WEBVTT 00:00:00.960 --> 00:00:02.440 Hi. My name is Lee Brinton. 00:00:02.440 --> 00:00:05.550 I'm an electrical engineering instructor at Salt Lake Community College. 00:00:05.550 --> 00:00:07.670 We're going to continue our introduction to electrical and 00:00:07.670 --> 00:00:11.360 computer engineering by considering now the principle of superposition. 00:00:13.040 --> 00:00:19.970 Depending on where you are in your math sequence, you may have already 00:00:19.970 --> 00:00:25.120 seen the principle of superposition as it applies to linear mathematical operators. 00:00:25.120 --> 00:00:30.110 For example, is L is a linear operator, and by that 00:00:30.110 --> 00:00:35.070 we mean it's something that performs operations that are linear in nature. 00:00:35.070 --> 00:00:36.670 I know that sounds a little bit circular, but 00:00:36.670 --> 00:00:39.380 let's just follow me through on this and you'll see what I mean. 00:00:39.380 --> 00:00:45.280 If it's a linear operator, that means that if you operate on 00:00:45.280 --> 00:00:50.240 two Different quantities, the sum of two different quantities you'll get the same 00:00:50.240 --> 00:00:55.220 result as you would if you operated on each of them individually 00:00:56.640 --> 00:01:00.750 and then added the individual results together. 00:01:01.840 --> 00:01:05.960 For example, Differentiation is a linear operator. 00:01:05.960 --> 00:01:11.083 We know that if you take the derivative with respect to time 00:01:11.083 --> 00:01:16.210 of two different functions u(t)+v(t) that's equal 00:01:16.210 --> 00:01:21.137 to the derivative with respect to time of the first one. 00:01:24.350 --> 00:01:29.520 Plus the derivative with respect to time of the second one. 00:01:31.130 --> 00:01:33.620 A little more concrete example. 00:01:33.620 --> 00:01:37.720 Let's say that u(t) equals, say, 00:01:37.720 --> 00:01:41.770 e to the a t, and v(t) equals the sine of omega t. 00:01:43.000 --> 00:01:46.150 And we want to take the derivative of the sum of those two things. 00:01:46.150 --> 00:01:53.930 In other words, the derivative of respect to t of e to the at plus sine omega t. 00:01:55.430 --> 00:02:01.120 Because the derivative is a linear operator, we know that. 00:02:02.980 --> 00:02:10.210 That is simply equal to the derivative with respect to t of e to the at. 00:02:10.210 --> 00:02:14.850 plus the derivative with respect to t of the sin of omega t. 00:02:16.340 --> 00:02:24.860 Which is equal to ae to the at plus omega Cosine of omega t. 00:02:26.990 --> 00:02:31.560 Thus the principle of linearity or the principle of superposition, rather, 00:02:31.560 --> 00:02:37.830 says that if we want to take the derivative of two things, 00:02:37.830 --> 00:02:43.240 the sum of two things, we can take the derivative of one thing And 00:02:43.240 --> 00:02:46.460 add to it the derivative of the other thing. 00:02:48.280 --> 00:02:52.230 When it comes to circuits and circuit analysis in linear circuits, 00:02:52.230 --> 00:02:55.380 there is a similar principle of superposition. 00:02:56.910 --> 00:03:00.190 In a circuit containing more than one independent source, 00:03:01.420 --> 00:03:04.560 More than one independent input to the circuit. 00:03:06.100 --> 00:03:11.030 The total response to the circuit can be found by combining the responses 00:03:11.030 --> 00:03:15.100 to the circuit to each independent source separately. 00:03:16.550 --> 00:03:17.810 What does that mean? 00:03:19.630 --> 00:03:22.300 Here we have a linear circuit. 00:03:22.300 --> 00:03:24.110 Consisting of two sources, 00:03:24.110 --> 00:03:28.840 and it's important to understand it's independent sources. 00:03:28.840 --> 00:03:32.550 Two independent sources, here we have a current source of 6 amps and 00:03:32.550 --> 00:03:35.550 over here we have a voltage source of 45 volts. 00:03:36.870 --> 00:03:42.380 Superposition says that we can determine some quantity within the circuit, 00:03:42.380 --> 00:03:45.839 say for example, the current flowing through this ten ohm resister. 00:03:46.930 --> 00:03:49.990 We can determine the total current resulting from 00:03:49.990 --> 00:03:53.970 both of these sources combined, 00:03:53.970 --> 00:03:59.370 by determining the current through this resistor due to the current source alone. 00:04:00.420 --> 00:04:04.960 And the current through this due to the voltage source alone. 00:04:07.070 --> 00:04:12.840 How do we then separate, or how we do turn off one source and then the other? 00:04:12.840 --> 00:04:18.839 Well, sometimes this turning off is referred to as deactivating the source, 00:04:18.839 --> 00:04:20.880 or turning the source to zero. 00:04:22.240 --> 00:04:26.430 In this first circuit here, we have just the current source. 00:04:26.430 --> 00:04:29.580 The voltage source has been turned to 0. 00:04:29.580 --> 00:04:33.830 Which means that we've got a voltage source here of 0 volts. 00:04:33.830 --> 00:04:38.730 So from this side of the circuit, this side of the source 00:04:38.730 --> 00:04:42.850 to this side of the source We increase zero volts. 00:04:42.850 --> 00:04:45.110 In other words, the voltage here is the same as the voltage there. 00:04:46.290 --> 00:04:50.030 Well that is as though we had taken that voltage source and 00:04:50.030 --> 00:04:55.390 replaced it with a wire or a short circuit. 00:04:55.390 --> 00:05:00.220 To deactivate a voltage source we replace it simply with. 00:05:00.220 --> 00:05:02.120 Of straight wire or short circuit. 00:05:04.230 --> 00:05:05.050 On the other hand, 00:05:05.050 --> 00:05:11.050 when it comes to analyzing the circuit with just the voltage source in play, 00:05:11.050 --> 00:05:13.810 we need to deactivate the current source and return it to zero. 00:05:14.810 --> 00:05:16.310 If we turn the current source to zero, 00:05:16.310 --> 00:05:20.640 that's saying that there are zero amps flowing through this branch. 00:05:20.640 --> 00:05:24.510 It's as though this branch were unplugged or open circuited. 00:05:25.540 --> 00:05:34.020 And we replace that current source then with an open circuit. 00:05:35.450 --> 00:05:40.520 LEt's go ahead now and calculate i the total i due to both of these two sources 00:05:40.520 --> 00:05:44.480 by calculating the current i one due to the current source. 00:05:44.480 --> 00:05:49.390 And the current I2 did the voltage source and then combine those two. 00:05:49.390 --> 00:05:56.950 All right, first of all, I1 can be calculated using a current divider, 00:05:56.950 --> 00:06:00.970 these two resistances are in series and we have a total current I coming in. 00:06:00.970 --> 00:06:09.690 I1 then is going to equal 6 amps times The five volts divided by five plus ten. 00:06:09.690 --> 00:06:13.510 Well, five over 15 is two thirds therefore two thirds of six, 00:06:13.510 --> 00:06:16.070 I'm sorry five over 15 is one third. 00:06:16.070 --> 00:06:18.900 One third of six is two amps. 00:06:20.590 --> 00:06:27.620 So the component of I due to independent current source Is two amps. 00:06:28.660 --> 00:06:35.350 Now lets determine the current over here, due to the voltage source. 00:06:35.350 --> 00:06:41.370 Well with the current source open circuited, their is no current coming this 00:06:41.370 --> 00:06:45.670 way and so any current that is flowing through the ten ohm resistor will also be 00:06:45.670 --> 00:06:49.139 flowing through the five ohm resistor, and those two resistors are in series. 00:06:50.280 --> 00:06:55.490 Thus we can say then that I2 is equal to, 00:06:55.490 --> 00:06:57.510 now you'll notice that I2 is referenced left or 00:06:57.510 --> 00:07:01.470 right but this voltage source we'll be putting the current going right to left, 00:07:01.470 --> 00:07:06.660 so I2, there's gonna be a minus sign here, I2 is equal 00:07:06.660 --> 00:07:12.950 to negative 45 volts divided by the series combination of those two, five plus Ten 00:07:14.120 --> 00:07:19.440 which gives us 45 divided by ten is three amps with a minus sign in front of it. 00:07:19.440 --> 00:07:24.330 So I2 then is equal to negative three amps. 00:07:26.330 --> 00:07:31.530 Now we can determine the total I, the total I is just equal to I1, 00:07:31.530 --> 00:07:33.398 the current due to the current source. 00:07:33.398 --> 00:07:40.250 Plus I2 the current digital voltage source, 00:07:40.250 --> 00:07:46.128 and that is equal to 2 amps minus 3 amps 00:07:46.128 --> 00:07:52.299 equals -1 amp.