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## Electric power and DC circuits

Current time:0:00Total duration:10:43

# Electric power

## Video transcript

- [Instructor] So if
you've ever run current through a resistor, you
might've noticed that that resistor warms up. And in fact if you run too much
current through the resistor that resistor can get
so hot it can burn you when you touch it, so
you have to be careful. So what I'm saying is
when you have current flowing through a resistor it warms up, and we want to explain why. Conceptually, why does current
moving through a resistor heat it up, and is
there way a to calculate exactly how much that current
would heat up that resistor over a given amount of time? There is a way to calculate it. We'll derive that in this video. But first we should just
explain conceptually why is it that current
moving through a resistor heats up the resistor. And so we'll explain
that with this current. Now the way I've got it drawn here, notice that I've got
these positive charges. Positive charges don't
actually move through a wire, but physicists always pretend like they do because positives moving
one way through a wire is equivalent to negatives
moving the other way through the wire, and if you
use the positive description, even though it's technically incorrect, you don't have to deal with
all those negative signs. It's easier to deal with,
and it's equivalent, so we may as well use it. But you can go through
this whole description with negative charges
moving the other way, and it works just as well. You just have to be very
careful with the negatives, and it kind of obscures
the conceptual meaning 'cause it's hiding behind
a bunch of negative signs. So we'll just use these positive charges, but know that it's really
negatives going the other way. So why do positive charges
flowing through a resistor cause that resistor to heat up? Well here's why. So we know that when current
flows through a resistor, there's a voltage across that resistor. In other words, between this
point here and this point here, there's a difference in
electrical potential. So there's a voltage across that resistor. Technically I'm gonna call it delta V 'cause it's a difference
in electric potential. In other words, V on this
side, the electric potential on this first side of the resistor is gonna have a different value
from the electric potential on this second side of the resistor. So why does this matter? It matters 'cause this final
side where the charges end up, is gonna have a lower electrical potential than the beginning. So these positive charges
are gonna be moving from a high potential region
to a low potential region. And that means they're gonna be changing their potential energy, so these charges have electric potential energy, and if they go from a region
of higher electric potential to a region of lower electric potential, they've started with
more potential energy, electrical potential
energy than they end with. So they're decreasing their
electric potential energy. And in case that's confusing,
remember that the definition of electric potential is the amount of electric potential energy per charge. So to just put a number in
here so it's not so abstract, let's say V two was
two joules per coulomb. That would mean for
every coulomb of charge at that point V two, there'd be two joules of electric potential energy. And if V one is at a
higher electric potential, maybe this is at six joules per coulomb, that would mean over here
at this position V one, for every coulomb of charge
there'd be six joules of electric potential energy. So as these charges move
through the resistor, they're gonna be decreasing
their electric potential energy, and so the obvious question is, well where does that energy go? If these charges are decreasing their electric potential energy, where's that potential energy going? My first guess is that they'd
increase their kinetic energy because I'd remember then
on Earth if you drop a ball and it decreases its potential energy, its gravitational potential energy, we know that when it decreases its gravitational potential
energy and falls down, it increases its kinetic
energy, it just speeds up. So the decrease in
gravitational potential energy just corresponds to an increase in kinetic energy of that object. And so maybe that's happening over here. Maybe as these charges
lose potential energy, they speed up, but that can't happen. Remember the current on
one side of a resistor has to be the same as the
current on the other side. These charges don't speed up. They're losing potential
energy, but they don't speed up. This is a little counterintuitive. We're used to things speeding up when they lose potential energy, but these charges aren't
going to speed up. What they do is they just
heat up the resistor. So as these charges fly
through this resistor, they strike the atoms and molecules in this lattice structure of this solid. So this resistor's made
out of atoms and molecules, and as these charges flow through here, and again it's really electrons
flowing the other way, but as the charges flow
through, same idea. They strike the atoms and molecules, they transfer energy into them. And as they pass through in their wake, they leave a resistor that's hotter, at a higher temperature. Which means these atoms
are jiggling around more than they were before. And since they're oscillating
more than they were before, they're jiggling, they've got more energy, the temperature of this
resistor increases. So these charges, rather
than keeping all the energy for themselves, they
actually just spread it out over that resistor as they pass through, and they spread it out
in the form of heat, or thermal energy. And they emerge with basically
the same kinetic energy that they started with, so this
change in potential energy, electrical potential energy,
corresponds to an increase in thermal energy of this resistor. And that's why the resistors heat up. But is there a way to
calculate exactly how much this resistor will heat up? How much energy it's gonna gain per time? There is, we just have to
use the definition of power. So we know the definition of
power is the work per time, or since work is the change in energy, or the energy transferred,
we can just write this as the amount of energy this
resistor's gaining per time. What we want is a formula
that tells us how much energy are these charges depositing
in the resistor per time. Well this energy gained by the resistor is coming from the loss
of potential energy of these charges. So these charges are
losing potential energy. They're losing electric potential energy, and that electric potential energy's turning into thermal energy, so the thermal energy this resistor gains is just equal to the amount
of electric potential energy that these charges lose. So I can just rewrite this. I can just say that the
power is gonna be equal to the change in
electrical potential energy of these charges per time. And so I'll just continue down here. Power's gonna be equal to,
how do we find the change in electric potential energy? Well remember, potential is defined to be the potential energy per charge. So that means the
electric potential energy is just the charge times
the electric potential. So if I want to find delta
U, I can just say that that's gonna be U when they
emerge, U two minus U one, and this is a way we
can find the U values. So the U at two, since it's Q times V, is just gonna be the
charge at two times V two. And then the U at one, so
we'll do minus the U at one, is the charge at one. But that's the same charge. Whatever charge enters this
resistor has to exit it, so it'd be the charge at
one times the V at one. This is the change in
electric potential energy. So I could rewrite this. I can pull out a common factor of Q in this expression right here. Then we get that the
power's going to be equal to this common factor of Q
times V two minus V one. So that's delta U, and
that's what we're plugging in right here for delta U. Delta U is just the difference
in these Q times V values. And then we still have to divide by time since we're talking about a power. But what is V two minus V one? That's simply the voltage
across this resistor. Delta V is V two minus V one. So I could rewrite this. I could just say that this is
Q times delta V, the voltage, across that resistor,
divided by the time it took for the charge to pass
through that resistor. And now something magical
happens, check this out. So we got power equals,
I've got charge the past through the resistor, divided
by the time that it took for that charge to pass
through the resistor, but charge per time is just
the definition of current. So we get this beautiful formula. If I just factor out this Q over T, I get Q over T times delta V, the voltage across the resistor. But Q over T is just the current, so I get that the power is going to equal the current through that resistor times the voltage across that resistor. And this is the formula
for electrical power. This tells you how many
joules of thermal energy are being created in
that resistor per time. So the units are joules per second, or in other words, watts. Because what this is telling you is the amount of thermal
energy generated per second. If the power value came
out to be 20 watts, that would mean there'd be 20 joules of thermal energy generated every second, which is a really useful thing to know. This formula's extremely useful
when you want to figure out how much energy's gonna
be used by a light bulb, or a toaster, or a TV, or
whatever electronic device you want to use. This tells you how much
energy that it's going to turn into either thermal
energy, or light, or sound, or whatever other kind of
energy that it's converting that electric potential energy into. 'Cause notice, we never really assumed that this was thermal energy. We just called it E, and then
we said that whatever energy got transformed came from the
change in potential energy of those charges, and
that's always gonna be true. It's always gonna be
electric potential energy converting into something,
whether it's heat, or light, or sound. So this doesn't just work for resistors. It works for almost all
electrical components that turn electric potential energy into some other kind of energy. And you could rewrite
this in different forms. Sometimes you'll see it a different way. There's this form, and then
I'm just gonna copy this. I'm gonna show you there's
a couple other forms you could see this in. Let me clean this up, we'll
put this formula down here. This gives us the power,
but we know from Ohm's law that delta V is equal to I R. So if both of these formulas are true, I can plug delta V as I R. So I could take this I R, I
could plug it in for delta V, and I get an alternate expression for the power used by
electrical component. I get that the power's
gonna be I times I times R, which is just I squared times R. So this one might be more
useful if you've got a situation where you don't know the voltage, but you happen to know the
current and the resistance. But there's one more. I could've solved this
Ohm's law formula for I, and I'd get that I equals delta V over R. And then if I plug delta
V over R in for I up here, I'd get an alternate
expression for the power. I'd get that the power equals
delta V over R times delta V. It's just gonna be delta V squared. The voltage across that resistor squared divided by the resistance
of that resistor. And this formula might be more useful if you know the voltages
and the resistance, but you don't know the current. So depending on what you know, these can get you the
power used by a resistor. They're all equivalent. They will all give you the
correct and the same power. It's just a matter of
what's more convenient for the actual problem
that you're dealing with. So recapping, when current
passes through a resistor, it converts electrical potential
energy into thermal energy, and you can calculate the amount of electrical potential
energy converted per second using current times voltage, current squared times the resistance, or voltage squared
divided by the resistance.