AP Physics 1 Review of Charge and Circuit | Physics | Khan Academy

- [Voiceover] Electric
Charge is a property

that some, but not all fundamental
particles in nature have.

The most commonly talked about

fundamentally charged
particles are the electrons,

which orbit the outside of the atom.

These are negatively charged.

There's also the protons, which reside

inside the nucleus, and
these are positively charged.

And the neutrons inside the nucleus

don't have any net charge.

It turns out that all
fundamentally charged particles

in the universe, have charges that come in

integer units of the elementary charge.

So if you find a particle in nature,

it's gonna have a charge
of one times this number,

two times this number,
three times this number,

and it could either be
positive or negative.

For instance, the electron
has a charge of negative 1.6

times 10 to the negative 19th Coulombs,

and the charge of the
proton is positive 1.6

times 10 to the negative 19th Coulombs.

However, most atoms in the universe

are electrically neutral overall,

since they'll have just
as many negative electrons

as they do positive protons.

But if an atom had too many electrons,

overall that atom would
be negatively charged,

and if an atom had too few electrons,

that atom would be overall
positively charged.

And something that's really
important to remember

is that the electric
charge is always conserved

for every process, in other words,

the total charge initially, is gonna equal

the total charge finally
after any process.

So what's an example problem involving

electric charge look like?

Let's say three identically
sized metal spheres

start off with the charges seen below.

Positive five Q, positive
three Q, and negative two Q.

If we touch sphere X to
sphere Y, and separate them,

and then touch sphere Y to
sphere Z, and separate them,

what will be the final
charge on each sphere?

Well first, when we touch X to Y,

the total charge has to be conserved.

There's a total charge
of eight Q amongst them,

and since they're identically sized,

they'll both share the total charge,

which means after they touch,

they'll both have positive four Q.

If one of the spheres were larger,

it would gain more of the charge,

but the total charge
would still be conserved.

And now when sphere Y
is touched to sphere Z,

the total charge amongst
them at that moment

would be positive four
Q plus negative two Q,

which is positive two Q.

They would share it equally,
so sphere Y would have

positive Q, and sphere Z
would also have positive Q.

So the answer here is C.

Opposite charges attract
and like charges repel,

and what Coulomb's Law does
is it gives you a way to find

the magnitude of the electric
force between two charges.

The formula for Coulomb's Law says

that the magnitude of the electric force

between two charges Q1
and Q2, is gonna equal the

electric constant K, which is
nine times 10 to the ninth,

times the product of the two
charges, measured in Coulombs

divided by the center to center distance

between those two charges, squared.

You can't forget to square this distance.

And it's gotta be in
meters if you want to find

SI units of Newtons for the force.

Also, don't rely on the
negative and positive signs

of the charges to tell you
which way the force points,

just use the fact that
opposite charges attract

and like charges repel,
and use Coulomb's Law

to get the magnitude of the force.

So what's an example problem involving

Coulomb's Law look like?

Let's say two charges
exert an electric force

of magnitude F on each other.

What would be the magnitude
of the new electric force

if the distance between
the charges is tripled

and the magnitude of one
of the charges is doubled?

Well we know the formula for Coulomb's Law

says that the force between two charges

is the electric constant
times one of the charges,

times the other charge
divided by the distance

between them squared, and now
once we triple the distance

and double a charge,
the new electric force

is gonna be the electric constant
times one of the charges,

multiplied by two times
one of the charges,

divided by three times the
distance, which is squared,

so I'm gonna get a factor of two on top,

and this three will get
squared, which gives me

a factor of nine on the bottom.

If I pull up those extra
factors I get that the new force

is gonna be two ninths
multiplied by K, Q1, Q2,

over D squared, but this entire quantity

was just the old force F, so the new force

is going to be two
ninths of the old force.

The electrical current I tells
you the amount of Coulombs

of charge that passes a
point in a wire per second.

So if you watch some point in a wire,

and you count how many Coulombs of charge

pass by that point per second,
that would be the current.

Or in equation form we
can see that the current I

is gonna be the amount
of charge that flows

past a point in a wire per time.

This gives the units of
I as Coulombs per second,

which we abbreviate as an Ampere.

And since charge and time aren't vectors,

current is not a vector either.

Something that's kind of strange

is that the so-called
conventional direction of current

would be the direction that
positive charges flow within a

wire, however positive charges
don't flow within a wire.

The only charges that
actually flow in a wire

are negative charges, but it turns out

that negative charges flowing
to the left is physically

the same as positive charges
flowing to the right.

So in physics problems
we pretend as if it were

the positive charges moving, however

it's really the electrons,
which are negative,

that are moving within the wire.

So what's an example problem

involving electrical current look like?

Let's say three amps
flows within a circuit.

How much charge would pass
by a point in that wire

during a time interval of five minutes?

Well we know the definition of current

is the charge per time,
that means the charge

is gonna be the amount of
current multiplied by the time,

so we take our current of three amps,

and we multiply by the time, but we can't

multiply by five because
that's in units of minutes,

since amps is Coulombs per second,

we've got to convert five
minutes into seconds,

which would be five minutes,
multiplied by 60 seconds

per minute, which would
give us a total amount

of charge of 900 Coulombs.

The resistance of a circuit
element measures how much

that element will restrict
the flow of current.

The larger the resistance,
the less current

there will be allowed to flow.

And this definition of
resistance is given by Ohm's Law.

Ohm's Law states that
the amount of current

that you'll get through
a portion of a circuit,

it's gonna be proportional to the voltage

across that portion,
divided by the resistance

of that portion of the circuit.

So between these two points,
the amount of current

that will flow, is gonna be equal to

the voltage between those two points,

divided by the resistance
between those two points.

So the larger the
resistance, the less current

will flow, but the greater
the voltage supplied,

the greater the current will be.

And this is what Ohm's Law says.

Even though Ohm's Law
gives you a way to define

the resistance, you can
determine the resistance

of a circuit element by knowing the size

and shape of that circuit element.

In other words, the resistance
of a cylindrical resistor,

is gonna be equal to the resistivity,

which is a measure of an
object's natural resistance

to current, multiplied by
the length of that resistor,

the longer the resistor,
the greater the resistance

and the more it will
resist the flow of current,

and then divide it by
the cross sectional area

of that resistor, which would
be this area right here,

the current is either
flowing into or out of.

If the resistor is cylindrical,
the area of this circle

would be Pi times r
squared, where little r

would be the radius of
this cross sectional area.

The units of resistance is
Ohms, and it is not a vector.

It is always positive or zero.

So what's an example
problem involving Ohm's Law,

or the resistance of a
cylindrical resistor look like?

Let's say a battery of
voltage V is hooked up

to a single cylindrical
resistor of length L

and radius little r, and when that's done,

a current I is flowing
through the battery.

What is the resistivity
Rho, of that resistor?

Well we know Ohm's Law states
that the current that flows

through a portion of a
circuit will be equal to

the voltage across that portion,

divided by the resistance of that portion.

And this means the resistance
of this resistor is gonna be

the voltage of the battery
divided by the current.

To factor resistivity
into this, we have to use

the formula for the resistance
of a cylindrical resistor,

which is Rho times L over A.

This gives us the
resistance of the resistor,

which is gotta equal V over I,

and now we can solve
for the resistivity Rho.

We get V times A over I
L, but since we're given

the radius little r,
we gotta write the area

in terms of that radius,
so this is gonna be

V times Pi, r squared,
divided by I times L,

which gives us an answer of C.

When dealing with complicated
circuits with many resistors,

you often have to reduce those resistors

into smaller, equivalent
amounts of resistors.

And the two ways you
do this are by finding

two resistors that are
in series or in parallel.

Resistors will be in
series if the same current

that flows through the same resistor,

flows through the next resistor.

If the current branched
off in between them,

these resistors would
no longer be in series,

but if they're in series you can find

the equivalent resistance
of this section of wire

by just adding up the two
individual resistances.

So the current for resistors in series

must be the same, but the
voltage might be different,

since they could have
different resistances.

Two resistors will be in parallel,

if a current comes in,
splits into two parts,

goes through one resistor
each, and then rejoins

before hitting anything
else in the circuit,

and if this is the case,
you can find the equivalent

resistance of this portion of the circuit,

i.e. between these two
points, by saying that one

over the equivalent
resistance is gonna equal

one over the resistance
of the first resistor,

plus one over the resistance
of the second resistor.

But be careful, one
over R1 plus one over R2

just gives you one over R equivalent.

If you want R equivalent,
you're gonna have to take

one over this entire side,
in order to get R equivalent.

So what's an example
problem involving resistors

in series and parallel look like?

Let's say we have this
circuit shown below,

and we want to know what current flows

through the eight Ohm resistor.

Now you might be tempted to say this,

since Ohm's Law says that the
current is delta V over R,

we can just plug in the
voltage of the battery,

which is 24 volts,
divided by the resistance

of the resistor, which is eight Ohms,

and that would give us three Amps.

But that's not right at all.

When using Ohm's Law,
the current that flows

through a resistor R, is gonna be equal to

the voltage across that
resistor divided by

the resistance of that resistor.

So if we plug eight Ohms into
the denominator, we've gotta

plug in the voltage across
that eight Ohm resistor.

But the voltage across
the eight Ohm resistor

is not gonna be the full
24 volts of the battery,

it's gonna be less than 24 volts.

In other words, the battery
provides a voltage between

this point and this point of 24 volts,

but there's gonna be voltage drops

across the six and 12 Ohm resistors,

which make it so that the voltage

across the eight Ohm resistor is not

gonna be the full 24 volts.

So we have to reduce these
resistors to a single resistance.

The six and the 12 are in parallel,

so we can say that one
over six, plus one over 12,

would equal one over the resistance

of that portion of the circuit.

This is gonna equal three
twelves, which is one fourth,

so that means that parallel
portion of the circuit

has an equivalent resistance of four Ohms.

So between this point and this point,

there are four Ohms of resistance,

and that equivalent
resistance is in series

with this eight Ohm resistor.

So we can add four and eight,

and get 12 Ohms of total resistance.

And now I can say that the
full 24 volts of the battery

is applied across this
entire equivalent resistance

of 12 Ohms, so if I come up
here and change this eight Ohms

to 12 Ohms of equivalent
resistance for the total circuit,

I'll get the correct current that flows

through the battery of two Amps.

And since that's the current
that's flowing through

the battery, that had to be the current

that's flowing through the
eight Ohm resistor as well.

Since this eight Ohm resistor
and the batter are in series.

Elements in a circuit
often use Electrical Power.

That is to say, when current
runs through a resistor,

the electrons moving through that resistor

turn some of their
electrical potential energy

into energies like heat, light, or sound.

And the rate at which these electrons

are turning their energy
into other forms of energy,

is called the electrical power.

So the rate at which a resistor is turning

electrical potential energy into heat,

is the electrical power
used by that resistor.

In other words, the amount of energy

converted into heat, divided by the time

it took to convert that
energy, is the definition

of the power, and there's
a way to determine

this number of Joules per
second, in terms of quantities

like the current, the
voltage, and the resistance.

The power used by a
resistor can be written as

the current through that resistor

multiplied by the voltage
across that resistor,

or if you substituted Ohm's
Law into this formula,

you see that this is equivalent to

the current through that resistor squared,

multiplied by the
resistance of the resistor,

or we could rearrange these formulas

to get that the power used by a resistor

would also be the voltage
across that resistor squared,

divided by the resistance
of that resistor.

All three of these, if used correctly,

will give you the same
number for the power used

by a resistor, and if
you wanted to determine

the number of Joules of
heat energy converted,

you could set any one of
these equal to the amount

of energy per time, and
solve for that energy.

The units of Electrical Power
are the same as the regular

units of power, which is
Watts, i.e. Joules per second,

and Electrical Power is not a vector.

So what's an example problem involving

Electrical Power look like?

Let's say a light bulb of resistance R

is hooked up to a source of voltage V,

and a second light bulb of resistance 2R,

is hooked up to a source of voltage 2V.

How does the power used
by the second light bulb

compare to the power used
by the first light bulb?

Since we have the
information about R and V,

I'll use the version of the power formula

that says that the
power used by a resistor

is gonna be delta V squared over R.

So in terms of quantities
given the power used

by the first light bulb is
gonna be V squared over R.

And the power used by the
second light bulb is gonna be

equal to the voltage across
the second light bulb,

which is two times the voltage
across the first light bulb,

and we square that,
divided by the resistance

of the second light
bulb, which is gonna be

two times the resistance
of the first light bulb.

The two squared on top is
gonna give me a factor of four,

and I'll have another
factor of two on the bottom.

So if I factor out this
four divided by two,

I get that the power used
by the second light bulb

is gonna be two times V squared over R,

but V squared over R was the power used

by the first light bulb, so the power used

by the second light bulb is gonna be two

times the power used by
the first light bulb,

and so if the light
bulb of resistance two R

has twice the power, and
that means it'll be brighter.

The quantity that
determines the brightness

of a light bulb, is the electrical
power of that light bulb.

It's not necessarily the
resistance or the voltage,

it's the combination of the two

in this formula that will
tell you the electrical power,

and therefore the brightness
of the light bulb.

Two of the most useful ideas in circuits

are referred to as Kirchhoff's Rules.

The first rule is called
the Junction rule,

and it states that all the
current entering a junction

must equal all the current
exiting that junction.

In other words, if you add
all the current that flows

into a junction, that has to equal

all the current that flows
out of that junction,

because current is just flowing charge,

and charge is conserved,
so charge can't be

created or destroyed at
any point in the circuit.

No more than water can
get created or destroyed

within a series of pipes.

And the second rule is
called the Loop rule,

which states that if you
add up all the changes

in electric potential, i.e.
voltages around any closed

loop in a circuit, it'll
always add up to zero.

So if you add up all
the voltages encountered

through a closed loop through a circuit,

it always adds up to zero.

And this is just a result
of conservation of energy.

The electrons will gain
energy when they flow

through the battery,
and they'll lose energy

every time they flow through a resistor,

but the total amount of energy they gain

from the battery, has to be
equal to the total amount

of energy they lose due to the resistors.

In other words, if we
consider a complicated circuit

that has a batter and three resistors,

the total current flowing
into a junction I1,

will have to be equal to the total current

coming out of that junction, I2 and I3.

Since no charge gets created or destroyed.

And that means when these
two currents combine again,

the total current flowing out

of that section is gonna again be I1.

And if we follow a closed
loop through this circuit,

the sum of all the
voltages around that loop

have to add up to zero, i.e.
the voltage of the battery

minus the voltage drop
across the first resistor,

minus the voltage drop
across the second resistor,

would have to equal zero.

So what's an example problem involving

Kirchhoff's Rules look like?

Let's say we have the circuit
below and we wanted to

determine the voltage
across the six Ohm resistor.

To do this, we could use the Loop rule,

I'll start behind the
battery, and I'll go through

the resistor, I want to
determine the voltage across.

I'll add up all the
voltages across that loop,

and set it equal to zero.

So the voltage across the battery

is gonna be positive 24 volts,

minus the voltage across
the six Ohm resistor,

and then minus the voltage across the

eight Ohm resistor has to equal zero.

But we're given this current,
so we know that two Amps

flows through the eight Ohm resistor,

and you can always determine the voltage

across the resistor using Ohm's Law,

so the voltage across
the eight Ohm resistor

is gonna be two Amps, which is flowing

through the eight Ohm resistor,

multiplied by eight Ohms,
and we get 16 volts.

Which I can plug into
here, and this gives me

24 volts minus the voltage
across the six Ohm resistor,

minus 16, has to equal zero.

And if I solve this for the voltage

across the six Ohm
resistor, I get 24 volts

minus 16 volts, which is eight volts.

So the voltage across the six Ohm resistor

would be eight volts.

Note, because the 12 Ohm
resistor and the six Ohm resistor

are in parallel, the voltage
across the 12 Ohm resistor

would also be eight volts,
because the voltage across

any two elements in parallel,
have to be the same.

Voltmeters are the device that you use

to measure the voltage between
two points in a circuit.

When hooking up a voltmeter you've gotta

hook it up in parallel
between the two points

you wanna find the voltage across.

In other words, to determine the voltage

between this point and
this point, which would be

the voltage across R3, you
would hook up the voltmeter

in parallel with R3.

Ammeters are the devices we use to measure

the electrical current that pass

through a point in a circuit, and ammeters

have to be hooked up in series
with the circuit element

you want to determine the current through.

In other words, if we wanted
to determine the current

through R1, we would hook up
the ammeter in series with R1.

Note that for these electrical
devices to work well,

the ammeter should have almost
zero internal resistance,

thereby not affecting
the current that flows

through the circuit, and
voltmeters should have

near infinite resistance,
so that it doesn't draw

any of the current from the resistor.

In reality, ammeters have a very small,

but non-zero internal resistance,

and voltmeters have a very high,

but not infinite internal resistance.

So what would an example problem involving

voltmeters and ammeters look like?

Let's say we have the circuit shown below,

and these numbered circles
represent possible places

we could stick a voltmeter
to measure the voltage

across the eight Ohm resistor.

Which two of these
voltmeters would correctly

give the voltage across
the eight Ohm resistor?

And you have to be
careful, some AP problems

are gonna require you to select

two correct answers for
the multiple choice,

so be sure to read the
instructions carefully.

Voltmeter number four
is a terrible choice,

you never hook up your
voltmeter in series,

but the circuit element
you're trying to find

the voltage across, and
voltmeter number one

is doing nothing really,
because its' measuring

the voltage between two points in a wire

with nothing in between that wire.

So the voltage measured by voltmeter one

should just be zero, since
the voltage across a wire

of zero resistance should
just give you zero volts.

So the correct choices would
be voltmeter number two,

which gives you the voltage
across the eight Ohm resistor,

and voltmeter number
three, which also gives you

an equivalent measurement of the voltage

across the resistor eight Ohms.