[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.47,0:00:01.69,Default,,0000,0000,0000,,- [Voiceover] Electric\NCharge is a property
Dialogue: 0,0:00:01.69,0:00:05.62,Default,,0000,0000,0000,,that some, but not all fundamental\Nparticles in nature have.
Dialogue: 0,0:00:05.62,0:00:06.76,Default,,0000,0000,0000,,The most commonly talked about
Dialogue: 0,0:00:06.76,0:00:09.56,Default,,0000,0000,0000,,fundamentally charged\Nparticles are the electrons,
Dialogue: 0,0:00:09.56,0:00:11.56,Default,,0000,0000,0000,,which orbit the outside of the atom.
Dialogue: 0,0:00:11.56,0:00:13.34,Default,,0000,0000,0000,,These are negatively charged.
Dialogue: 0,0:00:13.34,0:00:15.15,Default,,0000,0000,0000,,There's also the protons, which reside
Dialogue: 0,0:00:15.15,0:00:18.48,Default,,0000,0000,0000,,inside the nucleus, and\Nthese are positively charged.
Dialogue: 0,0:00:18.48,0:00:20.09,Default,,0000,0000,0000,,And the neutrons inside the nucleus
Dialogue: 0,0:00:20.09,0:00:21.86,Default,,0000,0000,0000,,don't have any net charge.
Dialogue: 0,0:00:21.86,0:00:24.21,Default,,0000,0000,0000,,It turns out that all\Nfundamentally charged particles
Dialogue: 0,0:00:24.21,0:00:26.31,Default,,0000,0000,0000,,in the universe, have charges that come in
Dialogue: 0,0:00:26.31,0:00:29.26,Default,,0000,0000,0000,,integer units of the elementary charge.
Dialogue: 0,0:00:29.26,0:00:30.80,Default,,0000,0000,0000,,So if you find a particle in nature,
Dialogue: 0,0:00:30.80,0:00:33.92,Default,,0000,0000,0000,,it's gonna have a charge\Nof one times this number,
Dialogue: 0,0:00:33.92,0:00:37.02,Default,,0000,0000,0000,,two times this number,\Nthree times this number,
Dialogue: 0,0:00:37.02,0:00:38.72,Default,,0000,0000,0000,,and it could either be\Npositive or negative.
Dialogue: 0,0:00:38.72,0:00:41.69,Default,,0000,0000,0000,,For instance, the electron\Nhas a charge of negative 1.6
Dialogue: 0,0:00:41.69,0:00:44.66,Default,,0000,0000,0000,,times 10 to the negative 19th Coulombs,
Dialogue: 0,0:00:44.66,0:00:47.54,Default,,0000,0000,0000,,and the charge of the\Nproton is positive 1.6
Dialogue: 0,0:00:47.54,0:00:49.89,Default,,0000,0000,0000,,times 10 to the negative 19th Coulombs.
Dialogue: 0,0:00:49.89,0:00:51.93,Default,,0000,0000,0000,,However, most atoms in the universe
Dialogue: 0,0:00:51.93,0:00:54.08,Default,,0000,0000,0000,,are electrically neutral overall,
Dialogue: 0,0:00:54.08,0:00:56.83,Default,,0000,0000,0000,,since they'll have just\Nas many negative electrons
Dialogue: 0,0:00:56.83,0:00:59.01,Default,,0000,0000,0000,,as they do positive protons.
Dialogue: 0,0:00:59.01,0:01:01.31,Default,,0000,0000,0000,,But if an atom had too many electrons,
Dialogue: 0,0:01:01.31,0:01:04.01,Default,,0000,0000,0000,,overall that atom would\Nbe negatively charged,
Dialogue: 0,0:01:04.01,0:01:06.50,Default,,0000,0000,0000,,and if an atom had too few electrons,
Dialogue: 0,0:01:06.50,0:01:08.84,Default,,0000,0000,0000,,that atom would be overall\Npositively charged.
Dialogue: 0,0:01:08.84,0:01:10.59,Default,,0000,0000,0000,,And something that's really\Nimportant to remember
Dialogue: 0,0:01:10.59,0:01:13.08,Default,,0000,0000,0000,,is that the electric\Ncharge is always conserved
Dialogue: 0,0:01:13.08,0:01:15.21,Default,,0000,0000,0000,,for every process, in other words,
Dialogue: 0,0:01:15.21,0:01:17.54,Default,,0000,0000,0000,,the total charge initially, is gonna equal
Dialogue: 0,0:01:17.54,0:01:20.59,Default,,0000,0000,0000,,the total charge finally\Nafter any process.
Dialogue: 0,0:01:20.59,0:01:22.09,Default,,0000,0000,0000,,So what's an example problem involving
Dialogue: 0,0:01:22.09,0:01:23.66,Default,,0000,0000,0000,,electric charge look like?
Dialogue: 0,0:01:23.66,0:01:26.19,Default,,0000,0000,0000,,Let's say three identically\Nsized metal spheres
Dialogue: 0,0:01:26.19,0:01:28.34,Default,,0000,0000,0000,,start off with the charges seen below.
Dialogue: 0,0:01:28.34,0:01:32.14,Default,,0000,0000,0000,,Positive five Q, positive\Nthree Q, and negative two Q.
Dialogue: 0,0:01:32.14,0:01:35.79,Default,,0000,0000,0000,,If we touch sphere X to\Nsphere Y, and separate them,
Dialogue: 0,0:01:35.79,0:01:39.61,Default,,0000,0000,0000,,and then touch sphere Y to\Nsphere Z, and separate them,
Dialogue: 0,0:01:39.61,0:01:42.58,Default,,0000,0000,0000,,what will be the final\Ncharge on each sphere?
Dialogue: 0,0:01:42.58,0:01:44.55,Default,,0000,0000,0000,,Well first, when we touch X to Y,
Dialogue: 0,0:01:44.55,0:01:46.69,Default,,0000,0000,0000,,the total charge has to be conserved.
Dialogue: 0,0:01:46.69,0:01:49.22,Default,,0000,0000,0000,,There's a total charge\Nof eight Q amongst them,
Dialogue: 0,0:01:49.22,0:01:50.93,Default,,0000,0000,0000,,and since they're identically sized,
Dialogue: 0,0:01:50.93,0:01:52.78,Default,,0000,0000,0000,,they'll both share the total charge,
Dialogue: 0,0:01:52.78,0:01:54.04,Default,,0000,0000,0000,,which means after they touch,
Dialogue: 0,0:01:54.04,0:01:56.01,Default,,0000,0000,0000,,they'll both have positive four Q.
Dialogue: 0,0:01:56.01,0:01:57.43,Default,,0000,0000,0000,,If one of the spheres were larger,
Dialogue: 0,0:01:57.43,0:01:59.03,Default,,0000,0000,0000,,it would gain more of the charge,
Dialogue: 0,0:01:59.03,0:02:01.38,Default,,0000,0000,0000,,but the total charge\Nwould still be conserved.
Dialogue: 0,0:02:01.38,0:02:04.19,Default,,0000,0000,0000,,And now when sphere Y\Nis touched to sphere Z,
Dialogue: 0,0:02:04.19,0:02:06.41,Default,,0000,0000,0000,,the total charge amongst\Nthem at that moment
Dialogue: 0,0:02:06.41,0:02:09.44,Default,,0000,0000,0000,,would be positive four\NQ plus negative two Q,
Dialogue: 0,0:02:09.44,0:02:10.98,Default,,0000,0000,0000,,which is positive two Q.
Dialogue: 0,0:02:10.98,0:02:13.49,Default,,0000,0000,0000,,They would share it equally,\Nso sphere Y would have
Dialogue: 0,0:02:13.49,0:02:17.03,Default,,0000,0000,0000,,positive Q, and sphere Z\Nwould also have positive Q.
Dialogue: 0,0:02:17.03,0:02:19.25,Default,,0000,0000,0000,,So the answer here is C.
Dialogue: 0,0:02:19.25,0:02:22.29,Default,,0000,0000,0000,,Opposite charges attract\Nand like charges repel,
Dialogue: 0,0:02:22.29,0:02:24.87,Default,,0000,0000,0000,,and what Coulomb's Law does\Nis it gives you a way to find
Dialogue: 0,0:02:24.87,0:02:28.50,Default,,0000,0000,0000,,the magnitude of the electric\Nforce between two charges.
Dialogue: 0,0:02:28.50,0:02:30.07,Default,,0000,0000,0000,,The formula for Coulomb's Law says
Dialogue: 0,0:02:30.07,0:02:32.12,Default,,0000,0000,0000,,that the magnitude of the electric force
Dialogue: 0,0:02:32.12,0:02:36.00,Default,,0000,0000,0000,,between two charges Q1\Nand Q2, is gonna equal the
Dialogue: 0,0:02:36.00,0:02:39.20,Default,,0000,0000,0000,,electric constant K, which is\Nnine times 10 to the ninth,
Dialogue: 0,0:02:39.20,0:02:42.62,Default,,0000,0000,0000,,times the product of the two\Ncharges, measured in Coulombs
Dialogue: 0,0:02:42.62,0:02:44.49,Default,,0000,0000,0000,,divided by the center to center distance
Dialogue: 0,0:02:44.49,0:02:47.26,Default,,0000,0000,0000,,between those two charges, squared.
Dialogue: 0,0:02:47.26,0:02:49.54,Default,,0000,0000,0000,,You can't forget to square this distance.
Dialogue: 0,0:02:49.54,0:02:51.61,Default,,0000,0000,0000,,And it's gotta be in\Nmeters if you want to find
Dialogue: 0,0:02:51.61,0:02:54.15,Default,,0000,0000,0000,,SI units of Newtons for the force.
Dialogue: 0,0:02:54.15,0:02:56.72,Default,,0000,0000,0000,,Also, don't rely on the\Nnegative and positive signs
Dialogue: 0,0:02:56.72,0:02:59.63,Default,,0000,0000,0000,,of the charges to tell you\Nwhich way the force points,
Dialogue: 0,0:02:59.63,0:03:02.06,Default,,0000,0000,0000,,just use the fact that\Nopposite charges attract
Dialogue: 0,0:03:02.06,0:03:04.99,Default,,0000,0000,0000,,and like charges repel,\Nand use Coulomb's Law
Dialogue: 0,0:03:04.99,0:03:07.18,Default,,0000,0000,0000,,to get the magnitude of the force.
Dialogue: 0,0:03:07.18,0:03:08.50,Default,,0000,0000,0000,,So what's an example problem involving
Dialogue: 0,0:03:08.50,0:03:09.79,Default,,0000,0000,0000,,Coulomb's Law look like?
Dialogue: 0,0:03:09.79,0:03:11.65,Default,,0000,0000,0000,,Let's say two charges\Nexert an electric force
Dialogue: 0,0:03:11.65,0:03:13.59,Default,,0000,0000,0000,,of magnitude F on each other.
Dialogue: 0,0:03:13.59,0:03:16.34,Default,,0000,0000,0000,,What would be the magnitude\Nof the new electric force
Dialogue: 0,0:03:16.34,0:03:18.75,Default,,0000,0000,0000,,if the distance between\Nthe charges is tripled
Dialogue: 0,0:03:18.75,0:03:21.69,Default,,0000,0000,0000,,and the magnitude of one\Nof the charges is doubled?
Dialogue: 0,0:03:21.69,0:03:23.38,Default,,0000,0000,0000,,Well we know the formula for Coulomb's Law
Dialogue: 0,0:03:23.38,0:03:25.26,Default,,0000,0000,0000,,says that the force between two charges
Dialogue: 0,0:03:25.26,0:03:28.03,Default,,0000,0000,0000,,is the electric constant\Ntimes one of the charges,
Dialogue: 0,0:03:28.03,0:03:30.35,Default,,0000,0000,0000,,times the other charge\Ndivided by the distance
Dialogue: 0,0:03:30.35,0:03:33.32,Default,,0000,0000,0000,,between them squared, and now\Nonce we triple the distance
Dialogue: 0,0:03:33.32,0:03:36.15,Default,,0000,0000,0000,,and double a charge,\Nthe new electric force
Dialogue: 0,0:03:36.15,0:03:39.18,Default,,0000,0000,0000,,is gonna be the electric constant\Ntimes one of the charges,
Dialogue: 0,0:03:39.18,0:03:42.72,Default,,0000,0000,0000,,multiplied by two times\None of the charges,
Dialogue: 0,0:03:42.72,0:03:46.40,Default,,0000,0000,0000,,divided by three times the\Ndistance, which is squared,
Dialogue: 0,0:03:46.40,0:03:48.15,Default,,0000,0000,0000,,so I'm gonna get a factor of two on top,
Dialogue: 0,0:03:48.15,0:03:49.93,Default,,0000,0000,0000,,and this three will get\Nsquared, which gives me
Dialogue: 0,0:03:49.93,0:03:51.76,Default,,0000,0000,0000,,a factor of nine on the bottom.
Dialogue: 0,0:03:51.76,0:03:54.15,Default,,0000,0000,0000,,If I pull up those extra\Nfactors I get that the new force
Dialogue: 0,0:03:54.15,0:03:58.59,Default,,0000,0000,0000,,is gonna be two ninths\Nmultiplied by K, Q1, Q2,
Dialogue: 0,0:03:58.59,0:04:00.99,Default,,0000,0000,0000,,over D squared, but this entire quantity
Dialogue: 0,0:04:00.99,0:04:03.97,Default,,0000,0000,0000,,was just the old force F, so the new force
Dialogue: 0,0:04:03.97,0:04:07.73,Default,,0000,0000,0000,,is going to be two\Nninths of the old force.
Dialogue: 0,0:04:07.73,0:04:10.64,Default,,0000,0000,0000,,The electrical current I tells\Nyou the amount of Coulombs
Dialogue: 0,0:04:10.64,0:04:14.06,Default,,0000,0000,0000,,of charge that passes a\Npoint in a wire per second.
Dialogue: 0,0:04:14.06,0:04:16.02,Default,,0000,0000,0000,,So if you watch some point in a wire,
Dialogue: 0,0:04:16.02,0:04:18.11,Default,,0000,0000,0000,,and you count how many Coulombs of charge
Dialogue: 0,0:04:18.11,0:04:21.33,Default,,0000,0000,0000,,pass by that point per second,\Nthat would be the current.
Dialogue: 0,0:04:21.33,0:04:23.66,Default,,0000,0000,0000,,Or in equation form we\Ncan see that the current I
Dialogue: 0,0:04:23.66,0:04:25.63,Default,,0000,0000,0000,,is gonna be the amount\Nof charge that flows
Dialogue: 0,0:04:25.63,0:04:27.90,Default,,0000,0000,0000,,past a point in a wire per time.
Dialogue: 0,0:04:27.90,0:04:30.59,Default,,0000,0000,0000,,This gives the units of\NI as Coulombs per second,
Dialogue: 0,0:04:30.59,0:04:32.85,Default,,0000,0000,0000,,which we abbreviate as an Ampere.
Dialogue: 0,0:04:32.85,0:04:34.68,Default,,0000,0000,0000,,And since charge and time aren't vectors,
Dialogue: 0,0:04:34.68,0:04:36.50,Default,,0000,0000,0000,,current is not a vector either.
Dialogue: 0,0:04:36.50,0:04:37.74,Default,,0000,0000,0000,,Something that's kind of strange
Dialogue: 0,0:04:37.74,0:04:40.91,Default,,0000,0000,0000,,is that the so-called\Nconventional direction of current
Dialogue: 0,0:04:40.91,0:04:43.48,Default,,0000,0000,0000,,would be the direction that\Npositive charges flow within a
Dialogue: 0,0:04:43.48,0:04:47.17,Default,,0000,0000,0000,,wire, however positive charges\Ndon't flow within a wire.
Dialogue: 0,0:04:47.17,0:04:49.46,Default,,0000,0000,0000,,The only charges that\Nactually flow in a wire
Dialogue: 0,0:04:49.46,0:04:51.66,Default,,0000,0000,0000,,are negative charges, but it turns out
Dialogue: 0,0:04:51.66,0:04:55.02,Default,,0000,0000,0000,,that negative charges flowing\Nto the left is physically
Dialogue: 0,0:04:55.02,0:04:57.93,Default,,0000,0000,0000,,the same as positive charges\Nflowing to the right.
Dialogue: 0,0:04:57.93,0:05:00.50,Default,,0000,0000,0000,,So in physics problems\Nwe pretend as if it were
Dialogue: 0,0:05:00.50,0:05:02.82,Default,,0000,0000,0000,,the positive charges moving, however
Dialogue: 0,0:05:02.82,0:05:04.89,Default,,0000,0000,0000,,it's really the electrons,\Nwhich are negative,
Dialogue: 0,0:05:04.89,0:05:06.46,Default,,0000,0000,0000,,that are moving within the wire.
Dialogue: 0,0:05:06.46,0:05:07.41,Default,,0000,0000,0000,,So what's an example problem
Dialogue: 0,0:05:07.41,0:05:09.30,Default,,0000,0000,0000,,involving electrical current look like?
Dialogue: 0,0:05:09.30,0:05:11.39,Default,,0000,0000,0000,,Let's say three amps\Nflows within a circuit.
Dialogue: 0,0:05:11.39,0:05:14.01,Default,,0000,0000,0000,,How much charge would pass\Nby a point in that wire
Dialogue: 0,0:05:14.01,0:05:16.50,Default,,0000,0000,0000,,during a time interval of five minutes?
Dialogue: 0,0:05:16.50,0:05:17.90,Default,,0000,0000,0000,,Well we know the definition of current
Dialogue: 0,0:05:17.90,0:05:20.34,Default,,0000,0000,0000,,is the charge per time,\Nthat means the charge
Dialogue: 0,0:05:20.34,0:05:23.28,Default,,0000,0000,0000,,is gonna be the amount of\Ncurrent multiplied by the time,
Dialogue: 0,0:05:23.28,0:05:25.30,Default,,0000,0000,0000,,so we take our current of three amps,
Dialogue: 0,0:05:25.30,0:05:27.30,Default,,0000,0000,0000,,and we multiply by the time, but we can't
Dialogue: 0,0:05:27.30,0:05:30.30,Default,,0000,0000,0000,,multiply by five because\Nthat's in units of minutes,
Dialogue: 0,0:05:30.30,0:05:32.49,Default,,0000,0000,0000,,since amps is Coulombs per second,
Dialogue: 0,0:05:32.49,0:05:34.54,Default,,0000,0000,0000,,we've got to convert five\Nminutes into seconds,
Dialogue: 0,0:05:34.54,0:05:37.93,Default,,0000,0000,0000,,which would be five minutes,\Nmultiplied by 60 seconds
Dialogue: 0,0:05:37.93,0:05:39.83,Default,,0000,0000,0000,,per minute, which would\Ngive us a total amount
Dialogue: 0,0:05:39.83,0:05:42.00,Default,,0000,0000,0000,,of charge of 900 Coulombs.
Dialogue: 0,0:05:42.89,0:05:45.79,Default,,0000,0000,0000,,The resistance of a circuit\Nelement measures how much
Dialogue: 0,0:05:45.79,0:05:48.70,Default,,0000,0000,0000,,that element will restrict\Nthe flow of current.
Dialogue: 0,0:05:48.70,0:05:51.13,Default,,0000,0000,0000,,The larger the resistance,\Nthe less current
Dialogue: 0,0:05:51.13,0:05:52.76,Default,,0000,0000,0000,,there will be allowed to flow.
Dialogue: 0,0:05:52.76,0:05:55.98,Default,,0000,0000,0000,,And this definition of\Nresistance is given by Ohm's Law.
Dialogue: 0,0:05:55.98,0:05:57.79,Default,,0000,0000,0000,,Ohm's Law states that\Nthe amount of current
Dialogue: 0,0:05:57.79,0:05:59.76,Default,,0000,0000,0000,,that you'll get through\Na portion of a circuit,
Dialogue: 0,0:05:59.76,0:06:01.48,Default,,0000,0000,0000,,it's gonna be proportional to the voltage
Dialogue: 0,0:06:01.48,0:06:04.37,Default,,0000,0000,0000,,across that portion,\Ndivided by the resistance
Dialogue: 0,0:06:04.37,0:06:06.01,Default,,0000,0000,0000,,of that portion of the circuit.
Dialogue: 0,0:06:06.01,0:06:08.33,Default,,0000,0000,0000,,So between these two points,\Nthe amount of current
Dialogue: 0,0:06:08.33,0:06:10.20,Default,,0000,0000,0000,,that will flow, is gonna be equal to
Dialogue: 0,0:06:10.20,0:06:12.54,Default,,0000,0000,0000,,the voltage between those two points,
Dialogue: 0,0:06:12.54,0:06:15.60,Default,,0000,0000,0000,,divided by the resistance\Nbetween those two points.
Dialogue: 0,0:06:15.60,0:06:17.89,Default,,0000,0000,0000,,So the larger the\Nresistance, the less current
Dialogue: 0,0:06:17.89,0:06:20.96,Default,,0000,0000,0000,,will flow, but the greater\Nthe voltage supplied,
Dialogue: 0,0:06:20.96,0:06:22.42,Default,,0000,0000,0000,,the greater the current will be.
Dialogue: 0,0:06:22.42,0:06:24.11,Default,,0000,0000,0000,,And this is what Ohm's Law says.
Dialogue: 0,0:06:24.11,0:06:26.32,Default,,0000,0000,0000,,Even though Ohm's Law\Ngives you a way to define
Dialogue: 0,0:06:26.32,0:06:29.08,Default,,0000,0000,0000,,the resistance, you can\Ndetermine the resistance
Dialogue: 0,0:06:29.08,0:06:31.04,Default,,0000,0000,0000,,of a circuit element by knowing the size
Dialogue: 0,0:06:31.04,0:06:32.95,Default,,0000,0000,0000,,and shape of that circuit element.
Dialogue: 0,0:06:32.95,0:06:35.57,Default,,0000,0000,0000,,In other words, the resistance\Nof a cylindrical resistor,
Dialogue: 0,0:06:35.57,0:06:37.87,Default,,0000,0000,0000,,is gonna be equal to the resistivity,
Dialogue: 0,0:06:37.87,0:06:40.21,Default,,0000,0000,0000,,which is a measure of an\Nobject's natural resistance
Dialogue: 0,0:06:40.21,0:06:43.13,Default,,0000,0000,0000,,to current, multiplied by\Nthe length of that resistor,
Dialogue: 0,0:06:43.13,0:06:45.77,Default,,0000,0000,0000,,the longer the resistor,\Nthe greater the resistance
Dialogue: 0,0:06:45.77,0:06:48.07,Default,,0000,0000,0000,,and the more it will\Nresist the flow of current,
Dialogue: 0,0:06:48.07,0:06:50.20,Default,,0000,0000,0000,,and then divide it by\Nthe cross sectional area
Dialogue: 0,0:06:50.20,0:06:52.86,Default,,0000,0000,0000,,of that resistor, which would\Nbe this area right here,
Dialogue: 0,0:06:52.86,0:06:55.58,Default,,0000,0000,0000,,the current is either\Nflowing into or out of.
Dialogue: 0,0:06:55.58,0:06:58.45,Default,,0000,0000,0000,,If the resistor is cylindrical,\Nthe area of this circle
Dialogue: 0,0:06:58.45,0:07:01.31,Default,,0000,0000,0000,,would be Pi times r\Nsquared, where little r
Dialogue: 0,0:07:01.31,0:07:04.36,Default,,0000,0000,0000,,would be the radius of\Nthis cross sectional area.
Dialogue: 0,0:07:04.36,0:07:07.52,Default,,0000,0000,0000,,The units of resistance is\NOhms, and it is not a vector.
Dialogue: 0,0:07:07.52,0:07:09.75,Default,,0000,0000,0000,,It is always positive or zero.
Dialogue: 0,0:07:09.75,0:07:11.89,Default,,0000,0000,0000,,So what's an example\Nproblem involving Ohm's Law,
Dialogue: 0,0:07:11.89,0:07:14.70,Default,,0000,0000,0000,,or the resistance of a\Ncylindrical resistor look like?
Dialogue: 0,0:07:14.70,0:07:16.78,Default,,0000,0000,0000,,Let's say a battery of\Nvoltage V is hooked up
Dialogue: 0,0:07:16.78,0:07:19.74,Default,,0000,0000,0000,,to a single cylindrical\Nresistor of length L
Dialogue: 0,0:07:19.74,0:07:21.81,Default,,0000,0000,0000,,and radius little r, and when that's done,
Dialogue: 0,0:07:21.81,0:07:24.48,Default,,0000,0000,0000,,a current I is flowing\Nthrough the battery.
Dialogue: 0,0:07:24.48,0:07:27.91,Default,,0000,0000,0000,,What is the resistivity\NRho, of that resistor?
Dialogue: 0,0:07:27.91,0:07:30.28,Default,,0000,0000,0000,,Well we know Ohm's Law states\Nthat the current that flows
Dialogue: 0,0:07:30.28,0:07:32.45,Default,,0000,0000,0000,,through a portion of a\Ncircuit will be equal to
Dialogue: 0,0:07:32.45,0:07:33.97,Default,,0000,0000,0000,,the voltage across that portion,
Dialogue: 0,0:07:33.97,0:07:36.28,Default,,0000,0000,0000,,divided by the resistance of that portion.
Dialogue: 0,0:07:36.28,0:07:38.60,Default,,0000,0000,0000,,And this means the resistance\Nof this resistor is gonna be
Dialogue: 0,0:07:38.60,0:07:40.88,Default,,0000,0000,0000,,the voltage of the battery\Ndivided by the current.
Dialogue: 0,0:07:40.88,0:07:43.27,Default,,0000,0000,0000,,To factor resistivity\Ninto this, we have to use
Dialogue: 0,0:07:43.27,0:07:46.15,Default,,0000,0000,0000,,the formula for the resistance\Nof a cylindrical resistor,
Dialogue: 0,0:07:46.15,0:07:48.22,Default,,0000,0000,0000,,which is Rho times L over A.
Dialogue: 0,0:07:48.22,0:07:50.64,Default,,0000,0000,0000,,This gives us the\Nresistance of the resistor,
Dialogue: 0,0:07:50.64,0:07:52.74,Default,,0000,0000,0000,,which is gotta equal V over I,
Dialogue: 0,0:07:52.74,0:07:55.02,Default,,0000,0000,0000,,and now we can solve\Nfor the resistivity Rho.
Dialogue: 0,0:07:55.02,0:07:58.38,Default,,0000,0000,0000,,We get V times A over I\NL, but since we're given
Dialogue: 0,0:07:58.38,0:08:01.12,Default,,0000,0000,0000,,the radius little r,\Nwe gotta write the area
Dialogue: 0,0:08:01.12,0:08:03.20,Default,,0000,0000,0000,,in terms of that radius,\Nso this is gonna be
Dialogue: 0,0:08:03.20,0:08:06.86,Default,,0000,0000,0000,,V times Pi, r squared,\Ndivided by I times L,
Dialogue: 0,0:08:06.86,0:08:09.21,Default,,0000,0000,0000,,which gives us an answer of C.
Dialogue: 0,0:08:09.21,0:08:12.28,Default,,0000,0000,0000,,When dealing with complicated\Ncircuits with many resistors,
Dialogue: 0,0:08:12.28,0:08:14.21,Default,,0000,0000,0000,,you often have to reduce those resistors
Dialogue: 0,0:08:14.21,0:08:17.10,Default,,0000,0000,0000,,into smaller, equivalent\Namounts of resistors.
Dialogue: 0,0:08:17.10,0:08:18.86,Default,,0000,0000,0000,,And the two ways you\Ndo this are by finding
Dialogue: 0,0:08:18.86,0:08:21.68,Default,,0000,0000,0000,,two resistors that are\Nin series or in parallel.
Dialogue: 0,0:08:21.68,0:08:24.16,Default,,0000,0000,0000,,Resistors will be in\Nseries if the same current
Dialogue: 0,0:08:24.16,0:08:25.89,Default,,0000,0000,0000,,that flows through the same resistor,
Dialogue: 0,0:08:25.89,0:08:27.69,Default,,0000,0000,0000,,flows through the next resistor.
Dialogue: 0,0:08:27.69,0:08:30.16,Default,,0000,0000,0000,,If the current branched\Noff in between them,
Dialogue: 0,0:08:30.16,0:08:32.79,Default,,0000,0000,0000,,these resistors would\Nno longer be in series,
Dialogue: 0,0:08:32.79,0:08:34.15,Default,,0000,0000,0000,,but if they're in series you can find
Dialogue: 0,0:08:34.15,0:08:37.04,Default,,0000,0000,0000,,the equivalent resistance\Nof this section of wire
Dialogue: 0,0:08:37.04,0:08:40.04,Default,,0000,0000,0000,,by just adding up the two\Nindividual resistances.
Dialogue: 0,0:08:40.04,0:08:41.78,Default,,0000,0000,0000,,So the current for resistors in series
Dialogue: 0,0:08:41.78,0:08:45.11,Default,,0000,0000,0000,,must be the same, but the\Nvoltage might be different,
Dialogue: 0,0:08:45.11,0:08:47.24,Default,,0000,0000,0000,,since they could have\Ndifferent resistances.
Dialogue: 0,0:08:47.24,0:08:49.10,Default,,0000,0000,0000,,Two resistors will be in parallel,
Dialogue: 0,0:08:49.10,0:08:52.00,Default,,0000,0000,0000,,if a current comes in,\Nsplits into two parts,
Dialogue: 0,0:08:52.00,0:08:54.80,Default,,0000,0000,0000,,goes through one resistor\Neach, and then rejoins
Dialogue: 0,0:08:54.80,0:08:56.82,Default,,0000,0000,0000,,before hitting anything\Nelse in the circuit,
Dialogue: 0,0:08:56.82,0:08:58.77,Default,,0000,0000,0000,,and if this is the case,\Nyou can find the equivalent
Dialogue: 0,0:08:58.77,0:09:00.88,Default,,0000,0000,0000,,resistance of this portion of the circuit,
Dialogue: 0,0:09:00.88,0:09:04.09,Default,,0000,0000,0000,,i.e. between these two\Npoints, by saying that one
Dialogue: 0,0:09:04.09,0:09:06.52,Default,,0000,0000,0000,,over the equivalent\Nresistance is gonna equal
Dialogue: 0,0:09:06.52,0:09:08.89,Default,,0000,0000,0000,,one over the resistance\Nof the first resistor,
Dialogue: 0,0:09:08.89,0:09:11.56,Default,,0000,0000,0000,,plus one over the resistance\Nof the second resistor.
Dialogue: 0,0:09:11.56,0:09:14.28,Default,,0000,0000,0000,,But be careful, one\Nover R1 plus one over R2
Dialogue: 0,0:09:14.28,0:09:16.54,Default,,0000,0000,0000,,just gives you one over R equivalent.
Dialogue: 0,0:09:16.54,0:09:18.66,Default,,0000,0000,0000,,If you want R equivalent,\Nyou're gonna have to take
Dialogue: 0,0:09:18.66,0:09:22.63,Default,,0000,0000,0000,,one over this entire side,\Nin order to get R equivalent.
Dialogue: 0,0:09:22.63,0:09:24.60,Default,,0000,0000,0000,,So what's an example\Nproblem involving resistors
Dialogue: 0,0:09:24.60,0:09:26.39,Default,,0000,0000,0000,,in series and parallel look like?
Dialogue: 0,0:09:26.39,0:09:28.36,Default,,0000,0000,0000,,Let's say we have this\Ncircuit shown below,
Dialogue: 0,0:09:28.36,0:09:30.16,Default,,0000,0000,0000,,and we want to know what current flows
Dialogue: 0,0:09:30.16,0:09:31.73,Default,,0000,0000,0000,,through the eight Ohm resistor.
Dialogue: 0,0:09:31.73,0:09:33.28,Default,,0000,0000,0000,,Now you might be tempted to say this,
Dialogue: 0,0:09:33.28,0:09:36.86,Default,,0000,0000,0000,,since Ohm's Law says that the\Ncurrent is delta V over R,
Dialogue: 0,0:09:36.86,0:09:38.73,Default,,0000,0000,0000,,we can just plug in the\Nvoltage of the battery,
Dialogue: 0,0:09:38.73,0:09:41.33,Default,,0000,0000,0000,,which is 24 volts,\Ndivided by the resistance
Dialogue: 0,0:09:41.33,0:09:43.47,Default,,0000,0000,0000,,of the resistor, which is eight Ohms,
Dialogue: 0,0:09:43.47,0:09:45.22,Default,,0000,0000,0000,,and that would give us three Amps.
Dialogue: 0,0:09:45.22,0:09:46.49,Default,,0000,0000,0000,,But that's not right at all.
Dialogue: 0,0:09:46.49,0:09:48.55,Default,,0000,0000,0000,,When using Ohm's Law,\Nthe current that flows
Dialogue: 0,0:09:48.55,0:09:51.51,Default,,0000,0000,0000,,through a resistor R, is gonna be equal to
Dialogue: 0,0:09:51.51,0:09:54.38,Default,,0000,0000,0000,,the voltage across that\Nresistor divided by
Dialogue: 0,0:09:54.38,0:09:56.41,Default,,0000,0000,0000,,the resistance of that resistor.
Dialogue: 0,0:09:56.41,0:09:59.38,Default,,0000,0000,0000,,So if we plug eight Ohms into\Nthe denominator, we've gotta
Dialogue: 0,0:09:59.38,0:10:02.57,Default,,0000,0000,0000,,plug in the voltage across\Nthat eight Ohm resistor.
Dialogue: 0,0:10:02.57,0:10:04.46,Default,,0000,0000,0000,,But the voltage across\Nthe eight Ohm resistor
Dialogue: 0,0:10:04.46,0:10:07.56,Default,,0000,0000,0000,,is not gonna be the full\N24 volts of the battery,
Dialogue: 0,0:10:07.56,0:10:09.60,Default,,0000,0000,0000,,it's gonna be less than 24 volts.
Dialogue: 0,0:10:09.60,0:10:12.06,Default,,0000,0000,0000,,In other words, the battery\Nprovides a voltage between
Dialogue: 0,0:10:12.06,0:10:15.14,Default,,0000,0000,0000,,this point and this point of 24 volts,
Dialogue: 0,0:10:15.14,0:10:16.61,Default,,0000,0000,0000,,but there's gonna be voltage drops
Dialogue: 0,0:10:16.61,0:10:18.86,Default,,0000,0000,0000,,across the six and 12 Ohm resistors,
Dialogue: 0,0:10:18.86,0:10:20.14,Default,,0000,0000,0000,,which make it so that the voltage
Dialogue: 0,0:10:20.14,0:10:21.78,Default,,0000,0000,0000,,across the eight Ohm resistor is not
Dialogue: 0,0:10:21.78,0:10:23.70,Default,,0000,0000,0000,,gonna be the full 24 volts.
Dialogue: 0,0:10:23.70,0:10:26.40,Default,,0000,0000,0000,,So we have to reduce these\Nresistors to a single resistance.
Dialogue: 0,0:10:26.40,0:10:28.34,Default,,0000,0000,0000,,The six and the 12 are in parallel,
Dialogue: 0,0:10:28.34,0:10:31.35,Default,,0000,0000,0000,,so we can say that one\Nover six, plus one over 12,
Dialogue: 0,0:10:31.35,0:10:32.91,Default,,0000,0000,0000,,would equal one over the resistance
Dialogue: 0,0:10:32.91,0:10:34.40,Default,,0000,0000,0000,,of that portion of the circuit.
Dialogue: 0,0:10:34.40,0:10:37.54,Default,,0000,0000,0000,,This is gonna equal three\Ntwelves, which is one fourth,
Dialogue: 0,0:10:37.54,0:10:39.28,Default,,0000,0000,0000,,so that means that parallel\Nportion of the circuit
Dialogue: 0,0:10:39.28,0:10:41.86,Default,,0000,0000,0000,,has an equivalent resistance of four Ohms.
Dialogue: 0,0:10:41.86,0:10:44.16,Default,,0000,0000,0000,,So between this point and this point,
Dialogue: 0,0:10:44.16,0:10:46.18,Default,,0000,0000,0000,,there are four Ohms of resistance,
Dialogue: 0,0:10:46.18,0:10:48.47,Default,,0000,0000,0000,,and that equivalent\Nresistance is in series
Dialogue: 0,0:10:48.47,0:10:50.14,Default,,0000,0000,0000,,with this eight Ohm resistor.
Dialogue: 0,0:10:50.14,0:10:51.65,Default,,0000,0000,0000,,So we can add four and eight,
Dialogue: 0,0:10:51.65,0:10:54.22,Default,,0000,0000,0000,,and get 12 Ohms of total resistance.
Dialogue: 0,0:10:54.22,0:10:57.72,Default,,0000,0000,0000,,And now I can say that the\Nfull 24 volts of the battery
Dialogue: 0,0:10:57.72,0:11:00.14,Default,,0000,0000,0000,,is applied across this\Nentire equivalent resistance
Dialogue: 0,0:11:00.14,0:11:03.68,Default,,0000,0000,0000,,of 12 Ohms, so if I come up\Nhere and change this eight Ohms
Dialogue: 0,0:11:03.68,0:11:06.98,Default,,0000,0000,0000,,to 12 Ohms of equivalent\Nresistance for the total circuit,
Dialogue: 0,0:11:06.98,0:11:08.44,Default,,0000,0000,0000,,I'll get the correct current that flows
Dialogue: 0,0:11:08.44,0:11:10.32,Default,,0000,0000,0000,,through the battery of two Amps.
Dialogue: 0,0:11:10.32,0:11:12.14,Default,,0000,0000,0000,,And since that's the current\Nthat's flowing through
Dialogue: 0,0:11:12.14,0:11:13.88,Default,,0000,0000,0000,,the battery, that had to be the current
Dialogue: 0,0:11:13.88,0:11:16.34,Default,,0000,0000,0000,,that's flowing through the\Neight Ohm resistor as well.
Dialogue: 0,0:11:16.34,0:11:20.06,Default,,0000,0000,0000,,Since this eight Ohm resistor\Nand the batter are in series.
Dialogue: 0,0:11:20.06,0:11:23.15,Default,,0000,0000,0000,,Elements in a circuit\Noften use Electrical Power.
Dialogue: 0,0:11:23.15,0:11:25.89,Default,,0000,0000,0000,,That is to say, when current\Nruns through a resistor,
Dialogue: 0,0:11:25.89,0:11:27.73,Default,,0000,0000,0000,,the electrons moving through that resistor
Dialogue: 0,0:11:27.73,0:11:30.43,Default,,0000,0000,0000,,turn some of their\Nelectrical potential energy
Dialogue: 0,0:11:30.43,0:11:33.77,Default,,0000,0000,0000,,into energies like heat, light, or sound.
Dialogue: 0,0:11:33.77,0:11:35.41,Default,,0000,0000,0000,,And the rate at which these electrons
Dialogue: 0,0:11:35.41,0:11:37.98,Default,,0000,0000,0000,,are turning their energy\Ninto other forms of energy,
Dialogue: 0,0:11:37.98,0:11:39.85,Default,,0000,0000,0000,,is called the electrical power.
Dialogue: 0,0:11:39.85,0:11:41.82,Default,,0000,0000,0000,,So the rate at which a resistor is turning
Dialogue: 0,0:11:41.82,0:11:43.97,Default,,0000,0000,0000,,electrical potential energy into heat,
Dialogue: 0,0:11:43.97,0:11:46.48,Default,,0000,0000,0000,,is the electrical power\Nused by that resistor.
Dialogue: 0,0:11:46.48,0:11:47.94,Default,,0000,0000,0000,,In other words, the amount of energy
Dialogue: 0,0:11:47.94,0:11:50.08,Default,,0000,0000,0000,,converted into heat, divided by the time
Dialogue: 0,0:11:50.08,0:11:52.38,Default,,0000,0000,0000,,it took to convert that\Nenergy, is the definition
Dialogue: 0,0:11:52.38,0:11:54.36,Default,,0000,0000,0000,,of the power, and there's\Na way to determine
Dialogue: 0,0:11:54.36,0:11:57.40,Default,,0000,0000,0000,,this number of Joules per\Nsecond, in terms of quantities
Dialogue: 0,0:11:57.40,0:12:00.20,Default,,0000,0000,0000,,like the current, the\Nvoltage, and the resistance.
Dialogue: 0,0:12:00.20,0:12:02.42,Default,,0000,0000,0000,,The power used by a\Nresistor can be written as
Dialogue: 0,0:12:02.42,0:12:04.41,Default,,0000,0000,0000,,the current through that resistor
Dialogue: 0,0:12:04.41,0:12:07.44,Default,,0000,0000,0000,,multiplied by the voltage\Nacross that resistor,
Dialogue: 0,0:12:07.44,0:12:10.66,Default,,0000,0000,0000,,or if you substituted Ohm's\NLaw into this formula,
Dialogue: 0,0:12:10.66,0:12:12.18,Default,,0000,0000,0000,,you see that this is equivalent to
Dialogue: 0,0:12:12.18,0:12:14.32,Default,,0000,0000,0000,,the current through that resistor squared,
Dialogue: 0,0:12:14.32,0:12:16.76,Default,,0000,0000,0000,,multiplied by the\Nresistance of the resistor,
Dialogue: 0,0:12:16.76,0:12:18.13,Default,,0000,0000,0000,,or we could rearrange these formulas
Dialogue: 0,0:12:18.13,0:12:20.02,Default,,0000,0000,0000,,to get that the power used by a resistor
Dialogue: 0,0:12:20.02,0:12:23.12,Default,,0000,0000,0000,,would also be the voltage\Nacross that resistor squared,
Dialogue: 0,0:12:23.12,0:12:25.52,Default,,0000,0000,0000,,divided by the resistance\Nof that resistor.
Dialogue: 0,0:12:25.52,0:12:27.54,Default,,0000,0000,0000,,All three of these, if used correctly,
Dialogue: 0,0:12:27.54,0:12:30.13,Default,,0000,0000,0000,,will give you the same\Nnumber for the power used
Dialogue: 0,0:12:30.13,0:12:32.07,Default,,0000,0000,0000,,by a resistor, and if\Nyou wanted to determine
Dialogue: 0,0:12:32.07,0:12:34.60,Default,,0000,0000,0000,,the number of Joules of\Nheat energy converted,
Dialogue: 0,0:12:34.60,0:12:37.05,Default,,0000,0000,0000,,you could set any one of\Nthese equal to the amount
Dialogue: 0,0:12:37.05,0:12:40.12,Default,,0000,0000,0000,,of energy per time, and\Nsolve for that energy.
Dialogue: 0,0:12:40.12,0:12:42.56,Default,,0000,0000,0000,,The units of Electrical Power\Nare the same as the regular
Dialogue: 0,0:12:42.56,0:12:46.26,Default,,0000,0000,0000,,units of power, which is\NWatts, i.e. Joules per second,
Dialogue: 0,0:12:46.26,0:12:48.33,Default,,0000,0000,0000,,and Electrical Power is not a vector.
Dialogue: 0,0:12:48.33,0:12:49.84,Default,,0000,0000,0000,,So what's an example problem involving
Dialogue: 0,0:12:49.84,0:12:51.45,Default,,0000,0000,0000,,Electrical Power look like?
Dialogue: 0,0:12:51.45,0:12:53.30,Default,,0000,0000,0000,,Let's say a light bulb of resistance R
Dialogue: 0,0:12:53.30,0:12:55.42,Default,,0000,0000,0000,,is hooked up to a source of voltage V,
Dialogue: 0,0:12:55.42,0:12:57.69,Default,,0000,0000,0000,,and a second light bulb of resistance 2R,
Dialogue: 0,0:12:57.69,0:13:00.25,Default,,0000,0000,0000,,is hooked up to a source of voltage 2V.
Dialogue: 0,0:13:00.25,0:13:02.79,Default,,0000,0000,0000,,How does the power used\Nby the second light bulb
Dialogue: 0,0:13:02.79,0:13:05.28,Default,,0000,0000,0000,,compare to the power used\Nby the first light bulb?
Dialogue: 0,0:13:05.28,0:13:07.55,Default,,0000,0000,0000,,Since we have the\Ninformation about R and V,
Dialogue: 0,0:13:07.55,0:13:09.38,Default,,0000,0000,0000,,I'll use the version of the power formula
Dialogue: 0,0:13:09.38,0:13:11.64,Default,,0000,0000,0000,,that says that the\Npower used by a resistor
Dialogue: 0,0:13:11.64,0:13:13.98,Default,,0000,0000,0000,,is gonna be delta V squared over R.
Dialogue: 0,0:13:13.98,0:13:15.94,Default,,0000,0000,0000,,So in terms of quantities\Ngiven the power used
Dialogue: 0,0:13:15.94,0:13:18.80,Default,,0000,0000,0000,,by the first light bulb is\Ngonna be V squared over R.
Dialogue: 0,0:13:18.80,0:13:21.22,Default,,0000,0000,0000,,And the power used by the\Nsecond light bulb is gonna be
Dialogue: 0,0:13:21.22,0:13:23.78,Default,,0000,0000,0000,,equal to the voltage across\Nthe second light bulb,
Dialogue: 0,0:13:23.78,0:13:26.85,Default,,0000,0000,0000,,which is two times the voltage\Nacross the first light bulb,
Dialogue: 0,0:13:26.85,0:13:28.95,Default,,0000,0000,0000,,and we square that,\Ndivided by the resistance
Dialogue: 0,0:13:28.95,0:13:30.68,Default,,0000,0000,0000,,of the second light\Nbulb, which is gonna be
Dialogue: 0,0:13:30.68,0:13:33.02,Default,,0000,0000,0000,,two times the resistance\Nof the first light bulb.
Dialogue: 0,0:13:33.02,0:13:35.55,Default,,0000,0000,0000,,The two squared on top is\Ngonna give me a factor of four,
Dialogue: 0,0:13:35.55,0:13:37.71,Default,,0000,0000,0000,,and I'll have another\Nfactor of two on the bottom.
Dialogue: 0,0:13:37.71,0:13:40.03,Default,,0000,0000,0000,,So if I factor out this\Nfour divided by two,
Dialogue: 0,0:13:40.03,0:13:42.27,Default,,0000,0000,0000,,I get that the power used\Nby the second light bulb
Dialogue: 0,0:13:42.27,0:13:44.67,Default,,0000,0000,0000,,is gonna be two times V squared over R,
Dialogue: 0,0:13:44.67,0:13:46.66,Default,,0000,0000,0000,,but V squared over R was the power used
Dialogue: 0,0:13:46.66,0:13:48.66,Default,,0000,0000,0000,,by the first light bulb, so the power used
Dialogue: 0,0:13:48.66,0:13:50.65,Default,,0000,0000,0000,,by the second light bulb is gonna be two
Dialogue: 0,0:13:50.65,0:13:52.84,Default,,0000,0000,0000,,times the power used by\Nthe first light bulb,
Dialogue: 0,0:13:52.84,0:13:54.68,Default,,0000,0000,0000,,and so if the light\Nbulb of resistance two R
Dialogue: 0,0:13:54.68,0:13:57.66,Default,,0000,0000,0000,,has twice the power, and\Nthat means it'll be brighter.
Dialogue: 0,0:13:57.66,0:13:59.59,Default,,0000,0000,0000,,The quantity that\Ndetermines the brightness
Dialogue: 0,0:13:59.59,0:14:03.12,Default,,0000,0000,0000,,of a light bulb, is the electrical\Npower of that light bulb.
Dialogue: 0,0:14:03.12,0:14:06.32,Default,,0000,0000,0000,,It's not necessarily the\Nresistance or the voltage,
Dialogue: 0,0:14:06.32,0:14:07.82,Default,,0000,0000,0000,,it's the combination of the two
Dialogue: 0,0:14:07.82,0:14:10.90,Default,,0000,0000,0000,,in this formula that will\Ntell you the electrical power,
Dialogue: 0,0:14:10.90,0:14:13.65,Default,,0000,0000,0000,,and therefore the brightness\Nof the light bulb.
Dialogue: 0,0:14:13.65,0:14:15.52,Default,,0000,0000,0000,,Two of the most useful ideas in circuits
Dialogue: 0,0:14:15.52,0:14:17.77,Default,,0000,0000,0000,,are referred to as Kirchhoff's Rules.
Dialogue: 0,0:14:17.77,0:14:19.85,Default,,0000,0000,0000,,The first rule is called\Nthe Junction rule,
Dialogue: 0,0:14:19.85,0:14:22.04,Default,,0000,0000,0000,,and it states that all the\Ncurrent entering a junction
Dialogue: 0,0:14:22.04,0:14:25.14,Default,,0000,0000,0000,,must equal all the current\Nexiting that junction.
Dialogue: 0,0:14:25.14,0:14:26.96,Default,,0000,0000,0000,,In other words, if you add\Nall the current that flows
Dialogue: 0,0:14:26.96,0:14:28.100,Default,,0000,0000,0000,,into a junction, that has to equal
Dialogue: 0,0:14:28.100,0:14:31.26,Default,,0000,0000,0000,,all the current that flows\Nout of that junction,
Dialogue: 0,0:14:31.26,0:14:33.26,Default,,0000,0000,0000,,because current is just flowing charge,
Dialogue: 0,0:14:33.26,0:14:35.90,Default,,0000,0000,0000,,and charge is conserved,\Nso charge can't be
Dialogue: 0,0:14:35.90,0:14:38.96,Default,,0000,0000,0000,,created or destroyed at\Nany point in the circuit.
Dialogue: 0,0:14:38.96,0:14:41.51,Default,,0000,0000,0000,,No more than water can\Nget created or destroyed
Dialogue: 0,0:14:41.51,0:14:43.21,Default,,0000,0000,0000,,within a series of pipes.
Dialogue: 0,0:14:43.21,0:14:45.08,Default,,0000,0000,0000,,And the second rule is\Ncalled the Loop rule,
Dialogue: 0,0:14:45.08,0:14:47.52,Default,,0000,0000,0000,,which states that if you\Nadd up all the changes
Dialogue: 0,0:14:47.52,0:14:51.24,Default,,0000,0000,0000,,in electric potential, i.e.\Nvoltages around any closed
Dialogue: 0,0:14:51.24,0:14:54.24,Default,,0000,0000,0000,,loop in a circuit, it'll\Nalways add up to zero.
Dialogue: 0,0:14:54.24,0:14:56.46,Default,,0000,0000,0000,,So if you add up all\Nthe voltages encountered
Dialogue: 0,0:14:56.46,0:14:58.19,Default,,0000,0000,0000,,through a closed loop through a circuit,
Dialogue: 0,0:14:58.19,0:14:59.77,Default,,0000,0000,0000,,it always adds up to zero.
Dialogue: 0,0:14:59.77,0:15:02.27,Default,,0000,0000,0000,,And this is just a result\Nof conservation of energy.
Dialogue: 0,0:15:02.27,0:15:04.41,Default,,0000,0000,0000,,The electrons will gain\Nenergy when they flow
Dialogue: 0,0:15:04.41,0:15:06.41,Default,,0000,0000,0000,,through the battery,\Nand they'll lose energy
Dialogue: 0,0:15:06.41,0:15:08.40,Default,,0000,0000,0000,,every time they flow through a resistor,
Dialogue: 0,0:15:08.40,0:15:09.92,Default,,0000,0000,0000,,but the total amount of energy they gain
Dialogue: 0,0:15:09.92,0:15:12.44,Default,,0000,0000,0000,,from the battery, has to be\Nequal to the total amount
Dialogue: 0,0:15:12.44,0:15:14.96,Default,,0000,0000,0000,,of energy they lose due to the resistors.
Dialogue: 0,0:15:14.96,0:15:17.34,Default,,0000,0000,0000,,In other words, if we\Nconsider a complicated circuit
Dialogue: 0,0:15:17.34,0:15:19.76,Default,,0000,0000,0000,,that has a batter and three resistors,
Dialogue: 0,0:15:19.76,0:15:23.12,Default,,0000,0000,0000,,the total current flowing\Ninto a junction I1,
Dialogue: 0,0:15:23.12,0:15:25.06,Default,,0000,0000,0000,,will have to be equal to the total current
Dialogue: 0,0:15:25.06,0:15:28.29,Default,,0000,0000,0000,,coming out of that junction, I2 and I3.
Dialogue: 0,0:15:28.29,0:15:30.95,Default,,0000,0000,0000,,Since no charge gets created or destroyed.
Dialogue: 0,0:15:30.95,0:15:32.99,Default,,0000,0000,0000,,And that means when these\Ntwo currents combine again,
Dialogue: 0,0:15:32.99,0:15:34.48,Default,,0000,0000,0000,,the total current flowing out
Dialogue: 0,0:15:34.48,0:15:37.19,Default,,0000,0000,0000,,of that section is gonna again be I1.
Dialogue: 0,0:15:37.19,0:15:39.99,Default,,0000,0000,0000,,And if we follow a closed\Nloop through this circuit,
Dialogue: 0,0:15:39.99,0:15:42.37,Default,,0000,0000,0000,,the sum of all the\Nvoltages around that loop
Dialogue: 0,0:15:42.37,0:15:45.92,Default,,0000,0000,0000,,have to add up to zero, i.e.\Nthe voltage of the battery
Dialogue: 0,0:15:45.92,0:15:48.94,Default,,0000,0000,0000,,minus the voltage drop\Nacross the first resistor,
Dialogue: 0,0:15:48.94,0:15:51.78,Default,,0000,0000,0000,,minus the voltage drop\Nacross the second resistor,
Dialogue: 0,0:15:51.78,0:15:53.48,Default,,0000,0000,0000,,would have to equal zero.
Dialogue: 0,0:15:53.48,0:15:54.89,Default,,0000,0000,0000,,So what's an example problem involving
Dialogue: 0,0:15:54.89,0:15:56.50,Default,,0000,0000,0000,,Kirchhoff's Rules look like?
Dialogue: 0,0:15:56.50,0:15:58.51,Default,,0000,0000,0000,,Let's say we have the circuit\Nbelow and we wanted to
Dialogue: 0,0:15:58.51,0:16:01.70,Default,,0000,0000,0000,,determine the voltage\Nacross the six Ohm resistor.
Dialogue: 0,0:16:01.70,0:16:03.41,Default,,0000,0000,0000,,To do this, we could use the Loop rule,
Dialogue: 0,0:16:03.41,0:16:05.63,Default,,0000,0000,0000,,I'll start behind the\Nbattery, and I'll go through
Dialogue: 0,0:16:05.63,0:16:08.39,Default,,0000,0000,0000,,the resistor, I want to\Ndetermine the voltage across.
Dialogue: 0,0:16:08.39,0:16:10.93,Default,,0000,0000,0000,,I'll add up all the\Nvoltages across that loop,
Dialogue: 0,0:16:10.93,0:16:12.36,Default,,0000,0000,0000,,and set it equal to zero.
Dialogue: 0,0:16:12.36,0:16:13.56,Default,,0000,0000,0000,,So the voltage across the battery
Dialogue: 0,0:16:13.56,0:16:15.70,Default,,0000,0000,0000,,is gonna be positive 24 volts,
Dialogue: 0,0:16:15.70,0:16:18.28,Default,,0000,0000,0000,,minus the voltage across\Nthe six Ohm resistor,
Dialogue: 0,0:16:18.28,0:16:19.74,Default,,0000,0000,0000,,and then minus the voltage across the
Dialogue: 0,0:16:19.74,0:16:22.22,Default,,0000,0000,0000,,eight Ohm resistor has to equal zero.
Dialogue: 0,0:16:22.22,0:16:24.68,Default,,0000,0000,0000,,But we're given this current,\Nso we know that two Amps
Dialogue: 0,0:16:24.68,0:16:26.71,Default,,0000,0000,0000,,flows through the eight Ohm resistor,
Dialogue: 0,0:16:26.71,0:16:27.85,Default,,0000,0000,0000,,and you can always determine the voltage
Dialogue: 0,0:16:27.85,0:16:29.94,Default,,0000,0000,0000,,across the resistor using Ohm's Law,
Dialogue: 0,0:16:29.94,0:16:31.87,Default,,0000,0000,0000,,so the voltage across\Nthe eight Ohm resistor
Dialogue: 0,0:16:31.87,0:16:33.87,Default,,0000,0000,0000,,is gonna be two Amps, which is flowing
Dialogue: 0,0:16:33.87,0:16:35.44,Default,,0000,0000,0000,,through the eight Ohm resistor,
Dialogue: 0,0:16:35.44,0:16:39.13,Default,,0000,0000,0000,,multiplied by eight Ohms,\Nand we get 16 volts.
Dialogue: 0,0:16:39.13,0:16:41.84,Default,,0000,0000,0000,,Which I can plug into\Nhere, and this gives me
Dialogue: 0,0:16:41.84,0:16:45.69,Default,,0000,0000,0000,,24 volts minus the voltage\Nacross the six Ohm resistor,
Dialogue: 0,0:16:45.69,0:16:48.39,Default,,0000,0000,0000,,minus 16, has to equal zero.
Dialogue: 0,0:16:48.39,0:16:49.58,Default,,0000,0000,0000,,And if I solve this for the voltage
Dialogue: 0,0:16:49.58,0:16:52.25,Default,,0000,0000,0000,,across the six Ohm\Nresistor, I get 24 volts
Dialogue: 0,0:16:52.25,0:16:54.87,Default,,0000,0000,0000,,minus 16 volts, which is eight volts.
Dialogue: 0,0:16:54.87,0:16:56.72,Default,,0000,0000,0000,,So the voltage across the six Ohm resistor
Dialogue: 0,0:16:56.72,0:16:58.31,Default,,0000,0000,0000,,would be eight volts.
Dialogue: 0,0:16:58.31,0:17:01.39,Default,,0000,0000,0000,,Note, because the 12 Ohm\Nresistor and the six Ohm resistor
Dialogue: 0,0:17:01.39,0:17:04.10,Default,,0000,0000,0000,,are in parallel, the voltage\Nacross the 12 Ohm resistor
Dialogue: 0,0:17:04.10,0:17:07.08,Default,,0000,0000,0000,,would also be eight volts,\Nbecause the voltage across
Dialogue: 0,0:17:07.08,0:17:11.63,Default,,0000,0000,0000,,any two elements in parallel,\Nhave to be the same.
Dialogue: 0,0:17:11.63,0:17:13.55,Default,,0000,0000,0000,,Voltmeters are the device that you use
Dialogue: 0,0:17:13.55,0:17:16.40,Default,,0000,0000,0000,,to measure the voltage between\Ntwo points in a circuit.
Dialogue: 0,0:17:16.40,0:17:18.11,Default,,0000,0000,0000,,When hooking up a voltmeter you've gotta
Dialogue: 0,0:17:18.11,0:17:20.58,Default,,0000,0000,0000,,hook it up in parallel\Nbetween the two points
Dialogue: 0,0:17:20.58,0:17:22.60,Default,,0000,0000,0000,,you wanna find the voltage across.
Dialogue: 0,0:17:22.60,0:17:24.02,Default,,0000,0000,0000,,In other words, to determine the voltage
Dialogue: 0,0:17:24.02,0:17:26.39,Default,,0000,0000,0000,,between this point and\Nthis point, which would be
Dialogue: 0,0:17:26.39,0:17:29.24,Default,,0000,0000,0000,,the voltage across R3, you\Nwould hook up the voltmeter
Dialogue: 0,0:17:29.24,0:17:31.46,Default,,0000,0000,0000,,in parallel with R3.
Dialogue: 0,0:17:31.46,0:17:33.35,Default,,0000,0000,0000,,Ammeters are the devices we use to measure
Dialogue: 0,0:17:33.35,0:17:34.91,Default,,0000,0000,0000,,the electrical current that pass
Dialogue: 0,0:17:34.91,0:17:36.78,Default,,0000,0000,0000,,through a point in a circuit, and ammeters
Dialogue: 0,0:17:36.78,0:17:39.60,Default,,0000,0000,0000,,have to be hooked up in series\Nwith the circuit element
Dialogue: 0,0:17:39.60,0:17:41.37,Default,,0000,0000,0000,,you want to determine the current through.
Dialogue: 0,0:17:41.37,0:17:43.26,Default,,0000,0000,0000,,In other words, if we wanted\Nto determine the current
Dialogue: 0,0:17:43.26,0:17:47.27,Default,,0000,0000,0000,,through R1, we would hook up\Nthe ammeter in series with R1.
Dialogue: 0,0:17:47.27,0:17:49.88,Default,,0000,0000,0000,,Note that for these electrical\Ndevices to work well,
Dialogue: 0,0:17:49.88,0:17:53.18,Default,,0000,0000,0000,,the ammeter should have almost\Nzero internal resistance,
Dialogue: 0,0:17:53.18,0:17:55.41,Default,,0000,0000,0000,,thereby not affecting\Nthe current that flows
Dialogue: 0,0:17:55.41,0:17:57.48,Default,,0000,0000,0000,,through the circuit, and\Nvoltmeters should have
Dialogue: 0,0:17:57.48,0:18:00.44,Default,,0000,0000,0000,,near infinite resistance,\Nso that it doesn't draw
Dialogue: 0,0:18:00.44,0:18:02.67,Default,,0000,0000,0000,,any of the current from the resistor.
Dialogue: 0,0:18:02.67,0:18:04.97,Default,,0000,0000,0000,,In reality, ammeters have a very small,
Dialogue: 0,0:18:04.97,0:18:07.15,Default,,0000,0000,0000,,but non-zero internal resistance,
Dialogue: 0,0:18:07.15,0:18:08.46,Default,,0000,0000,0000,,and voltmeters have a very high,
Dialogue: 0,0:18:08.46,0:18:10.65,Default,,0000,0000,0000,,but not infinite internal resistance.
Dialogue: 0,0:18:10.65,0:18:12.11,Default,,0000,0000,0000,,So what would an example problem involving
Dialogue: 0,0:18:12.11,0:18:14.11,Default,,0000,0000,0000,,voltmeters and ammeters look like?
Dialogue: 0,0:18:14.11,0:18:15.86,Default,,0000,0000,0000,,Let's say we have the circuit shown below,
Dialogue: 0,0:18:15.86,0:18:18.56,Default,,0000,0000,0000,,and these numbered circles\Nrepresent possible places
Dialogue: 0,0:18:18.56,0:18:21.17,Default,,0000,0000,0000,,we could stick a voltmeter\Nto measure the voltage
Dialogue: 0,0:18:21.17,0:18:23.16,Default,,0000,0000,0000,,across the eight Ohm resistor.
Dialogue: 0,0:18:23.16,0:18:25.38,Default,,0000,0000,0000,,Which two of these\Nvoltmeters would correctly
Dialogue: 0,0:18:25.38,0:18:28.08,Default,,0000,0000,0000,,give the voltage across\Nthe eight Ohm resistor?
Dialogue: 0,0:18:28.08,0:18:30.12,Default,,0000,0000,0000,,And you have to be\Ncareful, some AP problems
Dialogue: 0,0:18:30.12,0:18:31.84,Default,,0000,0000,0000,,are gonna require you to select
Dialogue: 0,0:18:31.84,0:18:33.84,Default,,0000,0000,0000,,two correct answers for\Nthe multiple choice,
Dialogue: 0,0:18:33.84,0:18:36.51,Default,,0000,0000,0000,,so be sure to read the\Ninstructions carefully.
Dialogue: 0,0:18:36.51,0:18:38.23,Default,,0000,0000,0000,,Voltmeter number four\Nis a terrible choice,
Dialogue: 0,0:18:38.23,0:18:40.37,Default,,0000,0000,0000,,you never hook up your\Nvoltmeter in series,
Dialogue: 0,0:18:40.37,0:18:42.33,Default,,0000,0000,0000,,but the circuit element\Nyou're trying to find
Dialogue: 0,0:18:42.33,0:18:44.38,Default,,0000,0000,0000,,the voltage across, and\Nvoltmeter number one
Dialogue: 0,0:18:44.38,0:18:46.52,Default,,0000,0000,0000,,is doing nothing really,\Nbecause its' measuring
Dialogue: 0,0:18:46.52,0:18:48.99,Default,,0000,0000,0000,,the voltage between two points in a wire
Dialogue: 0,0:18:48.99,0:18:51.21,Default,,0000,0000,0000,,with nothing in between that wire.
Dialogue: 0,0:18:51.21,0:18:52.95,Default,,0000,0000,0000,,So the voltage measured by voltmeter one
Dialogue: 0,0:18:52.95,0:18:56.03,Default,,0000,0000,0000,,should just be zero, since\Nthe voltage across a wire
Dialogue: 0,0:18:56.03,0:18:59.41,Default,,0000,0000,0000,,of zero resistance should\Njust give you zero volts.
Dialogue: 0,0:18:59.41,0:19:01.63,Default,,0000,0000,0000,,So the correct choices would\Nbe voltmeter number two,
Dialogue: 0,0:19:01.63,0:19:04.74,Default,,0000,0000,0000,,which gives you the voltage\Nacross the eight Ohm resistor,
Dialogue: 0,0:19:04.74,0:19:06.43,Default,,0000,0000,0000,,and voltmeter number\Nthree, which also gives you
Dialogue: 0,0:19:06.43,0:19:08.30,Default,,0000,0000,0000,,an equivalent measurement of the voltage
Dialogue: 0,0:19:08.30,0:19:10.87,Default,,0000,0000,0000,,across the resistor eight Ohms.