WEBVTT 00:00:00.468 --> 00:00:01.688 - [Voiceover] Electric Charge is a property 00:00:01.688 --> 00:00:05.622 that some, but not all fundamental particles in nature have. 00:00:05.622 --> 00:00:06.761 The most commonly talked about 00:00:06.761 --> 00:00:09.562 fundamentally charged particles are the electrons, 00:00:09.562 --> 00:00:11.564 which orbit the outside of the atom. 00:00:11.564 --> 00:00:13.345 These are negatively charged. 00:00:13.345 --> 00:00:15.146 There's also the protons, which reside 00:00:15.146 --> 00:00:18.484 inside the nucleus, and these are positively charged. 00:00:18.484 --> 00:00:20.088 And the neutrons inside the nucleus 00:00:20.088 --> 00:00:21.855 don't have any net charge. 00:00:21.855 --> 00:00:24.212 It turns out that all fundamentally charged particles 00:00:24.212 --> 00:00:26.310 in the universe, have charges that come in 00:00:26.310 --> 00:00:29.258 integer units of the elementary charge. 00:00:29.258 --> 00:00:30.804 So if you find a particle in nature, 00:00:30.804 --> 00:00:33.916 it's gonna have a charge of one times this number, 00:00:33.916 --> 00:00:37.016 two times this number, three times this number, 00:00:37.016 --> 00:00:38.717 and it could either be positive or negative. 00:00:38.717 --> 00:00:41.690 For instance, the electron has a charge of negative 1.6 00:00:41.690 --> 00:00:44.658 times 10 to the negative 19th Coulombs, 00:00:44.658 --> 00:00:47.542 and the charge of the proton is positive 1.6 00:00:47.542 --> 00:00:49.889 times 10 to the negative 19th Coulombs. 00:00:49.889 --> 00:00:51.928 However, most atoms in the universe 00:00:51.928 --> 00:00:54.076 are electrically neutral overall, 00:00:54.076 --> 00:00:56.828 since they'll have just as many negative electrons 00:00:56.828 --> 00:00:59.013 as they do positive protons. 00:00:59.013 --> 00:01:01.307 But if an atom had too many electrons, 00:01:01.307 --> 00:01:04.009 overall that atom would be negatively charged, 00:01:04.009 --> 00:01:06.501 and if an atom had too few electrons, 00:01:06.501 --> 00:01:08.844 that atom would be overall positively charged. 00:01:08.844 --> 00:01:10.593 And something that's really important to remember 00:01:10.593 --> 00:01:13.077 is that the electric charge is always conserved 00:01:13.077 --> 00:01:15.212 for every process, in other words, 00:01:15.212 --> 00:01:17.539 the total charge initially, is gonna equal 00:01:17.539 --> 00:01:20.586 the total charge finally after any process. 00:01:20.586 --> 00:01:22.086 So what's an example problem involving 00:01:22.086 --> 00:01:23.660 electric charge look like? 00:01:23.660 --> 00:01:26.187 Let's say three identically sized metal spheres 00:01:26.187 --> 00:01:28.344 start off with the charges seen below. 00:01:28.344 --> 00:01:32.136 Positive five Q, positive three Q, and negative two Q. 00:01:32.136 --> 00:01:35.793 If we touch sphere X to sphere Y, and separate them, 00:01:35.793 --> 00:01:39.609 and then touch sphere Y to sphere Z, and separate them, 00:01:39.609 --> 00:01:42.575 what will be the final charge on each sphere? 00:01:42.575 --> 00:01:44.547 Well first, when we touch X to Y, 00:01:44.547 --> 00:01:46.693 the total charge has to be conserved. 00:01:46.693 --> 00:01:49.224 There's a total charge of eight Q amongst them, 00:01:49.224 --> 00:01:50.934 and since they're identically sized, 00:01:50.934 --> 00:01:52.779 they'll both share the total charge, 00:01:52.779 --> 00:01:54.038 which means after they touch, 00:01:54.038 --> 00:01:56.008 they'll both have positive four Q. 00:01:56.008 --> 00:01:57.429 If one of the spheres were larger, 00:01:57.429 --> 00:01:59.028 it would gain more of the charge, 00:01:59.028 --> 00:02:01.378 but the total charge would still be conserved. 00:02:01.378 --> 00:02:04.193 And now when sphere Y is touched to sphere Z, 00:02:04.193 --> 00:02:06.414 the total charge amongst them at that moment 00:02:06.414 --> 00:02:09.443 would be positive four Q plus negative two Q, 00:02:09.443 --> 00:02:10.976 which is positive two Q. 00:02:10.976 --> 00:02:13.493 They would share it equally, so sphere Y would have 00:02:13.493 --> 00:02:17.029 positive Q, and sphere Z would also have positive Q. 00:02:17.029 --> 00:02:19.247 So the answer here is C. 00:02:19.247 --> 00:02:22.288 Opposite charges attract and like charges repel, 00:02:22.288 --> 00:02:24.874 and what Coulomb's Law does is it gives you a way to find 00:02:24.874 --> 00:02:28.499 the magnitude of the electric force between two charges. 00:02:28.499 --> 00:02:30.069 The formula for Coulomb's Law says 00:02:30.069 --> 00:02:32.123 that the magnitude of the electric force 00:02:32.123 --> 00:02:36.003 between two charges Q1 and Q2, is gonna equal the 00:02:36.003 --> 00:02:39.203 electric constant K, which is nine times 10 to the ninth, 00:02:39.203 --> 00:02:42.619 times the product of the two charges, measured in Coulombs 00:02:42.619 --> 00:02:44.489 divided by the center to center distance 00:02:44.489 --> 00:02:47.260 between those two charges, squared. 00:02:47.260 --> 00:02:49.542 You can't forget to square this distance. 00:02:49.542 --> 00:02:51.612 And it's gotta be in meters if you want to find 00:02:51.612 --> 00:02:54.151 SI units of Newtons for the force. 00:02:54.151 --> 00:02:56.716 Also, don't rely on the negative and positive signs 00:02:56.716 --> 00:02:59.626 of the charges to tell you which way the force points, 00:02:59.626 --> 00:03:02.064 just use the fact that opposite charges attract 00:03:02.064 --> 00:03:04.986 and like charges repel, and use Coulomb's Law 00:03:04.986 --> 00:03:07.175 to get the magnitude of the force. 00:03:07.175 --> 00:03:08.500 So what's an example problem involving 00:03:08.500 --> 00:03:09.790 Coulomb's Law look like? 00:03:09.790 --> 00:03:11.653 Let's say two charges exert an electric force 00:03:11.653 --> 00:03:13.586 of magnitude F on each other. 00:03:13.586 --> 00:03:16.336 What would be the magnitude of the new electric force 00:03:16.336 --> 00:03:18.749 if the distance between the charges is tripled 00:03:18.749 --> 00:03:21.687 and the magnitude of one of the charges is doubled? 00:03:21.687 --> 00:03:23.375 Well we know the formula for Coulomb's Law 00:03:23.375 --> 00:03:25.265 says that the force between two charges 00:03:25.265 --> 00:03:28.026 is the electric constant times one of the charges, 00:03:28.026 --> 00:03:30.354 times the other charge divided by the distance 00:03:30.354 --> 00:03:33.321 between them squared, and now once we triple the distance 00:03:33.321 --> 00:03:36.154 and double a charge, the new electric force 00:03:36.154 --> 00:03:39.180 is gonna be the electric constant times one of the charges, 00:03:39.180 --> 00:03:42.719 multiplied by two times one of the charges, 00:03:42.719 --> 00:03:46.402 divided by three times the distance, which is squared, 00:03:46.402 --> 00:03:48.148 so I'm gonna get a factor of two on top, 00:03:48.148 --> 00:03:49.926 and this three will get squared, which gives me 00:03:49.926 --> 00:03:51.761 a factor of nine on the bottom. 00:03:51.761 --> 00:03:54.148 If I pull up those extra factors I get that the new force 00:03:54.148 --> 00:03:58.588 is gonna be two ninths multiplied by K, Q1, Q2, 00:03:58.588 --> 00:04:00.993 over D squared, but this entire quantity 00:04:00.993 --> 00:04:03.971 was just the old force F, so the new force 00:04:03.971 --> 00:04:07.731 is going to be two ninths of the old force. 00:04:07.731 --> 00:04:10.644 The electrical current I tells you the amount of Coulombs 00:04:10.644 --> 00:04:14.061 of charge that passes a point in a wire per second. 00:04:14.061 --> 00:04:16.019 So if you watch some point in a wire, 00:04:16.019 --> 00:04:18.113 and you count how many Coulombs of charge 00:04:18.113 --> 00:04:21.328 pass by that point per second, that would be the current. 00:04:21.328 --> 00:04:23.659 Or in equation form we can see that the current I 00:04:23.659 --> 00:04:25.628 is gonna be the amount of charge that flows 00:04:25.628 --> 00:04:27.897 past a point in a wire per time. 00:04:27.897 --> 00:04:30.588 This gives the units of I as Coulombs per second, 00:04:30.588 --> 00:04:32.848 which we abbreviate as an Ampere. 00:04:32.848 --> 00:04:34.681 And since charge and time aren't vectors, 00:04:34.681 --> 00:04:36.499 current is not a vector either. 00:04:36.499 --> 00:04:37.736 Something that's kind of strange 00:04:37.736 --> 00:04:40.908 is that the so-called conventional direction of current 00:04:40.908 --> 00:04:43.475 would be the direction that positive charges flow within a 00:04:43.475 --> 00:04:47.171 wire, however positive charges don't flow within a wire. 00:04:47.171 --> 00:04:49.458 The only charges that actually flow in a wire 00:04:49.458 --> 00:04:51.658 are negative charges, but it turns out 00:04:51.658 --> 00:04:55.019 that negative charges flowing to the left is physically 00:04:55.019 --> 00:04:57.932 the same as positive charges flowing to the right. 00:04:57.932 --> 00:05:00.498 So in physics problems we pretend as if it were 00:05:00.498 --> 00:05:02.819 the positive charges moving, however 00:05:02.819 --> 00:05:04.889 it's really the electrons, which are negative, 00:05:04.889 --> 00:05:06.464 that are moving within the wire. 00:05:06.464 --> 00:05:07.410 So what's an example problem 00:05:07.410 --> 00:05:09.295 involving electrical current look like? 00:05:09.295 --> 00:05:11.387 Let's say three amps flows within a circuit. 00:05:11.387 --> 00:05:14.014 How much charge would pass by a point in that wire 00:05:14.014 --> 00:05:16.505 during a time interval of five minutes? 00:05:16.505 --> 00:05:17.904 Well we know the definition of current 00:05:17.904 --> 00:05:20.341 is the charge per time, that means the charge 00:05:20.341 --> 00:05:23.285 is gonna be the amount of current multiplied by the time, 00:05:23.285 --> 00:05:25.303 so we take our current of three amps, 00:05:25.303 --> 00:05:27.305 and we multiply by the time, but we can't 00:05:27.305 --> 00:05:30.295 multiply by five because that's in units of minutes, 00:05:30.295 --> 00:05:32.486 since amps is Coulombs per second, 00:05:32.486 --> 00:05:34.535 we've got to convert five minutes into seconds, 00:05:34.535 --> 00:05:37.934 which would be five minutes, multiplied by 60 seconds 00:05:37.934 --> 00:05:39.833 per minute, which would give us a total amount 00:05:39.833 --> 00:05:42.000 of charge of 900 Coulombs. 00:05:42.890 --> 00:05:45.794 The resistance of a circuit element measures how much 00:05:45.794 --> 00:05:48.703 that element will restrict the flow of current. 00:05:48.703 --> 00:05:51.132 The larger the resistance, the less current 00:05:51.132 --> 00:05:52.763 there will be allowed to flow. 00:05:52.763 --> 00:05:55.978 And this definition of resistance is given by Ohm's Law. 00:05:55.978 --> 00:05:57.794 Ohm's Law states that the amount of current 00:05:57.794 --> 00:05:59.759 that you'll get through a portion of a circuit, 00:05:59.759 --> 00:06:01.476 it's gonna be proportional to the voltage 00:06:01.476 --> 00:06:04.371 across that portion, divided by the resistance 00:06:04.371 --> 00:06:06.006 of that portion of the circuit. 00:06:06.006 --> 00:06:08.331 So between these two points, the amount of current 00:06:08.331 --> 00:06:10.198 that will flow, is gonna be equal to 00:06:10.198 --> 00:06:12.536 the voltage between those two points, 00:06:12.536 --> 00:06:15.604 divided by the resistance between those two points. 00:06:15.604 --> 00:06:17.889 So the larger the resistance, the less current 00:06:17.889 --> 00:06:20.958 will flow, but the greater the voltage supplied, 00:06:20.958 --> 00:06:22.418 the greater the current will be. 00:06:22.418 --> 00:06:24.107 And this is what Ohm's Law says. 00:06:24.107 --> 00:06:26.316 Even though Ohm's Law gives you a way to define 00:06:26.316 --> 00:06:29.082 the resistance, you can determine the resistance 00:06:29.082 --> 00:06:31.043 of a circuit element by knowing the size 00:06:31.043 --> 00:06:32.949 and shape of that circuit element. 00:06:32.949 --> 00:06:35.568 In other words, the resistance of a cylindrical resistor, 00:06:35.568 --> 00:06:37.872 is gonna be equal to the resistivity, 00:06:37.872 --> 00:06:40.214 which is a measure of an object's natural resistance 00:06:40.214 --> 00:06:43.126 to current, multiplied by the length of that resistor, 00:06:43.126 --> 00:06:45.768 the longer the resistor, the greater the resistance 00:06:45.768 --> 00:06:48.072 and the more it will resist the flow of current, 00:06:48.072 --> 00:06:50.199 and then divide it by the cross sectional area 00:06:50.199 --> 00:06:52.860 of that resistor, which would be this area right here, 00:06:52.860 --> 00:06:55.583 the current is either flowing into or out of. 00:06:55.583 --> 00:06:58.450 If the resistor is cylindrical, the area of this circle 00:06:58.450 --> 00:07:01.313 would be Pi times r squared, where little r 00:07:01.313 --> 00:07:04.356 would be the radius of this cross sectional area. 00:07:04.356 --> 00:07:07.521 The units of resistance is Ohms, and it is not a vector. 00:07:07.521 --> 00:07:09.746 It is always positive or zero. 00:07:09.746 --> 00:07:11.888 So what's an example problem involving Ohm's Law, 00:07:11.888 --> 00:07:14.699 or the resistance of a cylindrical resistor look like? 00:07:14.699 --> 00:07:16.776 Let's say a battery of voltage V is hooked up 00:07:16.776 --> 00:07:19.744 to a single cylindrical resistor of length L 00:07:19.744 --> 00:07:21.813 and radius little r, and when that's done, 00:07:21.813 --> 00:07:24.482 a current I is flowing through the battery. 00:07:24.482 --> 00:07:27.913 What is the resistivity Rho, of that resistor? 00:07:27.913 --> 00:07:30.278 Well we know Ohm's Law states that the current that flows 00:07:30.278 --> 00:07:32.446 through a portion of a circuit will be equal to 00:07:32.446 --> 00:07:33.972 the voltage across that portion, 00:07:33.972 --> 00:07:36.278 divided by the resistance of that portion. 00:07:36.278 --> 00:07:38.603 And this means the resistance of this resistor is gonna be 00:07:38.603 --> 00:07:40.883 the voltage of the battery divided by the current. 00:07:40.883 --> 00:07:43.268 To factor resistivity into this, we have to use 00:07:43.268 --> 00:07:46.153 the formula for the resistance of a cylindrical resistor, 00:07:46.153 --> 00:07:48.222 which is Rho times L over A. 00:07:48.222 --> 00:07:50.637 This gives us the resistance of the resistor, 00:07:50.637 --> 00:07:52.735 which is gotta equal V over I, 00:07:52.735 --> 00:07:55.023 and now we can solve for the resistivity Rho. 00:07:55.023 --> 00:07:58.381 We get V times A over I L, but since we're given 00:07:58.381 --> 00:08:01.117 the radius little r, we gotta write the area 00:08:01.117 --> 00:08:03.204 in terms of that radius, so this is gonna be 00:08:03.204 --> 00:08:06.858 V times Pi, r squared, divided by I times L, 00:08:06.858 --> 00:08:09.210 which gives us an answer of C. 00:08:09.210 --> 00:08:12.285 When dealing with complicated circuits with many resistors, 00:08:12.285 --> 00:08:14.214 you often have to reduce those resistors 00:08:14.214 --> 00:08:17.098 into smaller, equivalent amounts of resistors. 00:08:17.098 --> 00:08:18.862 And the two ways you do this are by finding 00:08:18.862 --> 00:08:21.680 two resistors that are in series or in parallel. 00:08:21.680 --> 00:08:24.164 Resistors will be in series if the same current 00:08:24.164 --> 00:08:25.891 that flows through the same resistor, 00:08:25.891 --> 00:08:27.688 flows through the next resistor. 00:08:27.688 --> 00:08:30.157 If the current branched off in between them, 00:08:30.157 --> 00:08:32.791 these resistors would no longer be in series, 00:08:32.791 --> 00:08:34.153 but if they're in series you can find 00:08:34.153 --> 00:08:37.044 the equivalent resistance of this section of wire 00:08:37.044 --> 00:08:40.036 by just adding up the two individual resistances. 00:08:40.036 --> 00:08:41.777 So the current for resistors in series 00:08:41.777 --> 00:08:45.112 must be the same, but the voltage might be different, 00:08:45.112 --> 00:08:47.244 since they could have different resistances. 00:08:47.244 --> 00:08:49.104 Two resistors will be in parallel, 00:08:49.104 --> 00:08:52.000 if a current comes in, splits into two parts, 00:08:52.000 --> 00:08:54.803 goes through one resistor each, and then rejoins 00:08:54.803 --> 00:08:56.818 before hitting anything else in the circuit, 00:08:56.818 --> 00:08:58.768 and if this is the case, you can find the equivalent 00:08:58.768 --> 00:09:00.882 resistance of this portion of the circuit, 00:09:00.882 --> 00:09:04.093 i.e. between these two points, by saying that one 00:09:04.093 --> 00:09:06.519 over the equivalent resistance is gonna equal 00:09:06.519 --> 00:09:08.893 one over the resistance of the first resistor, 00:09:08.893 --> 00:09:11.556 plus one over the resistance of the second resistor. 00:09:11.556 --> 00:09:14.275 But be careful, one over R1 plus one over R2 00:09:14.275 --> 00:09:16.544 just gives you one over R equivalent. 00:09:16.544 --> 00:09:18.663 If you want R equivalent, you're gonna have to take 00:09:18.663 --> 00:09:22.627 one over this entire side, in order to get R equivalent. 00:09:22.627 --> 00:09:24.604 So what's an example problem involving resistors 00:09:24.604 --> 00:09:26.386 in series and parallel look like? 00:09:26.386 --> 00:09:28.357 Let's say we have this circuit shown below, 00:09:28.357 --> 00:09:30.164 and we want to know what current flows 00:09:30.164 --> 00:09:31.726 through the eight Ohm resistor. 00:09:31.726 --> 00:09:33.280 Now you might be tempted to say this, 00:09:33.280 --> 00:09:36.861 since Ohm's Law says that the current is delta V over R, 00:09:36.861 --> 00:09:38.729 we can just plug in the voltage of the battery, 00:09:38.729 --> 00:09:41.330 which is 24 volts, divided by the resistance 00:09:41.330 --> 00:09:43.469 of the resistor, which is eight Ohms, 00:09:43.469 --> 00:09:45.220 and that would give us three Amps. 00:09:45.220 --> 00:09:46.487 But that's not right at all. 00:09:46.487 --> 00:09:48.549 When using Ohm's Law, the current that flows 00:09:48.549 --> 00:09:51.511 through a resistor R, is gonna be equal to 00:09:51.511 --> 00:09:54.382 the voltage across that resistor divided by 00:09:54.382 --> 00:09:56.406 the resistance of that resistor. 00:09:56.406 --> 00:09:59.382 So if we plug eight Ohms into the denominator, we've gotta 00:09:59.382 --> 00:10:02.567 plug in the voltage across that eight Ohm resistor. 00:10:02.567 --> 00:10:04.456 But the voltage across the eight Ohm resistor 00:10:04.456 --> 00:10:07.556 is not gonna be the full 24 volts of the battery, 00:10:07.556 --> 00:10:09.595 it's gonna be less than 24 volts. 00:10:09.595 --> 00:10:12.056 In other words, the battery provides a voltage between 00:10:12.056 --> 00:10:15.144 this point and this point of 24 volts, 00:10:15.144 --> 00:10:16.611 but there's gonna be voltage drops 00:10:16.611 --> 00:10:18.855 across the six and 12 Ohm resistors, 00:10:18.855 --> 00:10:20.138 which make it so that the voltage 00:10:20.138 --> 00:10:21.776 across the eight Ohm resistor is not 00:10:21.776 --> 00:10:23.701 gonna be the full 24 volts. 00:10:23.701 --> 00:10:26.396 So we have to reduce these resistors to a single resistance. 00:10:26.396 --> 00:10:28.335 The six and the 12 are in parallel, 00:10:28.335 --> 00:10:31.351 so we can say that one over six, plus one over 12, 00:10:31.351 --> 00:10:32.907 would equal one over the resistance 00:10:32.907 --> 00:10:34.400 of that portion of the circuit. 00:10:34.400 --> 00:10:37.540 This is gonna equal three twelves, which is one fourth, 00:10:37.540 --> 00:10:39.280 so that means that parallel portion of the circuit 00:10:39.280 --> 00:10:41.857 has an equivalent resistance of four Ohms. 00:10:41.857 --> 00:10:44.159 So between this point and this point, 00:10:44.159 --> 00:10:46.184 there are four Ohms of resistance, 00:10:46.184 --> 00:10:48.469 and that equivalent resistance is in series 00:10:48.469 --> 00:10:50.141 with this eight Ohm resistor. 00:10:50.141 --> 00:10:51.652 So we can add four and eight, 00:10:51.652 --> 00:10:54.217 and get 12 Ohms of total resistance. 00:10:54.217 --> 00:10:57.720 And now I can say that the full 24 volts of the battery 00:10:57.720 --> 00:11:00.135 is applied across this entire equivalent resistance 00:11:00.135 --> 00:11:03.685 of 12 Ohms, so if I come up here and change this eight Ohms 00:11:03.685 --> 00:11:06.980 to 12 Ohms of equivalent resistance for the total circuit, 00:11:06.980 --> 00:11:08.440 I'll get the correct current that flows 00:11:08.440 --> 00:11:10.322 through the battery of two Amps. 00:11:10.322 --> 00:11:12.143 And since that's the current that's flowing through 00:11:12.143 --> 00:11:13.879 the battery, that had to be the current 00:11:13.879 --> 00:11:16.343 that's flowing through the eight Ohm resistor as well. 00:11:16.343 --> 00:11:20.059 Since this eight Ohm resistor and the batter are in series. 00:11:20.059 --> 00:11:23.152 Elements in a circuit often use Electrical Power. 00:11:23.152 --> 00:11:25.892 That is to say, when current runs through a resistor, 00:11:25.892 --> 00:11:27.726 the electrons moving through that resistor 00:11:27.726 --> 00:11:30.428 turn some of their electrical potential energy 00:11:30.428 --> 00:11:33.766 into energies like heat, light, or sound. 00:11:33.766 --> 00:11:35.408 And the rate at which these electrons 00:11:35.408 --> 00:11:37.978 are turning their energy into other forms of energy, 00:11:37.978 --> 00:11:39.854 is called the electrical power. 00:11:39.854 --> 00:11:41.815 So the rate at which a resistor is turning 00:11:41.815 --> 00:11:43.974 electrical potential energy into heat, 00:11:43.974 --> 00:11:46.481 is the electrical power used by that resistor. 00:11:46.481 --> 00:11:47.940 In other words, the amount of energy 00:11:47.940 --> 00:11:50.084 converted into heat, divided by the time 00:11:50.084 --> 00:11:52.380 it took to convert that energy, is the definition 00:11:52.380 --> 00:11:54.358 of the power, and there's a way to determine 00:11:54.358 --> 00:11:57.404 this number of Joules per second, in terms of quantities 00:11:57.404 --> 00:12:00.199 like the current, the voltage, and the resistance. 00:12:00.199 --> 00:12:02.425 The power used by a resistor can be written as 00:12:02.425 --> 00:12:04.410 the current through that resistor 00:12:04.410 --> 00:12:07.436 multiplied by the voltage across that resistor, 00:12:07.436 --> 00:12:10.661 or if you substituted Ohm's Law into this formula, 00:12:10.661 --> 00:12:12.181 you see that this is equivalent to 00:12:12.181 --> 00:12:14.319 the current through that resistor squared, 00:12:14.319 --> 00:12:16.759 multiplied by the resistance of the resistor, 00:12:16.759 --> 00:12:18.132 or we could rearrange these formulas 00:12:18.132 --> 00:12:20.025 to get that the power used by a resistor 00:12:20.025 --> 00:12:23.124 would also be the voltage across that resistor squared, 00:12:23.124 --> 00:12:25.521 divided by the resistance of that resistor. 00:12:25.521 --> 00:12:27.540 All three of these, if used correctly, 00:12:27.540 --> 00:12:30.130 will give you the same number for the power used 00:12:30.130 --> 00:12:32.074 by a resistor, and if you wanted to determine 00:12:32.074 --> 00:12:34.605 the number of Joules of heat energy converted, 00:12:34.605 --> 00:12:37.054 you could set any one of these equal to the amount 00:12:37.054 --> 00:12:40.122 of energy per time, and solve for that energy. 00:12:40.122 --> 00:12:42.561 The units of Electrical Power are the same as the regular 00:12:42.561 --> 00:12:46.262 units of power, which is Watts, i.e. Joules per second, 00:12:46.262 --> 00:12:48.333 and Electrical Power is not a vector. 00:12:48.333 --> 00:12:49.836 So what's an example problem involving 00:12:49.836 --> 00:12:51.452 Electrical Power look like? 00:12:51.452 --> 00:12:53.296 Let's say a light bulb of resistance R 00:12:53.296 --> 00:12:55.416 is hooked up to a source of voltage V, 00:12:55.416 --> 00:12:57.694 and a second light bulb of resistance 2R, 00:12:57.694 --> 00:13:00.246 is hooked up to a source of voltage 2V. 00:13:00.246 --> 00:13:02.786 How does the power used by the second light bulb 00:13:02.786 --> 00:13:05.285 compare to the power used by the first light bulb? 00:13:05.285 --> 00:13:07.553 Since we have the information about R and V, 00:13:07.553 --> 00:13:09.382 I'll use the version of the power formula 00:13:09.382 --> 00:13:11.645 that says that the power used by a resistor 00:13:11.645 --> 00:13:13.980 is gonna be delta V squared over R. 00:13:13.980 --> 00:13:15.937 So in terms of quantities given the power used 00:13:15.937 --> 00:13:18.797 by the first light bulb is gonna be V squared over R. 00:13:18.797 --> 00:13:21.225 And the power used by the second light bulb is gonna be 00:13:21.225 --> 00:13:23.779 equal to the voltage across the second light bulb, 00:13:23.779 --> 00:13:26.852 which is two times the voltage across the first light bulb, 00:13:26.852 --> 00:13:28.953 and we square that, divided by the resistance 00:13:28.953 --> 00:13:30.678 of the second light bulb, which is gonna be 00:13:30.678 --> 00:13:33.015 two times the resistance of the first light bulb. 00:13:33.015 --> 00:13:35.547 The two squared on top is gonna give me a factor of four, 00:13:35.547 --> 00:13:37.708 and I'll have another factor of two on the bottom. 00:13:37.708 --> 00:13:40.032 So if I factor out this four divided by two, 00:13:40.032 --> 00:13:42.269 I get that the power used by the second light bulb 00:13:42.269 --> 00:13:44.666 is gonna be two times V squared over R, 00:13:44.666 --> 00:13:46.663 but V squared over R was the power used 00:13:46.663 --> 00:13:48.658 by the first light bulb, so the power used 00:13:48.658 --> 00:13:50.653 by the second light bulb is gonna be two 00:13:50.653 --> 00:13:52.838 times the power used by the first light bulb, 00:13:52.838 --> 00:13:54.681 and so if the light bulb of resistance two R 00:13:54.681 --> 00:13:57.662 has twice the power, and that means it'll be brighter. 00:13:57.662 --> 00:13:59.591 The quantity that determines the brightness 00:13:59.591 --> 00:14:03.124 of a light bulb, is the electrical power of that light bulb. 00:14:03.124 --> 00:14:06.318 It's not necessarily the resistance or the voltage, 00:14:06.318 --> 00:14:07.821 it's the combination of the two 00:14:07.821 --> 00:14:10.895 in this formula that will tell you the electrical power, 00:14:10.895 --> 00:14:13.649 and therefore the brightness of the light bulb. 00:14:13.649 --> 00:14:15.519 Two of the most useful ideas in circuits 00:14:15.519 --> 00:14:17.767 are referred to as Kirchhoff's Rules. 00:14:17.767 --> 00:14:19.854 The first rule is called the Junction rule, 00:14:19.854 --> 00:14:22.040 and it states that all the current entering a junction 00:14:22.040 --> 00:14:25.140 must equal all the current exiting that junction. 00:14:25.140 --> 00:14:26.961 In other words, if you add all the current that flows 00:14:26.961 --> 00:14:28.997 into a junction, that has to equal 00:14:28.997 --> 00:14:31.260 all the current that flows out of that junction, 00:14:31.260 --> 00:14:33.263 because current is just flowing charge, 00:14:33.263 --> 00:14:35.897 and charge is conserved, so charge can't be 00:14:35.897 --> 00:14:38.964 created or destroyed at any point in the circuit. 00:14:38.964 --> 00:14:41.506 No more than water can get created or destroyed 00:14:41.506 --> 00:14:43.213 within a series of pipes. 00:14:43.213 --> 00:14:45.080 And the second rule is called the Loop rule, 00:14:45.080 --> 00:14:47.515 which states that if you add up all the changes 00:14:47.515 --> 00:14:51.235 in electric potential, i.e. voltages around any closed 00:14:51.235 --> 00:14:54.238 loop in a circuit, it'll always add up to zero. 00:14:54.238 --> 00:14:56.464 So if you add up all the voltages encountered 00:14:56.464 --> 00:14:58.191 through a closed loop through a circuit, 00:14:58.191 --> 00:14:59.768 it always adds up to zero. 00:14:59.768 --> 00:15:02.273 And this is just a result of conservation of energy. 00:15:02.273 --> 00:15:04.408 The electrons will gain energy when they flow 00:15:04.408 --> 00:15:06.410 through the battery, and they'll lose energy 00:15:06.410 --> 00:15:08.396 every time they flow through a resistor, 00:15:08.396 --> 00:15:09.919 but the total amount of energy they gain 00:15:09.919 --> 00:15:12.443 from the battery, has to be equal to the total amount 00:15:12.443 --> 00:15:14.956 of energy they lose due to the resistors. 00:15:14.956 --> 00:15:17.345 In other words, if we consider a complicated circuit 00:15:17.345 --> 00:15:19.759 that has a batter and three resistors, 00:15:19.759 --> 00:15:23.125 the total current flowing into a junction I1, 00:15:23.125 --> 00:15:25.065 will have to be equal to the total current 00:15:25.065 --> 00:15:28.293 coming out of that junction, I2 and I3. 00:15:28.293 --> 00:15:30.951 Since no charge gets created or destroyed. 00:15:30.951 --> 00:15:32.986 And that means when these two currents combine again, 00:15:32.986 --> 00:15:34.482 the total current flowing out 00:15:34.482 --> 00:15:37.194 of that section is gonna again be I1. 00:15:37.194 --> 00:15:39.990 And if we follow a closed loop through this circuit, 00:15:39.990 --> 00:15:42.372 the sum of all the voltages around that loop 00:15:42.372 --> 00:15:45.919 have to add up to zero, i.e. the voltage of the battery 00:15:45.919 --> 00:15:48.939 minus the voltage drop across the first resistor, 00:15:48.939 --> 00:15:51.784 minus the voltage drop across the second resistor, 00:15:51.784 --> 00:15:53.480 would have to equal zero. 00:15:53.480 --> 00:15:54.894 So what's an example problem involving 00:15:54.894 --> 00:15:56.496 Kirchhoff's Rules look like? 00:15:56.496 --> 00:15:58.513 Let's say we have the circuit below and we wanted to 00:15:58.513 --> 00:16:01.702 determine the voltage across the six Ohm resistor. 00:16:01.702 --> 00:16:03.409 To do this, we could use the Loop rule, 00:16:03.409 --> 00:16:05.633 I'll start behind the battery, and I'll go through 00:16:05.633 --> 00:16:08.392 the resistor, I want to determine the voltage across. 00:16:08.392 --> 00:16:10.932 I'll add up all the voltages across that loop, 00:16:10.932 --> 00:16:12.364 and set it equal to zero. 00:16:12.364 --> 00:16:13.556 So the voltage across the battery 00:16:13.556 --> 00:16:15.703 is gonna be positive 24 volts, 00:16:15.703 --> 00:16:18.276 minus the voltage across the six Ohm resistor, 00:16:18.276 --> 00:16:19.745 and then minus the voltage across the 00:16:19.745 --> 00:16:22.220 eight Ohm resistor has to equal zero. 00:16:22.220 --> 00:16:24.676 But we're given this current, so we know that two Amps 00:16:24.676 --> 00:16:26.706 flows through the eight Ohm resistor, 00:16:26.706 --> 00:16:27.847 and you can always determine the voltage 00:16:27.847 --> 00:16:29.939 across the resistor using Ohm's Law, 00:16:29.939 --> 00:16:31.870 so the voltage across the eight Ohm resistor 00:16:31.870 --> 00:16:33.870 is gonna be two Amps, which is flowing 00:16:33.870 --> 00:16:35.436 through the eight Ohm resistor, 00:16:35.436 --> 00:16:39.129 multiplied by eight Ohms, and we get 16 volts. 00:16:39.129 --> 00:16:41.837 Which I can plug into here, and this gives me 00:16:41.837 --> 00:16:45.686 24 volts minus the voltage across the six Ohm resistor, 00:16:45.686 --> 00:16:48.389 minus 16, has to equal zero. 00:16:48.389 --> 00:16:49.583 And if I solve this for the voltage 00:16:49.583 --> 00:16:52.253 across the six Ohm resistor, I get 24 volts 00:16:52.253 --> 00:16:54.871 minus 16 volts, which is eight volts. 00:16:54.871 --> 00:16:56.721 So the voltage across the six Ohm resistor 00:16:56.721 --> 00:16:58.308 would be eight volts. 00:16:58.308 --> 00:17:01.392 Note, because the 12 Ohm resistor and the six Ohm resistor 00:17:01.392 --> 00:17:04.095 are in parallel, the voltage across the 12 Ohm resistor 00:17:04.095 --> 00:17:07.078 would also be eight volts, because the voltage across 00:17:07.078 --> 00:17:11.633 any two elements in parallel, have to be the same. 00:17:11.633 --> 00:17:13.549 Voltmeters are the device that you use 00:17:13.549 --> 00:17:16.398 to measure the voltage between two points in a circuit. 00:17:16.398 --> 00:17:18.113 When hooking up a voltmeter you've gotta 00:17:18.113 --> 00:17:20.580 hook it up in parallel between the two points 00:17:20.580 --> 00:17:22.600 you wanna find the voltage across. 00:17:22.600 --> 00:17:24.015 In other words, to determine the voltage 00:17:24.015 --> 00:17:26.388 between this point and this point, which would be 00:17:26.388 --> 00:17:29.235 the voltage across R3, you would hook up the voltmeter 00:17:29.235 --> 00:17:31.465 in parallel with R3. 00:17:31.465 --> 00:17:33.351 Ammeters are the devices we use to measure 00:17:33.351 --> 00:17:34.908 the electrical current that pass 00:17:34.908 --> 00:17:36.780 through a point in a circuit, and ammeters 00:17:36.780 --> 00:17:39.595 have to be hooked up in series with the circuit element 00:17:39.595 --> 00:17:41.373 you want to determine the current through. 00:17:41.373 --> 00:17:43.261 In other words, if we wanted to determine the current 00:17:43.261 --> 00:17:47.267 through R1, we would hook up the ammeter in series with R1. 00:17:47.267 --> 00:17:49.877 Note that for these electrical devices to work well, 00:17:49.877 --> 00:17:53.185 the ammeter should have almost zero internal resistance, 00:17:53.185 --> 00:17:55.407 thereby not affecting the current that flows 00:17:55.407 --> 00:17:57.479 through the circuit, and voltmeters should have 00:17:57.479 --> 00:18:00.440 near infinite resistance, so that it doesn't draw 00:18:00.440 --> 00:18:02.674 any of the current from the resistor. 00:18:02.674 --> 00:18:04.969 In reality, ammeters have a very small, 00:18:04.969 --> 00:18:07.148 but non-zero internal resistance, 00:18:07.148 --> 00:18:08.463 and voltmeters have a very high, 00:18:08.463 --> 00:18:10.653 but not infinite internal resistance. 00:18:10.653 --> 00:18:12.107 So what would an example problem involving 00:18:12.107 --> 00:18:14.108 voltmeters and ammeters look like? 00:18:14.108 --> 00:18:15.865 Let's say we have the circuit shown below, 00:18:15.865 --> 00:18:18.558 and these numbered circles represent possible places 00:18:18.558 --> 00:18:21.173 we could stick a voltmeter to measure the voltage 00:18:21.173 --> 00:18:23.165 across the eight Ohm resistor. 00:18:23.165 --> 00:18:25.380 Which two of these voltmeters would correctly 00:18:25.380 --> 00:18:28.079 give the voltage across the eight Ohm resistor? 00:18:28.079 --> 00:18:30.124 And you have to be careful, some AP problems 00:18:30.124 --> 00:18:31.844 are gonna require you to select 00:18:31.844 --> 00:18:33.838 two correct answers for the multiple choice, 00:18:33.838 --> 00:18:36.509 so be sure to read the instructions carefully. 00:18:36.509 --> 00:18:38.229 Voltmeter number four is a terrible choice, 00:18:38.229 --> 00:18:40.368 you never hook up your voltmeter in series, 00:18:40.368 --> 00:18:42.331 but the circuit element you're trying to find 00:18:42.331 --> 00:18:44.382 the voltage across, and voltmeter number one 00:18:44.382 --> 00:18:46.524 is doing nothing really, because its' measuring 00:18:46.524 --> 00:18:48.989 the voltage between two points in a wire 00:18:48.989 --> 00:18:51.206 with nothing in between that wire. 00:18:51.206 --> 00:18:52.952 So the voltage measured by voltmeter one 00:18:52.952 --> 00:18:56.034 should just be zero, since the voltage across a wire 00:18:56.034 --> 00:18:59.412 of zero resistance should just give you zero volts. 00:18:59.412 --> 00:19:01.627 So the correct choices would be voltmeter number two, 00:19:01.627 --> 00:19:04.736 which gives you the voltage across the eight Ohm resistor, 00:19:04.736 --> 00:19:06.429 and voltmeter number three, which also gives you 00:19:06.429 --> 00:19:08.304 an equivalent measurement of the voltage 00:19:08.304 --> 00:19:10.870 across the resistor eight Ohms.