[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:04.36,Default,,0000,0000,0000,,>> Hello, this is Dr. Cynthia Furse\Nat the University of Utah,
Dialogue: 0,0:00:04.36,0:00:07.100,Default,,0000,0000,0000,,and today we're going to talk\Nabout RC and RL circuits.
Dialogue: 0,0:00:07.100,0:00:09.33,Default,,0000,0000,0000,,The question of the day is,
Dialogue: 0,0:00:09.33,0:00:11.31,Default,,0000,0000,0000,,how can you calculate the\Nvoltage and current in
Dialogue: 0,0:00:11.31,0:00:13.100,Default,,0000,0000,0000,,a circuit that has\Ninductors and capacitors?
Dialogue: 0,0:00:13.100,0:00:15.72,Default,,0000,0000,0000,,In these cases, we're talking about
Dialogue: 0,0:00:15.72,0:00:18.40,Default,,0000,0000,0000,,a time domain circuit because\Nwe changed something,
Dialogue: 0,0:00:18.40,0:00:21.27,Default,,0000,0000,0000,,we move this switch from point\None to point two for example,
Dialogue: 0,0:00:21.27,0:00:23.84,Default,,0000,0000,0000,,at some point t equals zero.
Dialogue: 0,0:00:23.84,0:00:26.66,Default,,0000,0000,0000,,We talked about RC and\NRL circuits last time,
Dialogue: 0,0:00:26.66,0:00:28.26,Default,,0000,0000,0000,,and we're going to review those again,
Dialogue: 0,0:00:28.26,0:00:30.08,Default,,0000,0000,0000,,and I'm going to give you
Dialogue: 0,0:00:30.08,0:00:34.16,Default,,0000,0000,0000,,a general procedure for being\Nable to analyze these circuits.
Dialogue: 0,0:00:34.16,0:00:37.43,Default,,0000,0000,0000,,So let's review. Last time we talked about
Dialogue: 0,0:00:37.43,0:00:40.52,Default,,0000,0000,0000,,a circuit where the switch\Nclosed at time t equal to zero,
Dialogue: 0,0:00:40.52,0:00:44.06,Default,,0000,0000,0000,,connecting this resistor and\Ncapacitor to the voltage source.
Dialogue: 0,0:00:44.06,0:00:47.10,Default,,0000,0000,0000,,We discovered that the\Ncurrent went from zero,
Dialogue: 0,0:00:47.10,0:00:48.26,Default,,0000,0000,0000,,and suddenly jumped up,
Dialogue: 0,0:00:48.26,0:00:51.20,Default,,0000,0000,0000,,it changed instantly to Vs over R,
Dialogue: 0,0:00:51.20,0:00:53.88,Default,,0000,0000,0000,,and that the equation became Vs over R,
Dialogue: 0,0:00:53.88,0:00:55.56,Default,,0000,0000,0000,,e to the minus t over tau,
Dialogue: 0,0:00:55.56,0:00:57.30,Default,,0000,0000,0000,,where tau equals RC,
Dialogue: 0,0:00:57.30,0:00:58.91,Default,,0000,0000,0000,,and that is the time constant.
Dialogue: 0,0:00:58.91,0:01:00.64,Default,,0000,0000,0000,,At the time constant tau,
Dialogue: 0,0:01:00.64,0:01:05.44,Default,,0000,0000,0000,,this current has fallen down to\N37 percent of its original value.
Dialogue: 0,0:01:05.44,0:01:08.87,Default,,0000,0000,0000,,The voltage on the other hand\Nstarted out at zero and gradually
Dialogue: 0,0:01:08.87,0:01:13.00,Default,,0000,0000,0000,,rose to the value Vs. At\Nthe time constant tau,
Dialogue: 0,0:01:13.00,0:01:15.73,Default,,0000,0000,0000,,it had reached 63 percent\Nof its maximum value,
Dialogue: 0,0:01:15.73,0:01:17.68,Default,,0000,0000,0000,,and here's the equation.
Dialogue: 0,0:01:17.68,0:01:20.15,Default,,0000,0000,0000,,So here's the current and voltage.
Dialogue: 0,0:01:20.15,0:01:21.55,Default,,0000,0000,0000,,What had really happened?
Dialogue: 0,0:01:21.55,0:01:23.08,Default,,0000,0000,0000,,We know that for a capacitor,
Dialogue: 0,0:01:23.08,0:01:26.78,Default,,0000,0000,0000,,the current can change instantly\Nbut the voltage cannot,
Dialogue: 0,0:01:26.78,0:01:28.44,Default,,0000,0000,0000,,the voltage changes slowly.
Dialogue: 0,0:01:28.44,0:01:31.27,Default,,0000,0000,0000,,At time t equal to zero when\Nthere is a sudden change,
Dialogue: 0,0:01:31.27,0:01:34.25,Default,,0000,0000,0000,,the capacitor acts like a short\Ncircuit because it starts
Dialogue: 0,0:01:34.25,0:01:37.46,Default,,0000,0000,0000,,to charge quickly and the current
Dialogue: 0,0:01:37.46,0:01:40.79,Default,,0000,0000,0000,,continues to dump charges onto\Nthe capacitor until it reach
Dialogue: 0,0:01:40.79,0:01:45.21,Default,,0000,0000,0000,,steady state when it's all charged\Nup it acts like an open circuit.
Dialogue: 0,0:01:45.35,0:01:47.74,Default,,0000,0000,0000,,This is what it would look like.
Dialogue: 0,0:01:47.74,0:01:49.87,Default,,0000,0000,0000,,This is the voltage across the capacitor.
Dialogue: 0,0:01:49.87,0:01:52.84,Default,,0000,0000,0000,,If we put a square wave on this\Nand you'd see that it would
Dialogue: 0,0:01:52.84,0:01:56.81,Default,,0000,0000,0000,,charge and discharge and charge\Nand discharge and so on.
Dialogue: 0,0:01:56.97,0:02:01.98,Default,,0000,0000,0000,,So now let's consider a general procedure\Nfor analyzing this type of circuit.
Dialogue: 0,0:02:01.98,0:02:04.18,Default,,0000,0000,0000,,The first thing that we're\Ngoing to do is calculate
Dialogue: 0,0:02:04.18,0:02:06.74,Default,,0000,0000,0000,,the initial conditions before\Nanything has happened.
Dialogue: 0,0:02:06.74,0:02:08.42,Default,,0000,0000,0000,,This is when the switch is opened,
Dialogue: 0,0:02:08.42,0:02:10.78,Default,,0000,0000,0000,,we call it t equals zero minus,
Dialogue: 0,0:02:10.78,0:02:12.80,Default,,0000,0000,0000,,and this is before the switch closes,
Dialogue: 0,0:02:12.80,0:02:15.19,Default,,0000,0000,0000,,and we need to find\Nthe current and voltage there.
Dialogue: 0,0:02:15.19,0:02:18.23,Default,,0000,0000,0000,,At steady-state, we know that\Nthe capacitor is an open circuit,
Dialogue: 0,0:02:18.23,0:02:21.02,Default,,0000,0000,0000,,and in fact the current\Nbecause the switch is
Dialogue: 0,0:02:21.02,0:02:25.44,Default,,0000,0000,0000,,open is zero and the voltage\Nacross the capacitor is zero also.
Dialogue: 0,0:02:25.44,0:02:28.30,Default,,0000,0000,0000,,Then we need to consider\Nthe initial condition,
Dialogue: 0,0:02:28.30,0:02:29.92,Default,,0000,0000,0000,,this is when the switch is closed,
Dialogue: 0,0:02:29.92,0:02:31.72,Default,,0000,0000,0000,,that's time t equals zero plus,
Dialogue: 0,0:02:31.72,0:02:33.01,Default,,0000,0000,0000,,it's just barely closed,
Dialogue: 0,0:02:33.01,0:02:35.30,Default,,0000,0000,0000,,and we need to find the\Ncurrent and the voltage.
Dialogue: 0,0:02:35.30,0:02:36.83,Default,,0000,0000,0000,,When the current closes,
Dialogue: 0,0:02:36.83,0:02:39.35,Default,,0000,0000,0000,,this capacitor acts like a short circuit,
Dialogue: 0,0:02:39.35,0:02:42.29,Default,,0000,0000,0000,,the current can change instantly\Nbut the voltage cannot.
Dialogue: 0,0:02:42.29,0:02:49.92,Default,,0000,0000,0000,,So i of zero is not going to\Nbe the same thing as i0 minus,
Dialogue: 0,0:02:49.92,0:02:52.22,Default,,0000,0000,0000,,it's going to quickly go to Vs over
Dialogue: 0,0:02:52.22,0:02:56.15,Default,,0000,0000,0000,,R. The voltage on the other hand\Nis the same as zero minus,
Dialogue: 0,0:02:56.15,0:02:58.73,Default,,0000,0000,0000,,it's going to stay at zero.
Dialogue: 0,0:02:58.73,0:03:01.55,Default,,0000,0000,0000,,The next thing that we're\Ngoing to do is find
Dialogue: 0,0:03:01.55,0:03:03.98,Default,,0000,0000,0000,,the conditions at time t equal infinity,
Dialogue: 0,0:03:03.98,0:03:06.78,Default,,0000,0000,0000,,when the capacitor is fully charged up.
Dialogue: 0,0:03:06.78,0:03:10.20,Default,,0000,0000,0000,,In that case, the capacitor\Nacts like an open circuit.
Dialogue: 0,0:03:10.20,0:03:12.92,Default,,0000,0000,0000,,The current is now zero\Nand the voltage goes to
Dialogue: 0,0:03:12.92,0:03:16.44,Default,,0000,0000,0000,,the original voltage source\Nor 10 volts, in this case.
Dialogue: 0,0:03:16.44,0:03:19.78,Default,,0000,0000,0000,,The next thing that we do is\Nevaluate the time constant.
Dialogue: 0,0:03:19.78,0:03:21.11,Default,,0000,0000,0000,,For a circuit like this,
Dialogue: 0,0:03:21.11,0:03:23.08,Default,,0000,0000,0000,,the time constant is RC.
Dialogue: 0,0:03:23.08,0:03:24.77,Default,,0000,0000,0000,,I just made some notes here,
Dialogue: 0,0:03:24.77,0:03:29.01,Default,,0000,0000,0000,,because we're going to be using\Nthe form e to the minus t over tau,
Dialogue: 0,0:03:29.01,0:03:33.20,Default,,0000,0000,0000,,just remember that e to\Nthe negative one is 0.37,
Dialogue: 0,0:03:33.20,0:03:35.42,Default,,0000,0000,0000,,that's where the 37 percent comes from,
Dialogue: 0,0:03:35.42,0:03:37.98,Default,,0000,0000,0000,,and one minus e to the one is 0.63,
Dialogue: 0,0:03:37.98,0:03:40.76,Default,,0000,0000,0000,,that's where the 63 percent comes from.
Dialogue: 0,0:03:40.76,0:03:44.45,Default,,0000,0000,0000,,The next thing that we do is we\Nuse a general voltage equation,
Dialogue: 0,0:03:44.45,0:03:46.16,Default,,0000,0000,0000,,the general form of the voltage across
Dialogue: 0,0:03:46.16,0:03:49.22,Default,,0000,0000,0000,,the capacitor right here is where VC is,
Dialogue: 0,0:03:49.22,0:03:51.26,Default,,0000,0000,0000,,the general form of the voltage across
Dialogue: 0,0:03:51.26,0:03:53.58,Default,,0000,0000,0000,,the capacitor is the voltage at infinity,
Dialogue: 0,0:03:53.58,0:03:57.53,Default,,0000,0000,0000,,plus the difference of\Nvoltages at DC and infinity,
Dialogue: 0,0:03:57.53,0:04:01.85,Default,,0000,0000,0000,,and e to the minus t over tau\Nmultiplied by the step function.
Dialogue: 0,0:04:01.85,0:04:04.24,Default,,0000,0000,0000,,The step function just looks like this.
Dialogue: 0,0:04:04.24,0:04:06.39,Default,,0000,0000,0000,,Then if we plug in our values,
Dialogue: 0,0:04:06.39,0:04:10.07,Default,,0000,0000,0000,,we're going to get VS plus\Nzero minus VS times this exponent,
Dialogue: 0,0:04:10.07,0:04:13.65,Default,,0000,0000,0000,,which gives us the form\Nthat we derived yesterday.
Dialogue: 0,0:04:13.66,0:04:16.19,Default,,0000,0000,0000,,Now let's consider what would happen if
Dialogue: 0,0:04:16.19,0:04:17.93,Default,,0000,0000,0000,,the switch closed at some different time,
Dialogue: 0,0:04:17.93,0:04:21.26,Default,,0000,0000,0000,,it wasn't time t equals\Nzero, it was time t0.
Dialogue: 0,0:04:21.26,0:04:23.96,Default,,0000,0000,0000,,Everything would be the\Nsame except that we would
Dialogue: 0,0:04:23.96,0:04:28.23,Default,,0000,0000,0000,,substitute t minus zero\Nfor both the cases here.
Dialogue: 0,0:04:28.94,0:04:33.36,Default,,0000,0000,0000,,So in general, the first thing that we\Ndo is we find the initial conditions.
Dialogue: 0,0:04:33.36,0:04:35.51,Default,,0000,0000,0000,,We start before the switch closes and then
Dialogue: 0,0:04:35.51,0:04:37.92,Default,,0000,0000,0000,,we find the conditions\Nafter the switch closes.
Dialogue: 0,0:04:37.92,0:04:41.00,Default,,0000,0000,0000,,We then find the final conditions and we
Dialogue: 0,0:04:41.00,0:04:44.76,Default,,0000,0000,0000,,remember that the capacitor is\Na short circuit when it's charging,
Dialogue: 0,0:04:44.76,0:04:47.18,Default,,0000,0000,0000,,and an open circuit when\Nit's fully charged.
Dialogue: 0,0:04:47.18,0:04:50.33,Default,,0000,0000,0000,,We then find the time constant which is RC,
Dialogue: 0,0:04:50.33,0:04:53.09,Default,,0000,0000,0000,,and we plug everything into\Nthis equation right here,
Dialogue: 0,0:04:53.09,0:04:55.13,Default,,0000,0000,0000,,or if the switch closed\Nat a different time,
Dialogue: 0,0:04:55.13,0:04:56.82,Default,,0000,0000,0000,,this is the equation we use.
Dialogue: 0,0:04:56.82,0:04:59.05,Default,,0000,0000,0000,,Finally, if we want to find the current,
Dialogue: 0,0:04:59.05,0:05:01.73,Default,,0000,0000,0000,,just use this equation right\Nhere which shows that the
Dialogue: 0,0:05:01.73,0:05:04.70,Default,,0000,0000,0000,,current is capacitance times\Nthe derivative of voltage,
Dialogue: 0,0:05:04.70,0:05:07.76,Default,,0000,0000,0000,,and that will give you\Nthe equations shown here.
Dialogue: 0,0:05:07.76,0:05:11.06,Default,,0000,0000,0000,,Now let's review the RL circuit.
Dialogue: 0,0:05:11.06,0:05:14.48,Default,,0000,0000,0000,,This is again when the switch\Ncloses at t equal to zero,
Dialogue: 0,0:05:14.48,0:05:17.42,Default,,0000,0000,0000,,and we want to find out what\Nhappens to the voltage and current.
Dialogue: 0,0:05:17.42,0:05:20.50,Default,,0000,0000,0000,,If you recall, the current\Nhere changes gradually,
Dialogue: 0,0:05:20.50,0:05:24.65,Default,,0000,0000,0000,,it started out as zero and then it\Ngradually builds up to Vs over R,
Dialogue: 0,0:05:24.65,0:05:28.38,Default,,0000,0000,0000,,reaching 63 percent of\Nits value at the time constant,
Dialogue: 0,0:05:28.38,0:05:31.01,Default,,0000,0000,0000,,the time constant in this case is L over
Dialogue: 0,0:05:31.01,0:05:35.74,Default,,0000,0000,0000,,R. The voltage on the other hand does\Nchange quickly across an inductor,
Dialogue: 0,0:05:35.74,0:05:40.13,Default,,0000,0000,0000,,it started out at zero and it jumped\Nup very quickly to the value of Vs.
Dialogue: 0,0:05:40.13,0:05:42.11,Default,,0000,0000,0000,,It has reached and then it sags,
Dialogue: 0,0:05:42.11,0:05:46.28,Default,,0000,0000,0000,,and it reaches 37 percent of\Nits value at the time constant L over
Dialogue: 0,0:05:46.28,0:05:50.87,Default,,0000,0000,0000,,R. So this is what the current and\Nvoltage look like for an inductor.
Dialogue: 0,0:05:50.87,0:05:53.96,Default,,0000,0000,0000,,Oops, this is supposed to be\Nan inductor and not a capacitor.
Dialogue: 0,0:05:53.96,0:05:57.74,Default,,0000,0000,0000,,The voltage changes instantly\Nbut the current changes slowly.
Dialogue: 0,0:05:57.74,0:06:00.68,Default,,0000,0000,0000,,When there is a sudden change\Nacross an inductor,
Dialogue: 0,0:06:00.68,0:06:04.36,Default,,0000,0000,0000,,the inductor acts like an open circuit\Nand when it has reached steady-state,
Dialogue: 0,0:06:04.36,0:06:06.34,Default,,0000,0000,0000,,it acts like a short.
Dialogue: 0,0:06:06.34,0:06:08.36,Default,,0000,0000,0000,,So this is what it would look like.
Dialogue: 0,0:06:08.36,0:06:11.48,Default,,0000,0000,0000,,Here's the voltage that we would see across
Dialogue: 0,0:06:11.48,0:06:16.14,Default,,0000,0000,0000,,an inductor if we put\Na square wave across it.
Dialogue: 0,0:06:17.21,0:06:21.47,Default,,0000,0000,0000,,Now let's talk about\Nthe general solution procedure for RL.
Dialogue: 0,0:06:21.47,0:06:24.18,Default,,0000,0000,0000,,It works very much the same\Nway as a capacitor does.
Dialogue: 0,0:06:24.18,0:06:27.56,Default,,0000,0000,0000,,We start with the initial condition\Nbefore the switch has closed,
Dialogue: 0,0:06:27.56,0:06:29.98,Default,,0000,0000,0000,,we find out that the\Ncurrent and the voltage
Dialogue: 0,0:06:29.98,0:06:33.50,Default,,0000,0000,0000,,before the switch has closed\Nare both zero once again.
Dialogue: 0,0:06:33.50,0:06:36.41,Default,,0000,0000,0000,,Then we look to see what\Nhappens when we first
Dialogue: 0,0:06:36.41,0:06:39.00,Default,,0000,0000,0000,,close the switch at t equals zero plus,
Dialogue: 0,0:06:39.00,0:06:41.03,Default,,0000,0000,0000,,and we need to find\Nthe current and voltage.
Dialogue: 0,0:06:41.03,0:06:43.86,Default,,0000,0000,0000,,Remember that the current\Ndoes not change suddenly,
Dialogue: 0,0:06:43.86,0:06:47.03,Default,,0000,0000,0000,,so the current at time t equals\Nzero is going to be the current
Dialogue: 0,0:06:47.03,0:06:50.85,Default,,0000,0000,0000,,at t equals zero minus\Nor zero in this case.
Dialogue: 0,0:06:50.85,0:06:53.61,Default,,0000,0000,0000,,Then, we look at to see\Nwhat the voltage is.
Dialogue: 0,0:06:53.61,0:06:56.64,Default,,0000,0000,0000,,The inductor is open when the\Ncurrent first starts to flow,
Dialogue: 0,0:06:56.64,0:06:58.73,Default,,0000,0000,0000,,and so that makes the voltage the
Dialogue: 0,0:06:58.73,0:07:02.27,Default,,0000,0000,0000,,Vs. Now we need
Dialogue: 0,0:07:02.27,0:07:03.89,Default,,0000,0000,0000,,to go to find what's going to happen in
Dialogue: 0,0:07:03.89,0:07:05.94,Default,,0000,0000,0000,,the steady state or the final conditions,
Dialogue: 0,0:07:05.94,0:07:09.30,Default,,0000,0000,0000,,then we look to see what the current\Nis going to be, and in this case,
Dialogue: 0,0:07:09.30,0:07:12.74,Default,,0000,0000,0000,,the inductor is going to\Nbe a short circuit because
Dialogue: 0,0:07:12.74,0:07:16.49,Default,,0000,0000,0000,,it is now able to fully take all of\Nthe current that's going through it,
Dialogue: 0,0:07:16.49,0:07:19.46,Default,,0000,0000,0000,,so the current at infinity is Vs over R,
Dialogue: 0,0:07:19.46,0:07:21.62,Default,,0000,0000,0000,,and the voltage is zero.
Dialogue: 0,0:07:21.62,0:07:26.13,Default,,0000,0000,0000,,Next we need to consider the time constant\Nwhich for an inductor is L over R,
Dialogue: 0,0:07:26.13,0:07:29.52,Default,,0000,0000,0000,,and again remember that\Nthe same exponent apply.
Dialogue: 0,0:07:29.52,0:07:33.66,Default,,0000,0000,0000,,Then we use this equation for the current.
Dialogue: 0,0:07:33.66,0:07:37.08,Default,,0000,0000,0000,,It looks very similar to\Nthe voltage equation for the capacitor.
Dialogue: 0,0:07:37.08,0:07:39.01,Default,,0000,0000,0000,,Plugging all of our things in,
Dialogue: 0,0:07:39.01,0:07:45.37,Default,,0000,0000,0000,,we come up with exactly the same\Ncurrent equation that we had last time.
Dialogue: 0,0:07:45.37,0:07:48.22,Default,,0000,0000,0000,,So here's the general solution procedure.
Dialogue: 0,0:07:48.22,0:07:51.85,Default,,0000,0000,0000,,You start at t equals zero minus and\Nanalyze the voltage and current,
Dialogue: 0,0:07:51.85,0:07:55.20,Default,,0000,0000,0000,,and then you convert\Nthose over to t equals zero plus.
Dialogue: 0,0:07:55.20,0:07:57.28,Default,,0000,0000,0000,,You then find the final\Nconditions and the time
Dialogue: 0,0:07:57.28,0:08:00.07,Default,,0000,0000,0000,,constant and plug them\Ninto these equations.
Dialogue: 0,0:08:00.07,0:08:02.80,Default,,0000,0000,0000,,If we want to find the voltage\Ninstead of the current,
Dialogue: 0,0:08:02.80,0:08:04.15,Default,,0000,0000,0000,,remember that the voltage is
Dialogue: 0,0:08:04.15,0:08:08.04,Default,,0000,0000,0000,,the inductance times\Nthe derivative of the current.
Dialogue: 0,0:08:08.26,0:08:12.77,Default,,0000,0000,0000,,Now let's consider a general's\NThevenin solution procedure.
Dialogue: 0,0:08:12.77,0:08:15.68,Default,,0000,0000,0000,,In this case, we break this circuit\Nwhere we are looking at
Dialogue: 0,0:08:15.68,0:08:18.100,Default,,0000,0000,0000,,the capacitance or inductance\Nand we look into the circuit,
Dialogue: 0,0:08:18.100,0:08:21.80,Default,,0000,0000,0000,,and then we go back to our equations I just
Dialogue: 0,0:08:21.80,0:08:24.54,Default,,0000,0000,0000,,gave you and replace Vs with V Thevenin,
Dialogue: 0,0:08:24.54,0:08:28.98,Default,,0000,0000,0000,,and replace R with R Thevenin\Nin all of the equations.
Dialogue: 0,0:08:29.17,0:08:32.34,Default,,0000,0000,0000,,So that's an evaluation of\Nthe question of the day.
Dialogue: 0,0:08:32.34,0:08:34.25,Default,,0000,0000,0000,,How can you calculate the\Nvoltage and current in
Dialogue: 0,0:08:34.25,0:08:36.95,Default,,0000,0000,0000,,a circuit that has inductors or capacitors?
Dialogue: 0,0:08:36.95,0:08:39.77,Default,,0000,0000,0000,,Go back to the two general\Nsolution procedures that I gave
Dialogue: 0,0:08:39.77,0:08:43.56,Default,,0000,0000,0000,,you and practice those\Nwith some of the examples.