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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1 > Unit 12

Lesson 1: AP Physics 1 concept review- AP Physics 1 review of 1D motion
- AP Physics 1 review of 2D motion and vectors
- AP Physics 1 review of Forces and Newton's Laws
- AP Physics 1 review of Centripetal Forces
- AP Physics 1 review of Energy and Work
- AP Physics 1 review of Momentum and Impulse
- AP Physics 1 review of Torque and Angular momentum
- AP Physics 1 review of Waves and Harmonic motion
- AP Physics 1 Review of Charge and Circuits

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# AP Physics 1 Review of Charge and Circuits

In this video David quickly explains each charge and circuit concept and does a sample question for each one. Created by David SantoPietro.

## Want to join the conversation?

- why do you divide by area to find the resistance of a cylindrical resistor? wouldn't a wider resistor have higher resistance?(2 votes)
- No, the larger the area the leass the resistance. Unlike the lenght . As is increase resistance increases.(1 vote)

- at19:05isn't the voltmeter measuring the resistance between the parallel resistor and not the one 8 ohm one.(1 vote)
- Voltmeter or ohmmeter? Voltmeters don't measure resistance, they measure potential difference.(2 votes)

- At19:11we see three equations to solve for power. in equation one P=I^2R this indicates power is directly proportional to current and resistance. However, in the third one we see that P=V^2/R this indicates power is directly proportional to voltage and inversely proportional to resistance, but how is that possible. Did I miss something?(1 vote)
- V = IR so both of those are true.

P = I^2R indicates power proportional to R, if I is held constant. But if you just increase R and do nothing else, I will go down.

P = V^2/R indicates power inversely proportional to resistance if V is held constant.(2 votes)

- in the example circuit problem, the equivalent resistance of the first two in parallel is 4 ohms. The voltage going into them is 24 volts, so the current going through that equivalent resistance has to be 6. And since the parallel equivalent is in series with the 8ohm, wouldn't that make the 8 ohm also 6 amps, im confused(1 vote)
- What would happen in a circuit with no resistors?

Also, addressing the circuit that had 6 ohm and 12 ohm resistor in parallel and the 8 ohm resistor in series, does the current through the circuit decrease if you pop enough resistor in series between the current resistors?(1 vote) - At9:30how would you solve for the current going through the 6 ohm resistor and the 12 ohm resistor?(1 vote)
- One way to do this is by using the loop equation to find the voltage going through the parallel. Keep in mind, the voltage for the resistors is the voltage that comes out. Because the voltage is the same for both the resistors in the parallel, simply plug the found voltage into the equation and find the currents.(1 vote)

- 10:51How is the 8 in series with 6 and 12 and not a parellel?(1 vote)
- The 6 and 12 are in parallel, right? We are considering both resistors to be a single equivalent resistor of 4 ohms. If you were to remove the 2 resistors and replace them with a single equivalent resistor in the diagram it would be on series with the 8 ohm resistor.(1 vote)

- is there a difference on distribution of two different charges if it is charged with induction or contact?(1 vote)
- Is it necessary to know the charge of a proton and electron for the ap test? Or will that be given?(0 votes)
- Given, but you should be familiar with the relative sizes.(1 vote)

## Video transcript

- [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.