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- [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.