1 00:00:00,468 --> 00:00:01,688 - [Voiceover] Electric Charge is a property 2 00:00:01,688 --> 00:00:05,622 that some, but not all fundamental particles in nature have. 3 00:00:05,622 --> 00:00:06,761 The most commonly talked about 4 00:00:06,761 --> 00:00:09,562 fundamentally charged particles are the electrons, 5 00:00:09,562 --> 00:00:11,564 which orbit the outside of the atom. 6 00:00:11,564 --> 00:00:13,345 These are negatively charged. 7 00:00:13,345 --> 00:00:15,146 There's also the protons, which reside 8 00:00:15,146 --> 00:00:18,484 inside the nucleus, and these are positively charged. 9 00:00:18,484 --> 00:00:20,088 And the neutrons inside the nucleus 10 00:00:20,088 --> 00:00:21,855 don't have any net charge. 11 00:00:21,855 --> 00:00:24,212 It turns out that all fundamentally charged particles 12 00:00:24,212 --> 00:00:26,310 in the universe, have charges that come in 13 00:00:26,310 --> 00:00:29,258 integer units of the elementary charge. 14 00:00:29,258 --> 00:00:30,804 So if you find a particle in nature, 15 00:00:30,804 --> 00:00:33,916 it's gonna have a charge of one times this number, 16 00:00:33,916 --> 00:00:37,016 two times this number, three times this number, 17 00:00:37,016 --> 00:00:38,717 and it could either be positive or negative. 18 00:00:38,717 --> 00:00:41,690 For instance, the electron has a charge of negative 1.6 19 00:00:41,690 --> 00:00:44,658 times 10 to the negative 19th Coulombs, 20 00:00:44,658 --> 00:00:47,542 and the charge of the proton is positive 1.6 21 00:00:47,542 --> 00:00:49,889 times 10 to the negative 19th Coulombs. 22 00:00:49,889 --> 00:00:51,928 However, most atoms in the universe 23 00:00:51,928 --> 00:00:54,076 are electrically neutral overall, 24 00:00:54,076 --> 00:00:56,828 since they'll have just as many negative electrons 25 00:00:56,828 --> 00:00:59,013 as they do positive protons. 26 00:00:59,013 --> 00:01:01,307 But if an atom had too many electrons, 27 00:01:01,307 --> 00:01:04,009 overall that atom would be negatively charged, 28 00:01:04,009 --> 00:01:06,501 and if an atom had too few electrons, 29 00:01:06,501 --> 00:01:08,844 that atom would be overall positively charged. 30 00:01:08,844 --> 00:01:10,593 And something that's really important to remember 31 00:01:10,593 --> 00:01:13,077 is that the electric charge is always conserved 32 00:01:13,077 --> 00:01:15,212 for every process, in other words, 33 00:01:15,212 --> 00:01:17,539 the total charge initially, is gonna equal 34 00:01:17,539 --> 00:01:20,586 the total charge finally after any process. 35 00:01:20,586 --> 00:01:22,086 So what's an example problem involving 36 00:01:22,086 --> 00:01:23,660 electric charge look like? 37 00:01:23,660 --> 00:01:26,187 Let's say three identically sized metal spheres 38 00:01:26,187 --> 00:01:28,344 start off with the charges seen below. 39 00:01:28,344 --> 00:01:32,136 Positive five Q, positive three Q, and negative two Q. 40 00:01:32,136 --> 00:01:35,793 If we touch sphere X to sphere Y, and separate them, 41 00:01:35,793 --> 00:01:39,609 and then touch sphere Y to sphere Z, and separate them, 42 00:01:39,609 --> 00:01:42,575 what will be the final charge on each sphere? 43 00:01:42,575 --> 00:01:44,547 Well first, when we touch X to Y, 44 00:01:44,547 --> 00:01:46,693 the total charge has to be conserved. 45 00:01:46,693 --> 00:01:49,224 There's a total charge of eight Q amongst them, 46 00:01:49,224 --> 00:01:50,934 and since they're identically sized, 47 00:01:50,934 --> 00:01:52,779 they'll both share the total charge, 48 00:01:52,779 --> 00:01:54,038 which means after they touch, 49 00:01:54,038 --> 00:01:56,008 they'll both have positive four Q. 50 00:01:56,008 --> 00:01:57,429 If one of the spheres were larger, 51 00:01:57,429 --> 00:01:59,028 it would gain more of the charge, 52 00:01:59,028 --> 00:02:01,378 but the total charge would still be conserved. 53 00:02:01,378 --> 00:02:04,193 And now when sphere Y is touched to sphere Z, 54 00:02:04,193 --> 00:02:06,414 the total charge amongst them at that moment 55 00:02:06,414 --> 00:02:09,443 would be positive four Q plus negative two Q, 56 00:02:09,443 --> 00:02:10,976 which is positive two Q. 57 00:02:10,976 --> 00:02:13,493 They would share it equally, so sphere Y would have 58 00:02:13,493 --> 00:02:17,029 positive Q, and sphere Z would also have positive Q. 59 00:02:17,029 --> 00:02:19,247 So the answer here is C. 60 00:02:19,247 --> 00:02:22,288 Opposite charges attract and like charges repel, 61 00:02:22,288 --> 00:02:24,874 and what Coulomb's Law does is it gives you a way to find 62 00:02:24,874 --> 00:02:28,499 the magnitude of the electric force between two charges. 63 00:02:28,499 --> 00:02:30,069 The formula for Coulomb's Law says 64 00:02:30,069 --> 00:02:32,123 that the magnitude of the electric force 65 00:02:32,123 --> 00:02:36,003 between two charges Q1 and Q2, is gonna equal the 66 00:02:36,003 --> 00:02:39,203 electric constant K, which is nine times 10 to the ninth, 67 00:02:39,203 --> 00:02:42,619 times the product of the two charges, measured in Coulombs 68 00:02:42,619 --> 00:02:44,489 divided by the center to center distance 69 00:02:44,489 --> 00:02:47,260 between those two charges, squared. 70 00:02:47,260 --> 00:02:49,542 You can't forget to square this distance. 71 00:02:49,542 --> 00:02:51,612 And it's gotta be in meters if you want to find 72 00:02:51,612 --> 00:02:54,151 SI units of Newtons for the force. 73 00:02:54,151 --> 00:02:56,716 Also, don't rely on the negative and positive signs 74 00:02:56,716 --> 00:02:59,626 of the charges to tell you which way the force points, 75 00:02:59,626 --> 00:03:02,064 just use the fact that opposite charges attract 76 00:03:02,064 --> 00:03:04,986 and like charges repel, and use Coulomb's Law 77 00:03:04,986 --> 00:03:07,175 to get the magnitude of the force. 78 00:03:07,175 --> 00:03:08,500 So what's an example problem involving 79 00:03:08,500 --> 00:03:09,790 Coulomb's Law look like? 80 00:03:09,790 --> 00:03:11,653 Let's say two charges exert an electric force 81 00:03:11,653 --> 00:03:13,586 of magnitude F on each other. 82 00:03:13,586 --> 00:03:16,336 What would be the magnitude of the new electric force 83 00:03:16,336 --> 00:03:18,749 if the distance between the charges is tripled 84 00:03:18,749 --> 00:03:21,687 and the magnitude of one of the charges is doubled? 85 00:03:21,687 --> 00:03:23,375 Well we know the formula for Coulomb's Law 86 00:03:23,375 --> 00:03:25,265 says that the force between two charges 87 00:03:25,265 --> 00:03:28,026 is the electric constant times one of the charges, 88 00:03:28,026 --> 00:03:30,354 times the other charge divided by the distance 89 00:03:30,354 --> 00:03:33,321 between them squared, and now once we triple the distance 90 00:03:33,321 --> 00:03:36,154 and double a charge, the new electric force 91 00:03:36,154 --> 00:03:39,180 is gonna be the electric constant times one of the charges, 92 00:03:39,180 --> 00:03:42,719 multiplied by two times one of the charges, 93 00:03:42,719 --> 00:03:46,402 divided by three times the distance, which is squared, 94 00:03:46,402 --> 00:03:48,148 so I'm gonna get a factor of two on top, 95 00:03:48,148 --> 00:03:49,926 and this three will get squared, which gives me 96 00:03:49,926 --> 00:03:51,761 a factor of nine on the bottom. 97 00:03:51,761 --> 00:03:54,148 If I pull up those extra factors I get that the new force 98 00:03:54,148 --> 00:03:58,588 is gonna be two ninths multiplied by K, Q1, Q2, 99 00:03:58,588 --> 00:04:00,993 over D squared, but this entire quantity 100 00:04:00,993 --> 00:04:03,971 was just the old force F, so the new force 101 00:04:03,971 --> 00:04:07,731 is going to be two ninths of the old force. 102 00:04:07,731 --> 00:04:10,644 The electrical current I tells you the amount of Coulombs 103 00:04:10,644 --> 00:04:14,061 of charge that passes a point in a wire per second. 104 00:04:14,061 --> 00:04:16,019 So if you watch some point in a wire, 105 00:04:16,019 --> 00:04:18,113 and you count how many Coulombs of charge 106 00:04:18,113 --> 00:04:21,327 pass by that point per second, that would be the current. 107 00:04:21,327 --> 00:04:23,659 Or in equation form we can see that the current I 108 00:04:23,659 --> 00:04:25,628 is gonna be the amount of charge that flows 109 00:04:25,628 --> 00:04:27,897 past a point in a wire per time. 110 00:04:27,897 --> 00:04:30,588 This gives the units of I as Coulombs per second, 111 00:04:30,588 --> 00:04:32,848 which we abbreviate as an Ampere. 112 00:04:32,848 --> 00:04:34,681 And since charge and time aren't vectors, 113 00:04:34,681 --> 00:04:36,499 current is not a vector either. 114 00:04:36,499 --> 00:04:37,736 Something that's kind of strange 115 00:04:37,736 --> 00:04:40,908 is that the so-called conventional direction of current 116 00:04:40,908 --> 00:04:43,475 would be the direction that positive charges flow within a 117 00:04:43,475 --> 00:04:47,171 wire, however positive charges don't flow within a wire. 118 00:04:47,171 --> 00:04:49,458 The only charges that actually flow in a wire 119 00:04:49,458 --> 00:04:51,658 are negative charges, but it turns out 120 00:04:51,658 --> 00:04:55,019 that negative charges flowing to the left is physically 121 00:04:55,019 --> 00:04:57,932 the same as positive charges flowing to the right. 122 00:04:57,932 --> 00:05:00,498 So in physics problems we pretend as if it were 123 00:05:00,498 --> 00:05:02,819 the positive charges moving, however 124 00:05:02,819 --> 00:05:04,889 it's really the electrons, which are negative, 125 00:05:04,889 --> 00:05:06,464 that are moving within the wire. 126 00:05:06,464 --> 00:05:07,410 So what's an example problem 127 00:05:07,410 --> 00:05:09,295 involving electrical current look like? 128 00:05:09,295 --> 00:05:11,387 Let's say three amps flows within a circuit. 129 00:05:11,387 --> 00:05:14,014 How much charge would pass by a point in that wire 130 00:05:14,014 --> 00:05:16,505 during a time interval of five minutes? 131 00:05:16,505 --> 00:05:17,904 Well we know the definition of current 132 00:05:17,904 --> 00:05:20,341 is the charge per time, that means the charge 133 00:05:20,341 --> 00:05:23,285 is gonna be the amount of current multiplied by the time, 134 00:05:23,285 --> 00:05:25,303 so we take our current of three amps, 135 00:05:25,303 --> 00:05:27,305 and we multiply by the time, but we can't 136 00:05:27,305 --> 00:05:30,295 multiply by five because that's in units of minutes, 137 00:05:30,295 --> 00:05:32,486 since amps is Coulombs per second, 138 00:05:32,486 --> 00:05:34,535 we've got to convert five minutes into seconds, 139 00:05:34,535 --> 00:05:37,934 which would be five minutes, multiplied by 60 seconds 140 00:05:37,934 --> 00:05:39,833 per minute, which would give us a total amount 141 00:05:39,833 --> 00:05:42,000 of charge of 900 Coulombs. 142 00:05:42,890 --> 00:05:45,794 The resistance of a circuit element measures how much 143 00:05:45,794 --> 00:05:48,703 that element will restrict the flow of current. 144 00:05:48,703 --> 00:05:51,132 The larger the resistance, the less current 145 00:05:51,132 --> 00:05:52,763 there will be allowed to flow. 146 00:05:52,763 --> 00:05:55,978 And this definition of resistance is given by Ohm's Law. 147 00:05:55,978 --> 00:05:57,794 Ohm's Law states that the amount of current 148 00:05:57,794 --> 00:05:59,759 that you'll get through a portion of a circuit, 149 00:05:59,759 --> 00:06:01,476 it's gonna be proportional to the voltage 150 00:06:01,476 --> 00:06:04,371 across that portion, divided by the resistance 151 00:06:04,371 --> 00:06:06,006 of that portion of the circuit. 152 00:06:06,006 --> 00:06:08,331 So between these two points, the amount of current 153 00:06:08,331 --> 00:06:10,198 that will flow, is gonna be equal to 154 00:06:10,198 --> 00:06:12,536 the voltage between those two points, 155 00:06:12,536 --> 00:06:15,604 divided by the resistance between those two points. 156 00:06:15,604 --> 00:06:17,889 So the larger the resistance, the less current 157 00:06:17,889 --> 00:06:20,958 will flow, but the greater the voltage supplied, 158 00:06:20,958 --> 00:06:22,418 the greater the current will be. 159 00:06:22,418 --> 00:06:24,107 And this is what Ohm's Law says. 160 00:06:24,107 --> 00:06:26,316 Even though Ohm's Law gives you a way to define 161 00:06:26,316 --> 00:06:29,082 the resistance, you can determine the resistance 162 00:06:29,082 --> 00:06:31,043 of a circuit element by knowing the size 163 00:06:31,043 --> 00:06:32,949 and shape of that circuit element. 164 00:06:32,949 --> 00:06:35,568 In other words, the resistance of a cylindrical resistor, 165 00:06:35,568 --> 00:06:37,872 is gonna be equal to the resistivity, 166 00:06:37,872 --> 00:06:40,214 which is a measure of an object's natural resistance 167 00:06:40,214 --> 00:06:43,126 to current, multiplied by the length of that resistor, 168 00:06:43,126 --> 00:06:45,768 the longer the resistor, the greater the resistance 169 00:06:45,768 --> 00:06:48,072 and the more it will resist the flow of current, 170 00:06:48,072 --> 00:06:50,199 and then divide it by the cross sectional area 171 00:06:50,199 --> 00:06:52,860 of that resistor, which would be this area right here, 172 00:06:52,860 --> 00:06:55,583 the current is either flowing into or out of. 173 00:06:55,583 --> 00:06:58,450 If the resistor is cylindrical, the area of this circle 174 00:06:58,450 --> 00:07:01,313 would be Pi times r squared, where little r 175 00:07:01,313 --> 00:07:04,356 would be the radius of this cross sectional area. 176 00:07:04,356 --> 00:07:07,521 The units of resistance is Ohms, and it is not a vector. 177 00:07:07,521 --> 00:07:09,746 It is always positive or zero. 178 00:07:09,746 --> 00:07:11,888 So what's an example problem involving Ohm's Law, 179 00:07:11,888 --> 00:07:14,699 or the resistance of a cylindrical resistor look like? 180 00:07:14,699 --> 00:07:16,776 Let's say a battery of voltage V is hooked up 181 00:07:16,776 --> 00:07:19,744 to a single cylindrical resistor of length L 182 00:07:19,744 --> 00:07:21,813 and radius little r, and when that's done, 183 00:07:21,813 --> 00:07:24,482 a current I is flowing through the battery. 184 00:07:24,482 --> 00:07:27,913 What is the resistivity Rho, of that resistor? 185 00:07:27,913 --> 00:07:30,278 Well we know Ohm's Law states that the current that flows 186 00:07:30,278 --> 00:07:32,446 through a portion of a circuit will be equal to 187 00:07:32,446 --> 00:07:33,972 the voltage across that portion, 188 00:07:33,972 --> 00:07:36,278 divided by the resistance of that portion. 189 00:07:36,278 --> 00:07:38,603 And this means the resistance of this resistor is gonna be 190 00:07:38,603 --> 00:07:40,883 the voltage of the battery divided by the current. 191 00:07:40,883 --> 00:07:43,268 To factor resistivity into this, we have to use 192 00:07:43,268 --> 00:07:46,153 the formula for the resistance of a cylindrical resistor, 193 00:07:46,153 --> 00:07:48,222 which is Rho times L over A. 194 00:07:48,222 --> 00:07:50,637 This gives us the resistance of the resistor, 195 00:07:50,637 --> 00:07:52,735 which is gotta equal V over I, 196 00:07:52,735 --> 00:07:55,023 and now we can solve for the resistivity Rho. 197 00:07:55,023 --> 00:07:58,381 We get V times A over I L, but since we're given 198 00:07:58,381 --> 00:08:01,117 the radius little r, we gotta write the area 199 00:08:01,117 --> 00:08:03,204 in terms of that radius, so this is gonna be 200 00:08:03,204 --> 00:08:06,858 V times Pi, r squared, divided by I times L, 201 00:08:06,858 --> 00:08:09,210 which gives us an answer of C. 202 00:08:09,210 --> 00:08:12,285 When dealing with complicated circuits with many resistors, 203 00:08:12,285 --> 00:08:14,214 you often have to reduce those resistors 204 00:08:14,214 --> 00:08:17,098 into smaller, equivalent amounts of resistors. 205 00:08:17,098 --> 00:08:18,862 And the two ways you do this are by finding 206 00:08:18,862 --> 00:08:21,680 two resistors that are in series or in parallel. 207 00:08:21,680 --> 00:08:24,164 Resistors will be in series if the same current 208 00:08:24,164 --> 00:08:25,891 that flows through the same resistor, 209 00:08:25,891 --> 00:08:27,688 flows through the next resistor. 210 00:08:27,688 --> 00:08:30,157 If the current branched off in between them, 211 00:08:30,157 --> 00:08:32,791 these resistors would no longer be in series, 212 00:08:32,791 --> 00:08:34,153 but if they're in series you can find 213 00:08:34,153 --> 00:08:37,044 the equivalent resistance of this section of wire 214 00:08:37,044 --> 00:08:40,035 by just adding up the two individual resistances. 215 00:08:40,035 --> 00:08:41,776 So the current for resistors in series 216 00:08:41,777 --> 00:08:45,112 must be the same, but the voltage might be different, 217 00:08:45,112 --> 00:08:47,244 since they could have different resistances. 218 00:08:47,244 --> 00:08:49,104 Two resistors will be in parallel, 219 00:08:49,104 --> 00:08:52,000 if a current comes in, splits into two parts, 220 00:08:52,000 --> 00:08:54,803 goes through one resistor each, and then rejoins 221 00:08:54,803 --> 00:08:56,818 before hitting anything else in the circuit, 222 00:08:56,818 --> 00:08:58,768 and if this is the case, you can find the equivalent 223 00:08:58,768 --> 00:09:00,882 resistance of this portion of the circuit, 224 00:09:00,882 --> 00:09:04,093 i.e. between these two points, by saying that one 225 00:09:04,093 --> 00:09:06,519 over the equivalent resistance is gonna equal 226 00:09:06,519 --> 00:09:08,893 one over the resistance of the first resistor, 227 00:09:08,893 --> 00:09:11,556 plus one over the resistance of the second resistor. 228 00:09:11,556 --> 00:09:14,275 But be careful, one over R1 plus one over R2 229 00:09:14,275 --> 00:09:16,544 just gives you one over R equivalent. 230 00:09:16,544 --> 00:09:18,663 If you want R equivalent, you're gonna have to take 231 00:09:18,663 --> 00:09:22,627 one over this entire side, in order to get R equivalent. 232 00:09:22,627 --> 00:09:24,604 So what's an example problem involving resistors 233 00:09:24,604 --> 00:09:26,386 in series and parallel look like? 234 00:09:26,386 --> 00:09:28,357 Let's say we have this circuit shown below, 235 00:09:28,357 --> 00:09:30,164 and we want to know what current flows 236 00:09:30,164 --> 00:09:31,726 through the eight Ohm resistor. 237 00:09:31,726 --> 00:09:33,280 Now you might be tempted to say this, 238 00:09:33,280 --> 00:09:36,861 since Ohm's Law says that the current is delta V over R, 239 00:09:36,861 --> 00:09:38,729 we can just plug in the voltage of the battery, 240 00:09:38,729 --> 00:09:41,330 which is 24 volts, divided by the resistance 241 00:09:41,330 --> 00:09:43,469 of the resistor, which is eight Ohms, 242 00:09:43,469 --> 00:09:45,220 and that would give us three Amps. 243 00:09:45,220 --> 00:09:46,487 But that's not right at all. 244 00:09:46,487 --> 00:09:48,549 When using Ohm's Law, the current that flows 245 00:09:48,549 --> 00:09:51,511 through a resistor R, is gonna be equal to 246 00:09:51,511 --> 00:09:54,382 the voltage across that resistor divided by 247 00:09:54,382 --> 00:09:56,406 the resistance of that resistor. 248 00:09:56,406 --> 00:09:59,382 So if we plug eight Ohms into the denominator, we've gotta 249 00:09:59,382 --> 00:10:02,567 plug in the voltage across that eight Ohm resistor. 250 00:10:02,567 --> 00:10:04,456 But the voltage across the eight Ohm resistor 251 00:10:04,456 --> 00:10:07,556 is not gonna be the full 24 volts of the battery, 252 00:10:07,556 --> 00:10:09,595 it's gonna be less than 24 volts. 253 00:10:09,595 --> 00:10:12,056 In other words, the battery provides a voltage between 254 00:10:12,056 --> 00:10:15,144 this point and this point of 24 volts, 255 00:10:15,144 --> 00:10:16,611 but there's gonna be voltage drops 256 00:10:16,611 --> 00:10:18,855 across the six and 12 Ohm resistors, 257 00:10:18,855 --> 00:10:20,138 which make it so that the voltage 258 00:10:20,138 --> 00:10:21,776 across the eight Ohm resistor is not 259 00:10:21,776 --> 00:10:23,701 gonna be the full 24 volts. 260 00:10:23,701 --> 00:10:26,396 So we have to reduce these resistors to a single resistance. 261 00:10:26,396 --> 00:10:28,335 The six and the 12 are in parallel, 262 00:10:28,335 --> 00:10:31,351 so we can say that one over six, plus one over 12, 263 00:10:31,351 --> 00:10:32,907 would equal one over the resistance 264 00:10:32,907 --> 00:10:34,400 of that portion of the circuit. 265 00:10:34,400 --> 00:10:37,540 This is gonna equal three twelves, which is one fourth, 266 00:10:37,540 --> 00:10:39,280 so that means that parallel portion of the circuit 267 00:10:39,280 --> 00:10:41,857 has an equivalent resistance of four Ohms. 268 00:10:41,857 --> 00:10:44,159 So between this point and this point, 269 00:10:44,159 --> 00:10:46,184 there are four Ohms of resistance, 270 00:10:46,184 --> 00:10:48,469 and that equivalent resistance is in series 271 00:10:48,469 --> 00:10:50,141 with this eight Ohm resistor. 272 00:10:50,141 --> 00:10:51,652 So we can add four and eight, 273 00:10:51,652 --> 00:10:54,217 and get 12 Ohms of total resistance. 274 00:10:54,217 --> 00:10:57,720 And now I can say that the full 24 volts of the battery 275 00:10:57,720 --> 00:11:00,135 is applied across this entire equivalent resistance 276 00:11:00,135 --> 00:11:03,685 of 12 Ohms, so if I come up here and change this eight Ohms 277 00:11:03,685 --> 00:11:06,980 to 12 Ohms of equivalent resistance for the total circuit, 278 00:11:06,980 --> 00:11:08,440 I'll get the correct current that flows 279 00:11:08,440 --> 00:11:10,322 through the battery of two Amps. 280 00:11:10,322 --> 00:11:12,143 And since that's the current that's flowing through 281 00:11:12,143 --> 00:11:13,879 the battery, that had to be the current 282 00:11:13,879 --> 00:11:16,343 that's flowing through the eight Ohm resistor as well. 283 00:11:16,343 --> 00:11:20,059 Since this eight Ohm resistor and the batter are in series. 284 00:11:20,059 --> 00:11:23,152 Elements in a circuit often use Electrical Power. 285 00:11:23,152 --> 00:11:25,892 That is to say, when current runs through a resistor, 286 00:11:25,892 --> 00:11:27,726 the electrons moving through that resistor 287 00:11:27,726 --> 00:11:30,428 turn some of their electrical potential energy 288 00:11:30,428 --> 00:11:33,766 into energies like heat, light, or sound. 289 00:11:33,766 --> 00:11:35,408 And the rate at which these electrons 290 00:11:35,408 --> 00:11:37,978 are turning their energy into other forms of energy, 291 00:11:37,978 --> 00:11:39,854 is called the electrical power. 292 00:11:39,854 --> 00:11:41,815 So the rate at which a resistor is turning 293 00:11:41,815 --> 00:11:43,974 electrical potential energy into heat, 294 00:11:43,974 --> 00:11:46,481 is the electrical power used by that resistor. 295 00:11:46,481 --> 00:11:47,940 In other words, the amount of energy 296 00:11:47,940 --> 00:11:50,084 converted into heat, divided by the time 297 00:11:50,084 --> 00:11:52,380 it took to convert that energy, is the definition 298 00:11:52,380 --> 00:11:54,358 of the power, and there's a way to determine 299 00:11:54,358 --> 00:11:57,404 this number of Joules per second, in terms of quantities 300 00:11:57,404 --> 00:12:00,199 like the current, the voltage, and the resistance. 301 00:12:00,199 --> 00:12:02,425 The power used by a resistor can be written as 302 00:12:02,425 --> 00:12:04,410 the current through that resistor 303 00:12:04,410 --> 00:12:07,436 multiplied by the voltage across that resistor, 304 00:12:07,436 --> 00:12:10,661 or if you substituted Ohm's Law into this formula, 305 00:12:10,661 --> 00:12:12,181 you see that this is equivalent to 306 00:12:12,181 --> 00:12:14,319 the current through that resistor squared, 307 00:12:14,319 --> 00:12:16,759 multiplied by the resistance of the resistor, 308 00:12:16,759 --> 00:12:18,132 or we could rearrange these formulas 309 00:12:18,132 --> 00:12:20,025 to get that the power used by a resistor 310 00:12:20,025 --> 00:12:23,124 would also be the voltage across that resistor squared, 311 00:12:23,124 --> 00:12:25,521 divided by the resistance of that resistor. 312 00:12:25,521 --> 00:12:27,540 All three of these, if used correctly, 313 00:12:27,540 --> 00:12:30,130 will give you the same number for the power used 314 00:12:30,130 --> 00:12:32,074 by a resistor, and if you wanted to determine 315 00:12:32,074 --> 00:12:34,605 the number of Joules of heat energy converted, 316 00:12:34,605 --> 00:12:37,054 you could set any one of these equal to the amount 317 00:12:37,054 --> 00:12:40,122 of energy per time, and solve for that energy. 318 00:12:40,122 --> 00:12:42,561 The units of Electrical Power are the same as the regular 319 00:12:42,561 --> 00:12:46,262 units of power, which is Watts, i.e. Joules per second, 320 00:12:46,262 --> 00:12:48,333 and Electrical Power is not a vector. 321 00:12:48,333 --> 00:12:49,836 So what's an example problem involving 322 00:12:49,836 --> 00:12:51,452 Electrical Power look like? 323 00:12:51,452 --> 00:12:53,296 Let's say a light bulb of resistance R 324 00:12:53,296 --> 00:12:55,416 is hooked up to a source of voltage V, 325 00:12:55,416 --> 00:12:57,694 and a second light bulb of resistance 2R, 326 00:12:57,694 --> 00:13:00,246 is hooked up to a source of voltage 2V. 327 00:13:00,246 --> 00:13:02,786 How does the power used by the second light bulb 328 00:13:02,786 --> 00:13:05,285 compare to the power used by the first light bulb? 329 00:13:05,285 --> 00:13:07,553 Since we have the information about R and V, 330 00:13:07,553 --> 00:13:09,382 I'll use the version of the power formula 331 00:13:09,382 --> 00:13:11,645 that says that the power used by a resistor 332 00:13:11,645 --> 00:13:13,980 is gonna be delta V squared over R. 333 00:13:13,980 --> 00:13:15,937 So in terms of quantities given the power used 334 00:13:15,937 --> 00:13:18,797 by the first light bulb is gonna be V squared over R. 335 00:13:18,797 --> 00:13:21,225 And the power used by the second light bulb is gonna be 336 00:13:21,225 --> 00:13:23,779 equal to the voltage across the second light bulb, 337 00:13:23,779 --> 00:13:26,852 which is two times the voltage across the first light bulb, 338 00:13:26,852 --> 00:13:28,953 and we square that, divided by the resistance 339 00:13:28,953 --> 00:13:30,678 of the second light bulb, which is gonna be 340 00:13:30,678 --> 00:13:33,015 two times the resistance of the first light bulb. 341 00:13:33,015 --> 00:13:35,547 The two squared on top is gonna give me a factor of four, 342 00:13:35,547 --> 00:13:37,708 and I'll have another factor of two on the bottom. 343 00:13:37,708 --> 00:13:40,032 So if I factor out this four divided by two, 344 00:13:40,032 --> 00:13:42,269 I get that the power used by the second light bulb 345 00:13:42,269 --> 00:13:44,666 is gonna be two times V squared over R, 346 00:13:44,666 --> 00:13:46,663 but V squared over R was the power used 347 00:13:46,663 --> 00:13:48,658 by the first light bulb, so the power used 348 00:13:48,658 --> 00:13:50,653 by the second light bulb is gonna be two 349 00:13:50,653 --> 00:13:52,838 times the power used by the first light bulb, 350 00:13:52,838 --> 00:13:54,681 and so if the light bulb of resistance two R 351 00:13:54,681 --> 00:13:57,662 has twice the power, and that means it'll be brighter. 352 00:13:57,662 --> 00:13:59,591 The quantity that determines the brightness 353 00:13:59,591 --> 00:14:03,124 of a light bulb, is the electrical power of that light bulb. 354 00:14:03,124 --> 00:14:06,318 It's not necessarily the resistance or the voltage, 355 00:14:06,318 --> 00:14:07,821 it's the combination of the two 356 00:14:07,821 --> 00:14:10,895 in this formula that will tell you the electrical power, 357 00:14:10,895 --> 00:14:13,649 and therefore the brightness of the light bulb. 358 00:14:13,649 --> 00:14:15,519 Two of the most useful ideas in circuits 359 00:14:15,519 --> 00:14:17,767 are referred to as Kirchhoff's Rules. 360 00:14:17,767 --> 00:14:19,854 The first rule is called the Junction rule, 361 00:14:19,854 --> 00:14:22,040 and it states that all the current entering a junction 362 00:14:22,040 --> 00:14:25,140 must equal all the current exiting that junction. 363 00:14:25,140 --> 00:14:26,961 In other words, if you add all the current that flows 364 00:14:26,961 --> 00:14:28,997 into a junction, that has to equal 365 00:14:28,997 --> 00:14:31,260 all the current that flows out of that junction, 366 00:14:31,260 --> 00:14:33,263 because current is just flowing charge, 367 00:14:33,263 --> 00:14:35,897 and charge is conserved, so charge can't be 368 00:14:35,897 --> 00:14:38,964 created or destroyed at any point in the circuit. 369 00:14:38,964 --> 00:14:41,506 No more than water can get created or destroyed 370 00:14:41,506 --> 00:14:43,213 within a series of pipes. 371 00:14:43,213 --> 00:14:45,080 And the second rule is called the Loop rule, 372 00:14:45,080 --> 00:14:47,515 which states that if you add up all the changes 373 00:14:47,515 --> 00:14:51,235 in electric potential, i.e. voltages around any closed 374 00:14:51,235 --> 00:14:54,238 loop in a circuit, it'll always add up to zero. 375 00:14:54,238 --> 00:14:56,464 So if you add up all the voltages encountered 376 00:14:56,464 --> 00:14:58,191 through a closed loop through a circuit, 377 00:14:58,191 --> 00:14:59,768 it always adds up to zero. 378 00:14:59,768 --> 00:15:02,273 And this is just a result of conservation of energy. 379 00:15:02,273 --> 00:15:04,408 The electrons will gain energy when they flow 380 00:15:04,408 --> 00:15:06,410 through the battery, and they'll lose energy 381 00:15:06,410 --> 00:15:08,396 every time they flow through a resistor, 382 00:15:08,396 --> 00:15:09,919 but the total amount of energy they gain 383 00:15:09,919 --> 00:15:12,443 from the battery, has to be equal to the total amount 384 00:15:12,443 --> 00:15:14,956 of energy they lose due to the resistors. 385 00:15:14,956 --> 00:15:17,345 In other words, if we consider a complicated circuit 386 00:15:17,345 --> 00:15:19,759 that has a batter and three resistors, 387 00:15:19,759 --> 00:15:23,125 the total current flowing into a junction I1, 388 00:15:23,125 --> 00:15:25,065 will have to be equal to the total current 389 00:15:25,065 --> 00:15:28,293 coming out of that junction, I2 and I3. 390 00:15:28,293 --> 00:15:30,951 Since no charge gets created or destroyed. 391 00:15:30,951 --> 00:15:32,986 And that means when these two currents combine again, 392 00:15:32,986 --> 00:15:34,482 the total current flowing out 393 00:15:34,482 --> 00:15:37,194 of that section is gonna again be I1. 394 00:15:37,194 --> 00:15:39,990 And if we follow a closed loop through this circuit, 395 00:15:39,990 --> 00:15:42,372 the sum of all the voltages around that loop 396 00:15:42,372 --> 00:15:45,919 have to add up to zero, i.e. the voltage of the battery 397 00:15:45,919 --> 00:15:48,939 minus the voltage drop across the first resistor, 398 00:15:48,939 --> 00:15:51,784 minus the voltage drop across the second resistor, 399 00:15:51,784 --> 00:15:53,480 would have to equal zero. 400 00:15:53,480 --> 00:15:54,894 So what's an example problem involving 401 00:15:54,894 --> 00:15:56,496 Kirchhoff's Rules look like? 402 00:15:56,496 --> 00:15:58,513 Let's say we have the circuit below and we wanted to 403 00:15:58,513 --> 00:16:01,702 determine the voltage across the six Ohm resistor. 404 00:16:01,702 --> 00:16:03,409 To do this, we could use the Loop rule, 405 00:16:03,409 --> 00:16:05,633 I'll start behind the battery, and I'll go through 406 00:16:05,633 --> 00:16:08,392 the resistor, I want to determine the voltage across. 407 00:16:08,392 --> 00:16:10,932 I'll add up all the voltages across that loop, 408 00:16:10,932 --> 00:16:12,364 and set it equal to zero. 409 00:16:12,364 --> 00:16:13,556 So the voltage across the battery 410 00:16:13,556 --> 00:16:15,703 is gonna be positive 24 volts, 411 00:16:15,703 --> 00:16:18,276 minus the voltage across the six Ohm resistor, 412 00:16:18,276 --> 00:16:19,745 and then minus the voltage across the 413 00:16:19,745 --> 00:16:22,220 eight Ohm resistor has to equal zero. 414 00:16:22,220 --> 00:16:24,676 But we're given this current, so we know that two Amps 415 00:16:24,676 --> 00:16:26,706 flows through the eight Ohm resistor, 416 00:16:26,706 --> 00:16:27,847 and you can always determine the voltage 417 00:16:27,847 --> 00:16:29,939 across the resistor using Ohm's Law, 418 00:16:29,939 --> 00:16:31,870 so the voltage across the eight Ohm resistor 419 00:16:31,870 --> 00:16:33,870 is gonna be two Amps, which is flowing 420 00:16:33,870 --> 00:16:35,436 through the eight Ohm resistor, 421 00:16:35,436 --> 00:16:39,129 multiplied by eight Ohms, and we get 16 volts. 422 00:16:39,129 --> 00:16:41,837 Which I can plug into here, and this gives me 423 00:16:41,837 --> 00:16:45,686 24 volts minus the voltage across the six Ohm resistor, 424 00:16:45,686 --> 00:16:48,389 minus 16, has to equal zero. 425 00:16:48,389 --> 00:16:49,583 And if I solve this for the voltage 426 00:16:49,583 --> 00:16:52,253 across the six Ohm resistor, I get 24 volts 427 00:16:52,253 --> 00:16:54,871 minus 16 volts, which is eight volts. 428 00:16:54,871 --> 00:16:56,721 So the voltage across the six Ohm resistor 429 00:16:56,721 --> 00:16:58,308 would be eight volts. 430 00:16:58,308 --> 00:17:01,392 Note, because the 12 Ohm resistor and the six Ohm resistor 431 00:17:01,392 --> 00:17:04,095 are in parallel, the voltage across the 12 Ohm resistor 432 00:17:04,095 --> 00:17:07,078 would also be eight volts, because the voltage across 433 00:17:07,078 --> 00:17:11,633 any two elements in parallel, have to be the same. 434 00:17:11,633 --> 00:17:13,549 Voltmeters are the device that you use 435 00:17:13,549 --> 00:17:16,397 to measure the voltage between two points in a circuit. 436 00:17:16,397 --> 00:17:18,112 When hooking up a voltmeter you've gotta 437 00:17:18,113 --> 00:17:20,579 hook it up in parallel between the two points 438 00:17:20,579 --> 00:17:22,599 you wanna find the voltage across. 439 00:17:22,599 --> 00:17:24,014 In other words, to determine the voltage 440 00:17:24,015 --> 00:17:26,387 between this point and this point, which would be 441 00:17:26,387 --> 00:17:29,235 the voltage across R3, you would hook up the voltmeter 442 00:17:29,235 --> 00:17:31,465 in parallel with R3. 443 00:17:31,465 --> 00:17:33,351 Ammeters are the devices we use to measure 444 00:17:33,351 --> 00:17:34,908 the electrical current that pass 445 00:17:34,908 --> 00:17:36,780 through a point in a circuit, and ammeters 446 00:17:36,780 --> 00:17:39,595 have to be hooked up in series with the circuit element 447 00:17:39,595 --> 00:17:41,373 you want to determine the current through. 448 00:17:41,373 --> 00:17:43,261 In other words, if we wanted to determine the current 449 00:17:43,261 --> 00:17:47,267 through R1, we would hook up the ammeter in series with R1. 450 00:17:47,267 --> 00:17:49,877 Note that for these electrical devices to work well, 451 00:17:49,877 --> 00:17:53,185 the ammeter should have almost zero internal resistance, 452 00:17:53,185 --> 00:17:55,407 thereby not affecting the current that flows 453 00:17:55,407 --> 00:17:57,479 through the circuit, and voltmeters should have 454 00:17:57,479 --> 00:18:00,440 near infinite resistance, so that it doesn't draw 455 00:18:00,440 --> 00:18:02,674 any of the current from the resistor. 456 00:18:02,674 --> 00:18:04,969 In reality, ammeters have a very small, 457 00:18:04,969 --> 00:18:07,148 but non-zero internal resistance, 458 00:18:07,148 --> 00:18:08,463 and voltmeters have a very high, 459 00:18:08,463 --> 00:18:10,653 but not infinite internal resistance. 460 00:18:10,653 --> 00:18:12,107 So what would an example problem involving 461 00:18:12,107 --> 00:18:14,108 voltmeters and ammeters look like? 462 00:18:14,108 --> 00:18:15,865 Let's say we have the circuit shown below, 463 00:18:15,865 --> 00:18:18,558 and these numbered circles represent possible places 464 00:18:18,558 --> 00:18:21,173 we could stick a voltmeter to measure the voltage 465 00:18:21,173 --> 00:18:23,165 across the eight Ohm resistor. 466 00:18:23,165 --> 00:18:25,380 Which two of these voltmeters would correctly 467 00:18:25,380 --> 00:18:28,079 give the voltage across the eight Ohm resistor? 468 00:18:28,079 --> 00:18:30,124 And you have to be careful, some AP problems 469 00:18:30,124 --> 00:18:31,844 are gonna require you to select 470 00:18:31,844 --> 00:18:33,838 two correct answers for the multiple choice, 471 00:18:33,838 --> 00:18:36,509 so be sure to read the instructions carefully. 472 00:18:36,509 --> 00:18:38,229 Voltmeter number four is a terrible choice, 473 00:18:38,229 --> 00:18:40,368 you never hook up your voltmeter in series, 474 00:18:40,368 --> 00:18:42,331 but the circuit element you're trying to find 475 00:18:42,331 --> 00:18:44,382 the voltage across, and voltmeter number one 476 00:18:44,382 --> 00:18:46,524 is doing nothing really, because its' measuring 477 00:18:46,524 --> 00:18:48,989 the voltage between two points in a wire 478 00:18:48,989 --> 00:18:51,206 with nothing in between that wire. 479 00:18:51,206 --> 00:18:52,952 So the voltage measured by voltmeter one 480 00:18:52,952 --> 00:18:56,034 should just be zero, since the voltage across a wire 481 00:18:56,034 --> 00:18:59,412 of zero resistance should just give you zero volts. 482 00:18:59,412 --> 00:19:01,627 So the correct choices would be voltmeter number two, 483 00:19:01,627 --> 00:19:04,736 which gives you the voltage across the eight Ohm resistor, 484 00:19:04,736 --> 00:19:06,429 and voltmeter number three, which also gives you 485 00:19:06,429 --> 00:19:08,304 an equivalent measurement of the voltage 486 00:19:08,304 --> 00:00:00,000 across the resistor eight Ohms.