, the
voltage drop stays the same

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across both of those
parallel things.

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Ideally, that would be the ideal
scenario, but in real life,

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there is a slight drop across
the parallel setup because

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of the losses in the circuitry
and all the other things,

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but in in theory, when you
connect things in parallel,

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the voltage drop across both
of them will stay the same.

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And that's why the LED circuit
is connected in parallel

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to the 1.6 volts or
1.6 kiloohm resistor

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00:17:28,514 --> 00:17:31,742
and therefore the voltage
drop across that resistor

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00:17:31,742 --> 00:17:34,160
stays two volts only.

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Student: What would happen if
too much current run through,

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yeah, LED.

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Mujtaba: So the
LED is essentially

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a piece of material that has
electrons at certain transition

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states.

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So the way it emits emit
light or emits light

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is essentially activation
of those electrons.

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And that's how it
emits light and it

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requires certain energy level to
cause the activation to happen.

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If there's too much energy, and
most of the commercial ones,

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they just break.

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00:18:11,513 --> 00:18:14,974
So the LED just breaks and
if you put too much current

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through it, it just breaks
and they will not work again,

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00:18:17,884 --> 00:18:20,317
which essentially means that
the material whatever they're

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using in there just melts
down or the energy for it

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is too much.

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And some of the other cases,
if you put too much current

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through a thin wire,
especially when it's it

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doesn't have a heatsink
in any other way.

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It will just melt down.

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So in short answer
LED just breaks

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00:18:44,159 --> 00:18:46,990
and you will not be
able to use it again.

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Mujtaba: Yeah.

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00:18:47,870 --> 00:18:50,070
I hope that was clear.

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00:18:50,070 --> 00:18:53,590
You can use it in
your own projects.

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00:18:53,590 --> 00:18:57,780
One of the projects I
actually made was a clock.

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00:18:57,780 --> 00:19:00,650
I made a clock for my room.

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00:19:00,650 --> 00:19:04,334
I don't have it here
right now but it

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00:19:04,334 --> 00:19:08,832
was a clock that was controlled
by by an Arduino which

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00:19:08,832 --> 00:19:10,602
is just a microprocessor.

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00:19:10,602 --> 00:19:15,844
And each of the twelve numbers
on the clock were LEDs.

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00:19:15,844 --> 00:19:19,132
And it would turn on
depending on the hour

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and the minute of
the day it was so you

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00:19:21,897 --> 00:19:25,336
can use a simple calculations
like this like ohms law

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00:19:25,336 --> 00:19:27,803
and practice to
find the resistor

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00:19:27,803 --> 00:19:30,728
values for the
different LEDs that you

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00:19:30,728 --> 00:19:35,885
will be using in that case,
the clock for example.