0:00:00.468,0:00:01.688 - [Voiceover] Electric[br]Charge is a property 0:00:01.688,0:00:05.622 that some, but not all fundamental[br]particles in nature have. 0:00:05.622,0:00:06.761 The most commonly talked about 0:00:06.761,0:00:09.562 fundamentally charged[br]particles are the electrons, 0:00:09.562,0:00:11.564 which orbit the outside of the atom. 0:00:11.564,0:00:13.345 These are negatively charged. 0:00:13.345,0:00:15.146 There's also the protons, which reside 0:00:15.146,0:00:18.484 inside the nucleus, and[br]these are positively charged. 0:00:18.484,0:00:20.088 And the neutrons inside the nucleus 0:00:20.088,0:00:21.855 don't have any net charge. 0:00:21.855,0:00:24.212 It turns out that all[br]fundamentally charged particles 0:00:24.212,0:00:26.310 in the universe, have charges that come in 0:00:26.310,0:00:29.258 integer units of the elementary charge. 0:00:29.258,0:00:30.804 So if you find a particle in nature, 0:00:30.804,0:00:33.916 it's gonna have a charge[br]of one times this number, 0:00:33.916,0:00:37.016 two times this number,[br]three times this number, 0:00:37.016,0:00:38.717 and it could either be[br]positive or negative. 0:00:38.717,0:00:41.690 For instance, the electron[br]has a charge of negative 1.6 0:00:41.690,0:00:44.658 times 10 to the negative 19th Coulombs, 0:00:44.658,0:00:47.542 and the charge of the[br]proton is positive 1.6 0:00:47.542,0:00:49.889 times 10 to the negative 19th Coulombs. 0:00:49.889,0:00:51.928 However, most atoms in the universe 0:00:51.928,0:00:54.076 are electrically neutral overall, 0:00:54.076,0:00:56.828 since they'll have just[br]as many negative electrons 0:00:56.828,0:00:59.013 as they do positive protons. 0:00:59.013,0:01:01.307 But if an atom had too many electrons, 0:01:01.307,0:01:04.009 overall that atom would[br]be negatively charged, 0:01:04.009,0:01:06.501 and if an atom had too few electrons, 0:01:06.501,0:01:08.844 that atom would be overall[br]positively charged. 0:01:08.844,0:01:10.593 And something that's really[br]important to remember 0:01:10.593,0:01:13.077 is that the electric[br]charge is always conserved 0:01:13.077,0:01:15.212 for every process, in other words, 0:01:15.212,0:01:17.539 the total charge initially, is gonna equal 0:01:17.539,0:01:20.586 the total charge finally[br]after any process. 0:01:20.586,0:01:22.086 So what's an example problem involving 0:01:22.086,0:01:23.660 electric charge look like? 0:01:23.660,0:01:26.187 Let's say three identically[br]sized metal spheres 0:01:26.187,0:01:28.344 start off with the charges seen below. 0:01:28.344,0:01:32.136 Positive five Q, positive[br]three Q, and negative two Q. 0:01:32.136,0:01:35.793 If we touch sphere X to[br]sphere Y, and separate them, 0:01:35.793,0:01:39.609 and then touch sphere Y to[br]sphere Z, and separate them, 0:01:39.609,0:01:42.575 what will be the final[br]charge on each sphere? 0:01:42.575,0:01:44.547 Well first, when we touch X to Y, 0:01:44.547,0:01:46.693 the total charge has to be conserved. 0:01:46.693,0:01:49.224 There's a total charge[br]of eight Q amongst them, 0:01:49.224,0:01:50.934 and since they're identically sized, 0:01:50.934,0:01:52.779 they'll both share the total charge, 0:01:52.779,0:01:54.038 which means after they touch, 0:01:54.038,0:01:56.008 they'll both have positive four Q. 0:01:56.008,0:01:57.429 If one of the spheres were larger, 0:01:57.429,0:01:59.028 it would gain more of the charge, 0:01:59.028,0:02:01.378 but the total charge[br]would still be conserved. 0:02:01.378,0:02:04.193 And now when sphere Y[br]is touched to sphere Z, 0:02:04.193,0:02:06.414 the total charge amongst[br]them at that moment 0:02:06.414,0:02:09.443 would be positive four[br]Q plus negative two Q, 0:02:09.443,0:02:10.976 which is positive two Q. 0:02:10.976,0:02:13.493 They would share it equally,[br]so sphere Y would have 0:02:13.493,0:02:17.029 positive Q, and sphere Z[br]would also have positive Q. 0:02:17.029,0:02:19.247 So the answer here is C. 0:02:19.247,0:02:22.288 Opposite charges attract[br]and like charges repel, 0:02:22.288,0:02:24.874 and what Coulomb's Law does[br]is it gives you a way to find 0:02:24.874,0:02:28.499 the magnitude of the electric[br]force between two charges. 0:02:28.499,0:02:30.069 The formula for Coulomb's Law says 0:02:30.069,0:02:32.123 that the magnitude of the electric force 0:02:32.123,0:02:36.003 between two charges Q1[br]and Q2, is gonna equal the 0:02:36.003,0:02:39.203 electric constant K, which is[br]nine times 10 to the ninth, 0:02:39.203,0:02:42.619 times the product of the two[br]charges, measured in Coulombs 0:02:42.619,0:02:44.489 divided by the center to center distance 0:02:44.489,0:02:47.260 between those two charges, squared. 0:02:47.260,0:02:49.542 You can't forget to square this distance. 0:02:49.542,0:02:51.612 And it's gotta be in[br]meters if you want to find 0:02:51.612,0:02:54.151 SI units of Newtons for the force. 0:02:54.151,0:02:56.716 Also, don't rely on the[br]negative and positive signs 0:02:56.716,0:02:59.626 of the charges to tell you[br]which way the force points, 0:02:59.626,0:03:02.064 just use the fact that[br]opposite charges attract 0:03:02.064,0:03:04.986 and like charges repel,[br]and use Coulomb's Law 0:03:04.986,0:03:07.175 to get the magnitude of the force. 0:03:07.175,0:03:08.500 So what's an example problem involving 0:03:08.500,0:03:09.790 Coulomb's Law look like? 0:03:09.790,0:03:11.653 Let's say two charges[br]exert an electric force 0:03:11.653,0:03:13.586 of magnitude F on each other. 0:03:13.586,0:03:16.336 What would be the magnitude[br]of the new electric force 0:03:16.336,0:03:18.749 if the distance between[br]the charges is tripled 0:03:18.749,0:03:21.687 and the magnitude of one[br]of the charges is doubled? 0:03:21.687,0:03:23.375 Well we know the formula for Coulomb's Law 0:03:23.375,0:03:25.265 says that the force between two charges 0:03:25.265,0:03:28.026 is the electric constant[br]times one of the charges, 0:03:28.026,0:03:30.354 times the other charge[br]divided by the distance 0:03:30.354,0:03:33.321 between them squared, and now[br]once we triple the distance 0:03:33.321,0:03:36.154 and double a charge,[br]the new electric force 0:03:36.154,0:03:39.180 is gonna be the electric constant[br]times one of the charges, 0:03:39.180,0:03:42.719 multiplied by two times[br]one of the charges, 0:03:42.719,0:03:46.402 divided by three times the[br]distance, which is squared, 0:03:46.402,0:03:48.148 so I'm gonna get a factor of two on top, 0:03:48.148,0:03:49.926 and this three will get[br]squared, which gives me 0:03:49.926,0:03:51.761 a factor of nine on the bottom. 0:03:51.761,0:03:54.148 If I pull up those extra[br]factors I get that the new force 0:03:54.148,0:03:58.588 is gonna be two ninths[br]multiplied by K, Q1, Q2, 0:03:58.588,0:04:00.993 over D squared, but this entire quantity 0:04:00.993,0:04:03.971 was just the old force F, so the new force 0:04:03.971,0:04:07.731 is going to be two[br]ninths of the old force. 0:04:07.731,0:04:10.644 The electrical current I tells[br]you the amount of Coulombs 0:04:10.644,0:04:14.061 of charge that passes a[br]point in a wire per second. 0:04:14.061,0:04:16.019 So if you watch some point in a wire, 0:04:16.019,0:04:18.113 and you count how many Coulombs of charge 0:04:18.113,0:04:21.328 pass by that point per second,[br]that would be the current. 0:04:21.328,0:04:23.659 Or in equation form we[br]can see that the current I 0:04:23.659,0:04:25.628 is gonna be the amount[br]of charge that flows 0:04:25.628,0:04:27.897 past a point in a wire per time. 0:04:27.897,0:04:30.588 This gives the units of[br]I as Coulombs per second, 0:04:30.588,0:04:32.848 which we abbreviate as an Ampere. 0:04:32.848,0:04:34.681 And since charge and time aren't vectors, 0:04:34.681,0:04:36.499 current is not a vector either. 0:04:36.499,0:04:37.736 Something that's kind of strange 0:04:37.736,0:04:40.908 is that the so-called[br]conventional direction of current 0:04:40.908,0:04:43.475 would be the direction that[br]positive charges flow within a 0:04:43.475,0:04:47.171 wire, however positive charges[br]don't flow within a wire. 0:04:47.171,0:04:49.458 The only charges that[br]actually flow in a wire 0:04:49.458,0:04:51.658 are negative charges, but it turns out 0:04:51.658,0:04:55.019 that negative charges flowing[br]to the left is physically 0:04:55.019,0:04:57.932 the same as positive charges[br]flowing to the right. 0:04:57.932,0:05:00.498 So in physics problems[br]we pretend as if it were 0:05:00.498,0:05:02.819 the positive charges moving, however 0:05:02.819,0:05:04.889 it's really the electrons,[br]which are negative, 0:05:04.889,0:05:06.464 that are moving within the wire. 0:05:06.464,0:05:07.410 So what's an example problem 0:05:07.410,0:05:09.295 involving electrical current look like? 0:05:09.295,0:05:11.387 Let's say three amps[br]flows within a circuit. 0:05:11.387,0:05:14.014 How much charge would pass[br]by a point in that wire 0:05:14.014,0:05:16.505 during a time interval of five minutes? 0:05:16.505,0:05:17.904 Well we know the definition of current 0:05:17.904,0:05:20.341 is the charge per time,[br]that means the charge 0:05:20.341,0:05:23.285 is gonna be the amount of[br]current multiplied by the time, 0:05:23.285,0:05:25.303 so we take our current of three amps, 0:05:25.303,0:05:27.305 and we multiply by the time, but we can't 0:05:27.305,0:05:30.295 multiply by five because[br]that's in units of minutes, 0:05:30.295,0:05:32.486 since amps is Coulombs per second, 0:05:32.486,0:05:34.535 we've got to convert five[br]minutes into seconds, 0:05:34.535,0:05:37.934 which would be five minutes,[br]multiplied by 60 seconds 0:05:37.934,0:05:39.833 per minute, which would[br]give us a total amount 0:05:39.833,0:05:42.000 of charge of 900 Coulombs. 0:05:42.890,0:05:45.794 The resistance of a circuit[br]element measures how much 0:05:45.794,0:05:48.703 that element will restrict[br]the flow of current. 0:05:48.703,0:05:51.132 The larger the resistance,[br]the less current 0:05:51.132,0:05:52.763 there will be allowed to flow. 0:05:52.763,0:05:55.978 And this definition of[br]resistance is given by Ohm's Law. 0:05:55.978,0:05:57.794 Ohm's Law states that[br]the amount of current 0:05:57.794,0:05:59.759 that you'll get through[br]a portion of a circuit, 0:05:59.759,0:06:01.476 it's gonna be proportional to the voltage 0:06:01.476,0:06:04.371 across that portion,[br]divided by the resistance 0:06:04.371,0:06:06.006 of that portion of the circuit. 0:06:06.006,0:06:08.331 So between these two points,[br]the amount of current 0:06:08.331,0:06:10.198 that will flow, is gonna be equal to 0:06:10.198,0:06:12.536 the voltage between those two points, 0:06:12.536,0:06:15.604 divided by the resistance[br]between those two points. 0:06:15.604,0:06:17.889 So the larger the[br]resistance, the less current 0:06:17.889,0:06:20.958 will flow, but the greater[br]the voltage supplied, 0:06:20.958,0:06:22.418 the greater the current will be. 0:06:22.418,0:06:24.107 And this is what Ohm's Law says. 0:06:24.107,0:06:26.316 Even though Ohm's Law[br]gives you a way to define 0:06:26.316,0:06:29.082 the resistance, you can[br]determine the resistance 0:06:29.082,0:06:31.043 of a circuit element by knowing the size 0:06:31.043,0:06:32.949 and shape of that circuit element. 0:06:32.949,0:06:35.568 In other words, the resistance[br]of a cylindrical resistor, 0:06:35.568,0:06:37.872 is gonna be equal to the resistivity, 0:06:37.872,0:06:40.214 which is a measure of an[br]object's natural resistance 0:06:40.214,0:06:43.126 to current, multiplied by[br]the length of that resistor, 0:06:43.126,0:06:45.768 the longer the resistor,[br]the greater the resistance 0:06:45.768,0:06:48.072 and the more it will[br]resist the flow of current, 0:06:48.072,0:06:50.199 and then divide it by[br]the cross sectional area 0:06:50.199,0:06:52.860 of that resistor, which would[br]be this area right here, 0:06:52.860,0:06:55.583 the current is either[br]flowing into or out of. 0:06:55.583,0:06:58.450 If the resistor is cylindrical,[br]the area of this circle 0:06:58.450,0:07:01.313 would be Pi times r[br]squared, where little r 0:07:01.313,0:07:04.356 would be the radius of[br]this cross sectional area. 0:07:04.356,0:07:07.521 The units of resistance is[br]Ohms, and it is not a vector. 0:07:07.521,0:07:09.746 It is always positive or zero. 0:07:09.746,0:07:11.888 So what's an example[br]problem involving Ohm's Law, 0:07:11.888,0:07:14.699 or the resistance of a[br]cylindrical resistor look like? 0:07:14.699,0:07:16.776 Let's say a battery of[br]voltage V is hooked up 0:07:16.776,0:07:19.744 to a single cylindrical[br]resistor of length L 0:07:19.744,0:07:21.813 and radius little r, and when that's done, 0:07:21.813,0:07:24.482 a current I is flowing[br]through the battery. 0:07:24.482,0:07:27.913 What is the resistivity[br]Rho, of that resistor? 0:07:27.913,0:07:30.278 Well we know Ohm's Law states[br]that the current that flows 0:07:30.278,0:07:32.446 through a portion of a[br]circuit will be equal to 0:07:32.446,0:07:33.972 the voltage across that portion, 0:07:33.972,0:07:36.278 divided by the resistance of that portion. 0:07:36.278,0:07:38.603 And this means the resistance[br]of this resistor is gonna be 0:07:38.603,0:07:40.883 the voltage of the battery[br]divided by the current. 0:07:40.883,0:07:43.268 To factor resistivity[br]into this, we have to use 0:07:43.268,0:07:46.153 the formula for the resistance[br]of a cylindrical resistor, 0:07:46.153,0:07:48.222 which is Rho times L over A. 0:07:48.222,0:07:50.637 This gives us the[br]resistance of the resistor, 0:07:50.637,0:07:52.735 which is gotta equal V over I, 0:07:52.735,0:07:55.023 and now we can solve[br]for the resistivity Rho. 0:07:55.023,0:07:58.381 We get V times A over I[br]L, but since we're given 0:07:58.381,0:08:01.117 the radius little r,[br]we gotta write the area 0:08:01.117,0:08:03.204 in terms of that radius,[br]so this is gonna be 0:08:03.204,0:08:06.858 V times Pi, r squared,[br]divided by I times L, 0:08:06.858,0:08:09.210 which gives us an answer of C. 0:08:09.210,0:08:12.285 When dealing with complicated[br]circuits with many resistors, 0:08:12.285,0:08:14.214 you often have to reduce those resistors 0:08:14.214,0:08:17.098 into smaller, equivalent[br]amounts of resistors. 0:08:17.098,0:08:18.862 And the two ways you[br]do this are by finding 0:08:18.862,0:08:21.680 two resistors that are[br]in series or in parallel. 0:08:21.680,0:08:24.164 Resistors will be in[br]series if the same current 0:08:24.164,0:08:25.891 that flows through the same resistor, 0:08:25.891,0:08:27.688 flows through the next resistor. 0:08:27.688,0:08:30.157 If the current branched[br]off in between them, 0:08:30.157,0:08:32.791 these resistors would[br]no longer be in series, 0:08:32.791,0:08:34.153 but if they're in series you can find 0:08:34.153,0:08:37.044 the equivalent resistance[br]of this section of wire 0:08:37.044,0:08:40.036 by just adding up the two[br]individual resistances. 0:08:40.036,0:08:41.777 So the current for resistors in series 0:08:41.777,0:08:45.112 must be the same, but the[br]voltage might be different, 0:08:45.112,0:08:47.244 since they could have[br]different resistances. 0:08:47.244,0:08:49.104 Two resistors will be in parallel, 0:08:49.104,0:08:52.000 if a current comes in,[br]splits into two parts, 0:08:52.000,0:08:54.803 goes through one resistor[br]each, and then rejoins 0:08:54.803,0:08:56.818 before hitting anything[br]else in the circuit, 0:08:56.818,0:08:58.768 and if this is the case,[br]you can find the equivalent 0:08:58.768,0:09:00.882 resistance of this portion of the circuit, 0:09:00.882,0:09:04.093 i.e. between these two[br]points, by saying that one 0:09:04.093,0:09:06.519 over the equivalent[br]resistance is gonna equal 0:09:06.519,0:09:08.893 one over the resistance[br]of the first resistor, 0:09:08.893,0:09:11.556 plus one over the resistance[br]of the second resistor. 0:09:11.556,0:09:14.275 But be careful, one[br]over R1 plus one over R2 0:09:14.275,0:09:16.544 just gives you one over R equivalent. 0:09:16.544,0:09:18.663 If you want R equivalent,[br]you're gonna have to take 0:09:18.663,0:09:22.627 one over this entire side,[br]in order to get R equivalent. 0:09:22.627,0:09:24.604 So what's an example[br]problem involving resistors 0:09:24.604,0:09:26.386 in series and parallel look like? 0:09:26.386,0:09:28.357 Let's say we have this[br]circuit shown below, 0:09:28.357,0:09:30.164 and we want to know what current flows 0:09:30.164,0:09:31.726 through the eight Ohm resistor. 0:09:31.726,0:09:33.280 Now you might be tempted to say this, 0:09:33.280,0:09:36.861 since Ohm's Law says that the[br]current is delta V over R, 0:09:36.861,0:09:38.729 we can just plug in the[br]voltage of the battery, 0:09:38.729,0:09:41.330 which is 24 volts,[br]divided by the resistance 0:09:41.330,0:09:43.469 of the resistor, which is eight Ohms, 0:09:43.469,0:09:45.220 and that would give us three Amps. 0:09:45.220,0:09:46.487 But that's not right at all. 0:09:46.487,0:09:48.549 When using Ohm's Law,[br]the current that flows 0:09:48.549,0:09:51.511 through a resistor R, is gonna be equal to 0:09:51.511,0:09:54.382 the voltage across that[br]resistor divided by 0:09:54.382,0:09:56.406 the resistance of that resistor. 0:09:56.406,0:09:59.382 So if we plug eight Ohms into[br]the denominator, we've gotta 0:09:59.382,0:10:02.567 plug in the voltage across[br]that eight Ohm resistor. 0:10:02.567,0:10:04.456 But the voltage across[br]the eight Ohm resistor 0:10:04.456,0:10:07.556 is not gonna be the full[br]24 volts of the battery, 0:10:07.556,0:10:09.595 it's gonna be less than 24 volts. 0:10:09.595,0:10:12.056 In other words, the battery[br]provides a voltage between 0:10:12.056,0:10:15.144 this point and this point of 24 volts, 0:10:15.144,0:10:16.611 but there's gonna be voltage drops 0:10:16.611,0:10:18.855 across the six and 12 Ohm resistors, 0:10:18.855,0:10:20.138 which make it so that the voltage 0:10:20.138,0:10:21.776 across the eight Ohm resistor is not 0:10:21.776,0:10:23.701 gonna be the full 24 volts. 0:10:23.701,0:10:26.396 So we have to reduce these[br]resistors to a single resistance. 0:10:26.396,0:10:28.335 The six and the 12 are in parallel, 0:10:28.335,0:10:31.351 so we can say that one[br]over six, plus one over 12, 0:10:31.351,0:10:32.907 would equal one over the resistance 0:10:32.907,0:10:34.400 of that portion of the circuit. 0:10:34.400,0:10:37.540 This is gonna equal three[br]twelves, which is one fourth, 0:10:37.540,0:10:39.280 so that means that parallel[br]portion of the circuit 0:10:39.280,0:10:41.857 has an equivalent resistance of four Ohms. 0:10:41.857,0:10:44.159 So between this point and this point, 0:10:44.159,0:10:46.184 there are four Ohms of resistance, 0:10:46.184,0:10:48.469 and that equivalent[br]resistance is in series 0:10:48.469,0:10:50.141 with this eight Ohm resistor. 0:10:50.141,0:10:51.652 So we can add four and eight, 0:10:51.652,0:10:54.217 and get 12 Ohms of total resistance. 0:10:54.217,0:10:57.720 And now I can say that the[br]full 24 volts of the battery 0:10:57.720,0:11:00.135 is applied across this[br]entire equivalent resistance 0:11:00.135,0:11:03.685 of 12 Ohms, so if I come up[br]here and change this eight Ohms 0:11:03.685,0:11:06.980 to 12 Ohms of equivalent[br]resistance for the total circuit, 0:11:06.980,0:11:08.440 I'll get the correct current that flows 0:11:08.440,0:11:10.322 through the battery of two Amps. 0:11:10.322,0:11:12.143 And since that's the current[br]that's flowing through 0:11:12.143,0:11:13.879 the battery, that had to be the current 0:11:13.879,0:11:16.343 that's flowing through the[br]eight Ohm resistor as well. 0:11:16.343,0:11:20.059 Since this eight Ohm resistor[br]and the batter are in series. 0:11:20.059,0:11:23.152 Elements in a circuit[br]often use Electrical Power. 0:11:23.152,0:11:25.892 That is to say, when current[br]runs through a resistor, 0:11:25.892,0:11:27.726 the electrons moving through that resistor 0:11:27.726,0:11:30.428 turn some of their[br]electrical potential energy 0:11:30.428,0:11:33.766 into energies like heat, light, or sound. 0:11:33.766,0:11:35.408 And the rate at which these electrons 0:11:35.408,0:11:37.978 are turning their energy[br]into other forms of energy, 0:11:37.978,0:11:39.854 is called the electrical power. 0:11:39.854,0:11:41.815 So the rate at which a resistor is turning 0:11:41.815,0:11:43.974 electrical potential energy into heat, 0:11:43.974,0:11:46.481 is the electrical power[br]used by that resistor. 0:11:46.481,0:11:47.940 In other words, the amount of energy 0:11:47.940,0:11:50.084 converted into heat, divided by the time 0:11:50.084,0:11:52.380 it took to convert that[br]energy, is the definition 0:11:52.380,0:11:54.358 of the power, and there's[br]a way to determine 0:11:54.358,0:11:57.404 this number of Joules per[br]second, in terms of quantities 0:11:57.404,0:12:00.199 like the current, the[br]voltage, and the resistance. 0:12:00.199,0:12:02.425 The power used by a[br]resistor can be written as 0:12:02.425,0:12:04.410 the current through that resistor 0:12:04.410,0:12:07.436 multiplied by the voltage[br]across that resistor, 0:12:07.436,0:12:10.661 or if you substituted Ohm's[br]Law into this formula, 0:12:10.661,0:12:12.181 you see that this is equivalent to 0:12:12.181,0:12:14.319 the current through that resistor squared, 0:12:14.319,0:12:16.759 multiplied by the[br]resistance of the resistor, 0:12:16.759,0:12:18.132 or we could rearrange these formulas 0:12:18.132,0:12:20.025 to get that the power used by a resistor 0:12:20.025,0:12:23.124 would also be the voltage[br]across that resistor squared, 0:12:23.124,0:12:25.521 divided by the resistance[br]of that resistor. 0:12:25.521,0:12:27.540 All three of these, if used correctly, 0:12:27.540,0:12:30.130 will give you the same[br]number for the power used 0:12:30.130,0:12:32.074 by a resistor, and if[br]you wanted to determine 0:12:32.074,0:12:34.605 the number of Joules of[br]heat energy converted, 0:12:34.605,0:12:37.054 you could set any one of[br]these equal to the amount 0:12:37.054,0:12:40.122 of energy per time, and[br]solve for that energy. 0:12:40.122,0:12:42.561 The units of Electrical Power[br]are the same as the regular 0:12:42.561,0:12:46.262 units of power, which is[br]Watts, i.e. Joules per second, 0:12:46.262,0:12:48.333 and Electrical Power is not a vector. 0:12:48.333,0:12:49.836 So what's an example problem involving 0:12:49.836,0:12:51.452 Electrical Power look like? 0:12:51.452,0:12:53.296 Let's say a light bulb of resistance R 0:12:53.296,0:12:55.416 is hooked up to a source of voltage V, 0:12:55.416,0:12:57.694 and a second light bulb of resistance 2R, 0:12:57.694,0:13:00.246 is hooked up to a source of voltage 2V. 0:13:00.246,0:13:02.786 How does the power used[br]by the second light bulb 0:13:02.786,0:13:05.285 compare to the power used[br]by the first light bulb? 0:13:05.285,0:13:07.553 Since we have the[br]information about R and V, 0:13:07.553,0:13:09.382 I'll use the version of the power formula 0:13:09.382,0:13:11.645 that says that the[br]power used by a resistor 0:13:11.645,0:13:13.980 is gonna be delta V squared over R. 0:13:13.980,0:13:15.937 So in terms of quantities[br]given the power used 0:13:15.937,0:13:18.797 by the first light bulb is[br]gonna be V squared over R. 0:13:18.797,0:13:21.225 And the power used by the[br]second light bulb is gonna be 0:13:21.225,0:13:23.779 equal to the voltage across[br]the second light bulb, 0:13:23.779,0:13:26.852 which is two times the voltage[br]across the first light bulb, 0:13:26.852,0:13:28.953 and we square that,[br]divided by the resistance 0:13:28.953,0:13:30.678 of the second light[br]bulb, which is gonna be 0:13:30.678,0:13:33.015 two times the resistance[br]of the first light bulb. 0:13:33.015,0:13:35.547 The two squared on top is[br]gonna give me a factor of four, 0:13:35.547,0:13:37.708 and I'll have another[br]factor of two on the bottom. 0:13:37.708,0:13:40.032 So if I factor out this[br]four divided by two, 0:13:40.032,0:13:42.269 I get that the power used[br]by the second light bulb 0:13:42.269,0:13:44.666 is gonna be two times V squared over R, 0:13:44.666,0:13:46.663 but V squared over R was the power used 0:13:46.663,0:13:48.658 by the first light bulb, so the power used 0:13:48.658,0:13:50.653 by the second light bulb is gonna be two 0:13:50.653,0:13:52.838 times the power used by[br]the first light bulb, 0:13:52.838,0:13:54.681 and so if the light[br]bulb of resistance two R 0:13:54.681,0:13:57.662 has twice the power, and[br]that means it'll be brighter. 0:13:57.662,0:13:59.591 The quantity that[br]determines the brightness 0:13:59.591,0:14:03.124 of a light bulb, is the electrical[br]power of that light bulb. 0:14:03.124,0:14:06.318 It's not necessarily the[br]resistance or the voltage, 0:14:06.318,0:14:07.821 it's the combination of the two 0:14:07.821,0:14:10.895 in this formula that will[br]tell you the electrical power, 0:14:10.895,0:14:13.649 and therefore the brightness[br]of the light bulb. 0:14:13.649,0:14:15.519 Two of the most useful ideas in circuits 0:14:15.519,0:14:17.767 are referred to as Kirchhoff's Rules. 0:14:17.767,0:14:19.854 The first rule is called[br]the Junction rule, 0:14:19.854,0:14:22.040 and it states that all the[br]current entering a junction 0:14:22.040,0:14:25.140 must equal all the current[br]exiting that junction. 0:14:25.140,0:14:26.961 In other words, if you add[br]all the current that flows 0:14:26.961,0:14:28.997 into a junction, that has to equal 0:14:28.997,0:14:31.260 all the current that flows[br]out of that junction, 0:14:31.260,0:14:33.263 because current is just flowing charge, 0:14:33.263,0:14:35.897 and charge is conserved,[br]so charge can't be 0:14:35.897,0:14:38.964 created or destroyed at[br]any point in the circuit. 0:14:38.964,0:14:41.506 No more than water can[br]get created or destroyed 0:14:41.506,0:14:43.213 within a series of pipes. 0:14:43.213,0:14:45.080 And the second rule is[br]called the Loop rule, 0:14:45.080,0:14:47.515 which states that if you[br]add up all the changes 0:14:47.515,0:14:51.235 in electric potential, i.e.[br]voltages around any closed 0:14:51.235,0:14:54.238 loop in a circuit, it'll[br]always add up to zero. 0:14:54.238,0:14:56.464 So if you add up all[br]the voltages encountered 0:14:56.464,0:14:58.191 through a closed loop through a circuit, 0:14:58.191,0:14:59.768 it always adds up to zero. 0:14:59.768,0:15:02.273 And this is just a result[br]of conservation of energy. 0:15:02.273,0:15:04.408 The electrons will gain[br]energy when they flow 0:15:04.408,0:15:06.410 through the battery,[br]and they'll lose energy 0:15:06.410,0:15:08.396 every time they flow through a resistor, 0:15:08.396,0:15:09.919 but the total amount of energy they gain 0:15:09.919,0:15:12.443 from the battery, has to be[br]equal to the total amount 0:15:12.443,0:15:14.956 of energy they lose due to the resistors. 0:15:14.956,0:15:17.345 In other words, if we[br]consider a complicated circuit 0:15:17.345,0:15:19.759 that has a batter and three resistors, 0:15:19.759,0:15:23.125 the total current flowing[br]into a junction I1, 0:15:23.125,0:15:25.065 will have to be equal to the total current 0:15:25.065,0:15:28.293 coming out of that junction, I2 and I3. 0:15:28.293,0:15:30.951 Since no charge gets created or destroyed. 0:15:30.951,0:15:32.986 And that means when these[br]two currents combine again, 0:15:32.986,0:15:34.482 the total current flowing out 0:15:34.482,0:15:37.194 of that section is gonna again be I1. 0:15:37.194,0:15:39.990 And if we follow a closed[br]loop through this circuit, 0:15:39.990,0:15:42.372 the sum of all the[br]voltages around that loop 0:15:42.372,0:15:45.919 have to add up to zero, i.e.[br]the voltage of the battery 0:15:45.919,0:15:48.939 minus the voltage drop[br]across the first resistor, 0:15:48.939,0:15:51.784 minus the voltage drop[br]across the second resistor, 0:15:51.784,0:15:53.480 would have to equal zero. 0:15:53.480,0:15:54.894 So what's an example problem involving 0:15:54.894,0:15:56.496 Kirchhoff's Rules look like? 0:15:56.496,0:15:58.513 Let's say we have the circuit[br]below and we wanted to 0:15:58.513,0:16:01.702 determine the voltage[br]across the six Ohm resistor. 0:16:01.702,0:16:03.409 To do this, we could use the Loop rule, 0:16:03.409,0:16:05.633 I'll start behind the[br]battery, and I'll go through 0:16:05.633,0:16:08.392 the resistor, I want to[br]determine the voltage across. 0:16:08.392,0:16:10.932 I'll add up all the[br]voltages across that loop, 0:16:10.932,0:16:12.364 and set it equal to zero. 0:16:12.364,0:16:13.556 So the voltage across the battery 0:16:13.556,0:16:15.703 is gonna be positive 24 volts, 0:16:15.703,0:16:18.276 minus the voltage across[br]the six Ohm resistor, 0:16:18.276,0:16:19.745 and then minus the voltage across the 0:16:19.745,0:16:22.220 eight Ohm resistor has to equal zero. 0:16:22.220,0:16:24.676 But we're given this current,[br]so we know that two Amps 0:16:24.676,0:16:26.706 flows through the eight Ohm resistor, 0:16:26.706,0:16:27.847 and you can always determine the voltage 0:16:27.847,0:16:29.939 across the resistor using Ohm's Law, 0:16:29.939,0:16:31.870 so the voltage across[br]the eight Ohm resistor 0:16:31.870,0:16:33.870 is gonna be two Amps, which is flowing 0:16:33.870,0:16:35.436 through the eight Ohm resistor, 0:16:35.436,0:16:39.129 multiplied by eight Ohms,[br]and we get 16 volts. 0:16:39.129,0:16:41.837 Which I can plug into[br]here, and this gives me 0:16:41.837,0:16:45.686 24 volts minus the voltage[br]across the six Ohm resistor, 0:16:45.686,0:16:48.389 minus 16, has to equal zero. 0:16:48.389,0:16:49.583 And if I solve this for the voltage 0:16:49.583,0:16:52.253 across the six Ohm[br]resistor, I get 24 volts 0:16:52.253,0:16:54.871 minus 16 volts, which is eight volts. 0:16:54.871,0:16:56.721 So the voltage across the six Ohm resistor 0:16:56.721,0:16:58.308 would be eight volts. 0:16:58.308,0:17:01.392 Note, because the 12 Ohm[br]resistor and the six Ohm resistor 0:17:01.392,0:17:04.095 are in parallel, the voltage[br]across the 12 Ohm resistor 0:17:04.095,0:17:07.078 would also be eight volts,[br]because the voltage across 0:17:07.078,0:17:11.633 any two elements in parallel,[br]have to be the same. 0:17:11.633,0:17:13.549 Voltmeters are the device that you use 0:17:13.549,0:17:16.398 to measure the voltage between[br]two points in a circuit. 0:17:16.398,0:17:18.113 When hooking up a voltmeter you've gotta 0:17:18.113,0:17:20.580 hook it up in parallel[br]between the two points 0:17:20.580,0:17:22.600 you wanna find the voltage across. 0:17:22.600,0:17:24.015 In other words, to determine the voltage 0:17:24.015,0:17:26.388 between this point and[br]this point, which would be 0:17:26.388,0:17:29.235 the voltage across R3, you[br]would hook up the voltmeter 0:17:29.235,0:17:31.465 in parallel with R3. 0:17:31.465,0:17:33.351 Ammeters are the devices we use to measure 0:17:33.351,0:17:34.908 the electrical current that pass 0:17:34.908,0:17:36.780 through a point in a circuit, and ammeters 0:17:36.780,0:17:39.595 have to be hooked up in series[br]with the circuit element 0:17:39.595,0:17:41.373 you want to determine the current through. 0:17:41.373,0:17:43.261 In other words, if we wanted[br]to determine the current 0:17:43.261,0:17:47.267 through R1, we would hook up[br]the ammeter in series with R1. 0:17:47.267,0:17:49.877 Note that for these electrical[br]devices to work well, 0:17:49.877,0:17:53.185 the ammeter should have almost[br]zero internal resistance, 0:17:53.185,0:17:55.407 thereby not affecting[br]the current that flows 0:17:55.407,0:17:57.479 through the circuit, and[br]voltmeters should have 0:17:57.479,0:18:00.440 near infinite resistance,[br]so that it doesn't draw 0:18:00.440,0:18:02.674 any of the current from the resistor. 0:18:02.674,0:18:04.969 In reality, ammeters have a very small, 0:18:04.969,0:18:07.148 but non-zero internal resistance, 0:18:07.148,0:18:08.463 and voltmeters have a very high, 0:18:08.463,0:18:10.653 but not infinite internal resistance. 0:18:10.653,0:18:12.107 So what would an example problem involving 0:18:12.107,0:18:14.108 voltmeters and ammeters look like? 0:18:14.108,0:18:15.865 Let's say we have the circuit shown below, 0:18:15.865,0:18:18.558 and these numbered circles[br]represent possible places 0:18:18.558,0:18:21.173 we could stick a voltmeter[br]to measure the voltage 0:18:21.173,0:18:23.165 across the eight Ohm resistor. 0:18:23.165,0:18:25.380 Which two of these[br]voltmeters would correctly 0:18:25.380,0:18:28.079 give the voltage across[br]the eight Ohm resistor? 0:18:28.079,0:18:30.124 And you have to be[br]careful, some AP problems 0:18:30.124,0:18:31.844 are gonna require you to select 0:18:31.844,0:18:33.838 two correct answers for[br]the multiple choice, 0:18:33.838,0:18:36.509 so be sure to read the[br]instructions carefully. 0:18:36.509,0:18:38.229 Voltmeter number four[br]is a terrible choice, 0:18:38.229,0:18:40.368 you never hook up your[br]voltmeter in series, 0:18:40.368,0:18:42.331 but the circuit element[br]you're trying to find 0:18:42.331,0:18:44.382 the voltage across, and[br]voltmeter number one 0:18:44.382,0:18:46.524 is doing nothing really,[br]because its' measuring 0:18:46.524,0:18:48.989 the voltage between two points in a wire 0:18:48.989,0:18:51.206 with nothing in between that wire. 0:18:51.206,0:18:52.952 So the voltage measured by voltmeter one 0:18:52.952,0:18:56.034 should just be zero, since[br]the voltage across a wire 0:18:56.034,0:18:59.412 of zero resistance should[br]just give you zero volts. 0:18:59.412,0:19:01.627 So the correct choices would[br]be voltmeter number two, 0:19:01.627,0:19:04.736 which gives you the voltage[br]across the eight Ohm resistor, 0:19:04.736,0:19:06.429 and voltmeter number[br]three, which also gives you 0:19:06.429,0:19:08.304 an equivalent measurement of the voltage 0:19:08.304,0:19:10.870 across the resistor eight Ohms.