[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.96,0:00:02.44,Default,,0000,0000,0000,,Hi.\NMy name is Lee Brinton.
Dialogue: 0,0:00:02.44,0:00:05.55,Default,,0000,0000,0000,,I'm an electrical engineering instructor\Nat Salt Lake Community College.
Dialogue: 0,0:00:05.55,0:00:07.67,Default,,0000,0000,0000,,We're going to continue our\Nintroduction to electrical and
Dialogue: 0,0:00:07.67,0:00:11.36,Default,,0000,0000,0000,,computer engineering by considering\Nnow the principle of superposition.
Dialogue: 0,0:00:13.04,0:00:19.97,Default,,0000,0000,0000,,Depending on where you are in your\Nmath sequence, you may have already
Dialogue: 0,0:00:19.97,0:00:25.12,Default,,0000,0000,0000,,seen the principle of superposition as it\Napplies to linear mathematical operators.
Dialogue: 0,0:00:25.12,0:00:30.11,Default,,0000,0000,0000,,For example,\Nis L is a linear operator, and by that
Dialogue: 0,0:00:30.11,0:00:35.07,Default,,0000,0000,0000,,we mean it's something that performs\Noperations that are linear in nature.
Dialogue: 0,0:00:35.07,0:00:36.67,Default,,0000,0000,0000,,I know that sounds a little bit circular,\Nbut
Dialogue: 0,0:00:36.67,0:00:39.38,Default,,0000,0000,0000,,let's just follow me through on this and\Nyou'll see what I mean.
Dialogue: 0,0:00:39.38,0:00:45.28,Default,,0000,0000,0000,,If it's a linear operator,\Nthat means that if you operate on
Dialogue: 0,0:00:45.28,0:00:50.24,Default,,0000,0000,0000,,two Different quantities, the sum of two\Ndifferent quantities you'll get the same
Dialogue: 0,0:00:50.24,0:00:55.22,Default,,0000,0000,0000,,result as you would if you operated\Non each of them individually
Dialogue: 0,0:00:56.64,0:01:00.75,Default,,0000,0000,0000,,and then added the individual\Nresults together.
Dialogue: 0,0:01:01.84,0:01:05.96,Default,,0000,0000,0000,,For example,\NDifferentiation is a linear operator.
Dialogue: 0,0:01:05.96,0:01:11.08,Default,,0000,0000,0000,,We know that if you take\Nthe derivative with respect to time
Dialogue: 0,0:01:11.08,0:01:16.21,Default,,0000,0000,0000,,of two different functions\Nu(t)+v(t) that's equal
Dialogue: 0,0:01:16.21,0:01:21.14,Default,,0000,0000,0000,,to the derivative with respect\Nto time of the first one.
Dialogue: 0,0:01:24.35,0:01:29.52,Default,,0000,0000,0000,,Plus the derivative with respect\Nto time of the second one.
Dialogue: 0,0:01:31.13,0:01:33.62,Default,,0000,0000,0000,,A little more concrete example.
Dialogue: 0,0:01:33.62,0:01:37.72,Default,,0000,0000,0000,,Let's say that u(t) equals, say,
Dialogue: 0,0:01:37.72,0:01:41.77,Default,,0000,0000,0000,,e to the a t, and\Nv(t) equals the sine of omega t.
Dialogue: 0,0:01:43.00,0:01:46.15,Default,,0000,0000,0000,,And we want to take the derivative\Nof the sum of those two things.
Dialogue: 0,0:01:46.15,0:01:53.93,Default,,0000,0000,0000,,In other words, the derivative of respect\Nto t of e to the at plus sine omega t.
Dialogue: 0,0:01:55.43,0:02:01.12,Default,,0000,0000,0000,,Because the derivative is\Na linear operator, we know that.
Dialogue: 0,0:02:02.98,0:02:10.21,Default,,0000,0000,0000,,That is simply equal to the derivative\Nwith respect to t of e to the at.
Dialogue: 0,0:02:10.21,0:02:14.85,Default,,0000,0000,0000,,plus the derivative with respect\Nto t of the sin of omega t.
Dialogue: 0,0:02:16.34,0:02:24.86,Default,,0000,0000,0000,,Which is equal to ae to the at\Nplus omega Cosine of omega t.
Dialogue: 0,0:02:26.99,0:02:31.56,Default,,0000,0000,0000,,Thus the principle of linearity or\Nthe principle of superposition, rather,
Dialogue: 0,0:02:31.56,0:02:37.83,Default,,0000,0000,0000,,says that if we want to take\Nthe derivative of two things,
Dialogue: 0,0:02:37.83,0:02:43.24,Default,,0000,0000,0000,,the sum of two things, we can take\Nthe derivative of one thing And
Dialogue: 0,0:02:43.24,0:02:46.46,Default,,0000,0000,0000,,add to it the derivative\Nof the other thing.
Dialogue: 0,0:02:48.28,0:02:52.23,Default,,0000,0000,0000,,When it comes to circuits and\Ncircuit analysis in linear circuits,
Dialogue: 0,0:02:52.23,0:02:55.38,Default,,0000,0000,0000,,there is a similar\Nprinciple of superposition.
Dialogue: 0,0:02:56.91,0:03:00.19,Default,,0000,0000,0000,,In a circuit containing more\Nthan one independent source,
Dialogue: 0,0:03:01.42,0:03:04.56,Default,,0000,0000,0000,,More than one independent\Ninput to the circuit.
Dialogue: 0,0:03:06.10,0:03:11.03,Default,,0000,0000,0000,,The total response to the circuit can\Nbe found by combining the responses
Dialogue: 0,0:03:11.03,0:03:15.10,Default,,0000,0000,0000,,to the circuit to each\Nindependent source separately.
Dialogue: 0,0:03:16.55,0:03:17.81,Default,,0000,0000,0000,,What does that mean?
Dialogue: 0,0:03:19.63,0:03:22.30,Default,,0000,0000,0000,,Here we have a linear circuit.
Dialogue: 0,0:03:22.30,0:03:24.11,Default,,0000,0000,0000,,Consisting of two sources,
Dialogue: 0,0:03:24.11,0:03:28.84,Default,,0000,0000,0000,,and it's important to understand\Nit's independent sources.
Dialogue: 0,0:03:28.84,0:03:32.55,Default,,0000,0000,0000,,Two independent sources, here we\Nhave a current source of 6 amps and
Dialogue: 0,0:03:32.55,0:03:35.55,Default,,0000,0000,0000,,over here we have a voltage\Nsource of 45 volts.
Dialogue: 0,0:03:36.87,0:03:42.38,Default,,0000,0000,0000,,Superposition says that we can determine\Nsome quantity within the circuit,
Dialogue: 0,0:03:42.38,0:03:45.84,Default,,0000,0000,0000,,say for example, the current flowing\Nthrough this ten ohm resister.
Dialogue: 0,0:03:46.93,0:03:49.99,Default,,0000,0000,0000,,We can determine the total\Ncurrent resulting from
Dialogue: 0,0:03:49.99,0:03:53.97,Default,,0000,0000,0000,,both of these sources combined,
Dialogue: 0,0:03:53.97,0:03:59.37,Default,,0000,0000,0000,,by determining the current through this\Nresistor due to the current source alone.
Dialogue: 0,0:04:00.42,0:04:04.96,Default,,0000,0000,0000,,And the current through this due\Nto the voltage source alone.
Dialogue: 0,0:04:07.07,0:04:12.84,Default,,0000,0000,0000,,How do we then separate, or how we do\Nturn off one source and then the other?
Dialogue: 0,0:04:12.84,0:04:18.84,Default,,0000,0000,0000,,Well, sometimes this turning off is\Nreferred to as deactivating the source,
Dialogue: 0,0:04:18.84,0:04:20.88,Default,,0000,0000,0000,,or turning the source to zero.
Dialogue: 0,0:04:22.24,0:04:26.43,Default,,0000,0000,0000,,In this first circuit here,\Nwe have just the current source.
Dialogue: 0,0:04:26.43,0:04:29.58,Default,,0000,0000,0000,,The voltage source has been turned to 0.
Dialogue: 0,0:04:29.58,0:04:33.83,Default,,0000,0000,0000,,Which means that we've got\Na voltage source here of 0 volts.
Dialogue: 0,0:04:33.83,0:04:38.73,Default,,0000,0000,0000,,So from this side of the circuit,\Nthis side of the source
Dialogue: 0,0:04:38.73,0:04:42.85,Default,,0000,0000,0000,,to this side of the source\NWe increase zero volts.
Dialogue: 0,0:04:42.85,0:04:45.11,Default,,0000,0000,0000,,In other words, the voltage here\Nis the same as the voltage there.
Dialogue: 0,0:04:46.29,0:04:50.03,Default,,0000,0000,0000,,Well that is as though we had\Ntaken that voltage source and
Dialogue: 0,0:04:50.03,0:04:55.39,Default,,0000,0000,0000,,replaced it with a wire or\Na short circuit.
Dialogue: 0,0:04:55.39,0:05:00.22,Default,,0000,0000,0000,,To deactivate a voltage source\Nwe replace it simply with.
Dialogue: 0,0:05:00.22,0:05:02.12,Default,,0000,0000,0000,,Of straight wire or short circuit.
Dialogue: 0,0:05:04.23,0:05:05.05,Default,,0000,0000,0000,,On the other hand,
Dialogue: 0,0:05:05.05,0:05:11.05,Default,,0000,0000,0000,,when it comes to analyzing the circuit\Nwith just the voltage source in play,
Dialogue: 0,0:05:11.05,0:05:13.81,Default,,0000,0000,0000,,we need to deactivate the current\Nsource and return it to zero.
Dialogue: 0,0:05:14.81,0:05:16.31,Default,,0000,0000,0000,,If we turn the current source to zero,
Dialogue: 0,0:05:16.31,0:05:20.64,Default,,0000,0000,0000,,that's saying that there are zero\Namps flowing through this branch.
Dialogue: 0,0:05:20.64,0:05:24.51,Default,,0000,0000,0000,,It's as though this branch were\Nunplugged or open circuited.
Dialogue: 0,0:05:25.54,0:05:34.02,Default,,0000,0000,0000,,And we replace that current\Nsource then with an open circuit.
Dialogue: 0,0:05:35.45,0:05:40.52,Default,,0000,0000,0000,,LEt's go ahead now and calculate i the\Ntotal i due to both of these two sources
Dialogue: 0,0:05:40.52,0:05:44.48,Default,,0000,0000,0000,,by calculating the current i\None due to the current source.
Dialogue: 0,0:05:44.48,0:05:49.39,Default,,0000,0000,0000,,And the current I2 did the voltage\Nsource and then combine those two.
Dialogue: 0,0:05:49.39,0:05:56.95,Default,,0000,0000,0000,,All right, first of all, I1 can be\Ncalculated using a current divider,
Dialogue: 0,0:05:56.95,0:06:00.97,Default,,0000,0000,0000,,these two resistances are in series and\Nwe have a total current I coming in.
Dialogue: 0,0:06:00.97,0:06:09.69,Default,,0000,0000,0000,,I1 then is going to equal 6 amps times\NThe five volts divided by five plus ten.
Dialogue: 0,0:06:09.69,0:06:13.51,Default,,0000,0000,0000,,Well, five over 15 is two thirds\Ntherefore two thirds of six,
Dialogue: 0,0:06:13.51,0:06:16.07,Default,,0000,0000,0000,,I'm sorry five over 15 is one third.
Dialogue: 0,0:06:16.07,0:06:18.90,Default,,0000,0000,0000,,One third of six is two amps.
Dialogue: 0,0:06:20.59,0:06:27.62,Default,,0000,0000,0000,,So the component of I due to\Nindependent current source Is two amps.
Dialogue: 0,0:06:28.66,0:06:35.35,Default,,0000,0000,0000,,Now lets determine the current over here,\Ndue to the voltage source.
Dialogue: 0,0:06:35.35,0:06:41.37,Default,,0000,0000,0000,,Well with the current source open\Ncircuited, their is no current coming this
Dialogue: 0,0:06:41.37,0:06:45.67,Default,,0000,0000,0000,,way and so any current that is flowing\Nthrough the ten ohm resistor will also be
Dialogue: 0,0:06:45.67,0:06:49.14,Default,,0000,0000,0000,,flowing through the five ohm resistor,\Nand those two resistors are in series.
Dialogue: 0,0:06:50.28,0:06:55.49,Default,,0000,0000,0000,,Thus we can say then that I2 is equal to,
Dialogue: 0,0:06:55.49,0:06:57.51,Default,,0000,0000,0000,,now you'll notice that\NI2 is referenced left or
Dialogue: 0,0:06:57.51,0:07:01.47,Default,,0000,0000,0000,,right but this voltage source we'll be\Nputting the current going right to left,
Dialogue: 0,0:07:01.47,0:07:06.66,Default,,0000,0000,0000,,so I2, there's gonna be a minus sign here,\NI2 is equal
Dialogue: 0,0:07:06.66,0:07:12.95,Default,,0000,0000,0000,,to negative 45 volts divided by the series\Ncombination of those two, five plus Ten
Dialogue: 0,0:07:14.12,0:07:19.44,Default,,0000,0000,0000,,which gives us 45 divided by ten is three\Namps with a minus sign in front of it.
Dialogue: 0,0:07:19.44,0:07:24.33,Default,,0000,0000,0000,,So I2 then is equal to\Nnegative three amps.
Dialogue: 0,0:07:26.33,0:07:31.53,Default,,0000,0000,0000,,Now we can determine the total I,\Nthe total I is just equal to I1,
Dialogue: 0,0:07:31.53,0:07:33.40,Default,,0000,0000,0000,,the current due to the current source.
Dialogue: 0,0:07:33.40,0:07:40.25,Default,,0000,0000,0000,,Plus I2 the current\Ndigital voltage source,
Dialogue: 0,0:07:40.25,0:07:46.13,Default,,0000,0000,0000,,and that is equal to 2 amps minus 3 amps
Dialogue: 0,0:07:46.13,0:07:52.30,Default,,0000,0000,0000,,equals -1 amp.