1 00:00:06,170 --> 00:00:09,734 We're continuing on with the lesson on Systematic Solution Methods, so this is 2 00:00:09,734 --> 00:00:16,224 part two of the two part lesson. Just to remind you where we are in this 3 00:00:16,224 --> 00:00:18,732 module, we've covered all the basic material and now we're doing the 4 00:00:18,732 --> 00:00:24,456 Systematic Solution Methods. The lesson objective, is to demonstrate 5 00:00:24,456 --> 00:00:27,606 Thevenin equivalent and Norton equivalent circuits. 6 00:00:27,606 --> 00:00:31,116 And also source transformations, which are a way of applying the Thevenin and 7 00:00:31,116 --> 00:00:36,700 Norton equivalent circuits. Looking back at this overview of the 8 00:00:36,700 --> 00:00:40,396 different methods that we have, recall that mesh analysis, and node analysis, 9 00:00:40,396 --> 00:00:45,575 are primarily algebraic methods. Thevenin and Norton equivalents are 10 00:00:45,575 --> 00:00:50,758 primarily graphical methods where we find simple equivalent circuits. 11 00:00:50,758 --> 00:00:55,098 And we do source transformations to reduce our, our schematic to a simpler 12 00:00:55,098 --> 00:00:59,844 form. We've talked about when to apply mesh and 13 00:00:59,844 --> 00:01:03,555 node Analysis in the first part of this lesson. 14 00:01:03,555 --> 00:01:07,724 Now I wanted to mention when we would apply Thevenin and Norton equivalent. 15 00:01:07,724 --> 00:01:11,189 And that's when we don't care about intermediate values, and maybe only the 16 00:01:11,189 --> 00:01:17,950 output voltage or current is important. Let's look at the Thevenin equivalent 17 00:01:17,950 --> 00:01:21,940 fundamental concepts here. I've got a circuit, and this circuit 18 00:01:21,940 --> 00:01:26,615 could contain a lot of elements in it, it could be very messy. 19 00:01:26,615 --> 00:01:30,971 It has to contain linear elements, so just no diodes or transistors, but linear 20 00:01:30,971 --> 00:01:34,038 elements. It can be very messy, and I want to 21 00:01:34,038 --> 00:01:37,460 simplify it and say if I'm only interested in what's happening at these 22 00:01:37,460 --> 00:01:41,620 terminals, then I can reduce this entire circuit. 23 00:01:42,760 --> 00:01:47,740 To something much easier, an equivalent voltage source and resistance. 24 00:01:47,740 --> 00:01:53,520 And we call this Thevenin voltage, and the Thevenin resistance. 25 00:01:53,520 --> 00:01:57,426 And then notice the current going, flowing through this resistance is i sub 26 00:01:57,426 --> 00:02:00,686 sc. So then if I look at applying or putting 27 00:02:00,686 --> 00:02:04,512 a resistor or some other load at this terminal. 28 00:02:04,512 --> 00:02:08,607 The voltage and resistance, the voltage and current through this point right 29 00:02:08,607 --> 00:02:12,387 here, these points a and b, is going to be the same as when I went back to the 30 00:02:12,387 --> 00:02:22,700 original circuit. So, the process is based on this formula, 31 00:02:22,700 --> 00:02:29,191 right here. The Thevenin Equivalent voltage is equal 32 00:02:29,191 --> 00:02:33,370 to the Thevenin Equivalent resistance times Is of sc. 33 00:02:33,370 --> 00:02:37,304 So I've gotta solve for these quantities. Since I've got an equation, I really only 34 00:02:37,304 --> 00:02:46,889 have to solve for two of the three. [NOISE] So if I wanted to solve for the 35 00:02:46,889 --> 00:02:52,036 Thevenin voltage I open circuit across the terminals ab so these ab terminals. 36 00:02:52,036 --> 00:02:58,460 I open circuit across it, and then I find that open circuit voltage. 37 00:02:58,460 --> 00:03:04,160 The voltage v sub a minus v sub b. And that's my Thevenin voltage. 38 00:03:04,160 --> 00:03:08,000 Now to find the i sub sc, I take those same terminals, and short circuit across 39 00:03:08,000 --> 00:03:11,514 them, and I find the current through them. 40 00:03:11,514 --> 00:03:20,830 R Thevenin, I can find by zeroing out all my sources. 41 00:03:20,830 --> 00:03:25,480 I zero out all my independent sources, so voltage sources if I zero them out. 42 00:03:25,480 --> 00:03:28,488 Becomes zero, they're shorted out, there's no voltage, so that means a short 43 00:03:28,488 --> 00:03:31,442 circuit. Current sources, when there's no current, 44 00:03:31,442 --> 00:03:36,264 that's equivalent to an open circuit. I can only apply, I can only find 45 00:03:36,264 --> 00:03:40,296 Thevenin equivalent resistance in this method when there are no dependent 46 00:03:40,296 --> 00:03:46,005 sources present in my circuit. So I only have to use two of the three of 47 00:03:46,005 --> 00:03:51,025 these to find these variables, and then use the equation to find the third one. 48 00:03:51,025 --> 00:03:56,225 Let's look like an example of Thevenin's equivalent method to solve a circuit 49 00:03:56,225 --> 00:04:01,240 problem. In this particular case, I'm going to 50 00:04:01,240 --> 00:04:05,630 select part of the circuit to reduce down. 51 00:04:05,630 --> 00:04:08,920 Now let's suppose I'm interested in this output. 52 00:04:08,920 --> 00:04:13,373 I'm going to call them a and b, the two nodes that will determine, that will 53 00:04:13,373 --> 00:04:17,753 separate my circuit into the part that I want to reduce, and the part that I 54 00:04:17,753 --> 00:04:23,748 want to keep. So let's say everything to the left of 55 00:04:23,748 --> 00:04:28,190 this line, I want to reduce down. And maybe the reason that I don't want to 56 00:04:28,190 --> 00:04:31,270 reduce is, part of the circuit down is because maybe why, I want to try 57 00:04:31,270 --> 00:04:35,050 different values in here and see how it works. 58 00:04:35,050 --> 00:04:39,274 So, that way I can reduce down everything else in the circuit, and then just plug 59 00:04:39,274 --> 00:04:42,852 in these values. And it's easier to solve them 60 00:04:42,852 --> 00:04:47,910 individually, rather than having to re-solve the entire circuit each time. 61 00:04:49,430 --> 00:04:54,258 In this case, I don't have any dependent sources, so it's usually easier to solve, 62 00:04:54,258 --> 00:04:58,802 Thevenin's equivalent circuit by finding the R resistance, the Thevinin's 63 00:04:58,802 --> 00:05:08,330 resistance. By zeroing out all of the sources. 64 00:05:08,330 --> 00:05:13,331 I'm going to redraw this schematic, zeroing out the sources. 65 00:05:13,331 --> 00:05:18,479 And that means, that I'm going to short circuit my voltages, so along to the left 66 00:05:18,479 --> 00:05:30,850 there. I shorted out that voltage source. 67 00:05:30,850 --> 00:05:34,630 And in here I've got a current source, and if I want zero current, that means I 68 00:05:34,630 --> 00:05:44,914 open circuit it. Again, I short circuit across that 69 00:05:44,914 --> 00:05:49,270 voltage, and then I leave off that R sub 0, because that's the part of the circuit 70 00:05:49,270 --> 00:05:54,550 that I'm not worried about. I'm trying to reduce everything else. 71 00:05:54,550 --> 00:05:58,674 So now I've got a simple circuit. Simple schematic with resistors and 72 00:05:58,674 --> 00:06:03,556 series in a parallel. So, the resistors in parallel, I'm going 73 00:06:03,556 --> 00:06:12,531 to replace over here with. Resistors in parallel will be, 2 times 4 74 00:06:12,531 --> 00:06:21,006 or 8 over 2 plus 4 or 6. And that's going to be in series with a 75 00:06:21,006 --> 00:06:30,242 ten ohm resistor. And that is equivalent to this circuit 76 00:06:30,242 --> 00:06:48,538 here, which is 11 and 1 3rds. So R Thevenin is equal to 11 and 1 3rds 77 00:06:48,538 --> 00:06:56,737 Ohms. Now, I'm going to find the Thevenin 78 00:06:56,737 --> 00:07:05,430 voltage. And here I have a choice, I could have 79 00:07:05,430 --> 00:07:10,290 found the I sub SC or I could find the Thevenin voltage, either one is fine. 80 00:07:11,790 --> 00:07:16,000 So in this particular method to find V Thevenin. 81 00:07:16,000 --> 00:07:20,350 I open circuit across the terminals a and b, and find that voltage. 82 00:07:20,350 --> 00:07:26,519 So, let me redraw the schematic. And I leave the sources in. 83 00:07:46,070 --> 00:07:50,694 But I open circuit here. So this circuit is fairly easy to analyze 84 00:07:50,694 --> 00:07:54,334 because of this open circuit, all the current that flows through this leg here 85 00:07:54,334 --> 00:08:00,198 has to continue on and flow through here. So the current in here exactly matches 86 00:08:00,198 --> 00:08:04,422 the current there. So looking at Kirchoff's Current Law, 87 00:08:04,422 --> 00:08:08,062 that means that all the current goes, that flows into this leg is exact current 88 00:08:08,062 --> 00:08:12,730 that flows through that. So I've got this loop where I know the 89 00:08:12,730 --> 00:08:17,760 voltage and I know the resistance, so I can use Ohm's law to find this current. 90 00:08:17,760 --> 00:08:22,688 And this current, because I've got one volt and a total of six Ohms the current 91 00:08:22,688 --> 00:08:31,970 has got to be 1 over 6. And that would be in Amps. 92 00:08:31,970 --> 00:08:36,600 Now, I still need to find the voltage drop between a and b. 93 00:08:39,300 --> 00:08:45,340 And that'll be the Thevenin voltage. So, I know this current is 1 6th, and I 94 00:08:45,340 --> 00:08:51,550 know this current has got to be 0.2 going in that direction. 95 00:08:51,550 --> 00:08:55,556 So if I try a voltage, a Kirchoff's voltage law around this. 96 00:08:55,556 --> 00:09:02,693 Oh, I need to add in this, I have a, I have a voltage source here that I should 97 00:09:02,693 --> 00:09:08,790 not have gotten rid of. It's still there. 98 00:09:08,790 --> 00:09:12,190 So, let's go ahead and do a Kirchoff's voltage law around here. 99 00:09:12,190 --> 00:09:15,360 Let me start from this direction. I'm going in this direction. 100 00:09:15,360 --> 00:09:20,220 I hit the plus first, so I add a plus 2 and then going in this direction, I'm 101 00:09:20,220 --> 00:09:29,190 going against the current flow that I've defined, so that's minus 4 times 1 6th. 102 00:09:29,190 --> 00:09:34,131 And if I'm following along this way, again, I'm going against the current 103 00:09:34,131 --> 00:09:41,500 flow, so it's a minus 10 times 0.2 equals 0 plus, plus v sub s equals 0. 104 00:09:41,500 --> 00:09:53,310 Plus v sub Thevenin equals 0. If I solve for vThevenin I get 4 6th. 105 00:09:53,310 --> 00:09:59,736 So the final picture is, I'm going to erase this here, because I've run out of 106 00:09:59,736 --> 00:10:05,820 room, and I'll just show the final picture. 107 00:10:05,820 --> 00:10:10,870 And it's based on this Thevenin. Voltage in that Thevenin resistance. 108 00:10:13,070 --> 00:10:25,292 So this circuit, looking from the left this way, is equivalent to this circuit. 109 00:10:25,292 --> 00:10:40,900 And that's 2 3rds volts. 11 and 1 3rd ohm. 110 00:10:40,900 --> 00:10:43,990 Terminals a and b. And then I can then attach, if I want, 111 00:10:43,990 --> 00:10:47,069 whatever resistances, that's the load I want to. 112 00:10:57,180 --> 00:11:00,280 So in this particular case. We found the resistors. 113 00:11:00,280 --> 00:11:06,443 The, the R Thevenin and V Thevenin. I could have also found the i sub sc. 114 00:11:07,890 --> 00:11:11,876 And I applied Thevenins that way. There's a, there's actually another way 115 00:11:11,876 --> 00:11:22,310 to, to do this. There's actually another way to do this. 116 00:11:22,310 --> 00:11:25,605 And that's the use of the Norton equivalent circuit. 117 00:11:25,605 --> 00:11:30,451 I have that basic relationship, the V and R and i relationship there. 118 00:11:30,451 --> 00:11:34,675 And I can apply it either as a resistor with in series with the voltage source, 119 00:11:34,675 --> 00:11:38,950 or as resistor in series with a current source. 120 00:11:38,950 --> 00:11:46,026 And these two, these two. Circuits are equivalent. 121 00:11:46,026 --> 00:11:49,386 Thévenin, the Thévenin equivalent circuit is equivalent to Norton equivalent 122 00:11:49,386 --> 00:11:52,746 circuit. And what I mean by equivalent is if I 123 00:11:52,746 --> 00:11:57,702 look at, if I look at the voltage drop across here, and the current going 124 00:11:57,702 --> 00:12:02,775 through here. They're the same in the Norton equivalent 125 00:12:02,775 --> 00:12:06,640 circuit as they are in the Thévenin equivalent circuit. 126 00:12:06,640 --> 00:12:11,378 They both satisfy this relationship. So, the interesting thing is that I can 127 00:12:11,378 --> 00:12:16,470 substitute these two configurations in a schematic, at will. 128 00:12:20,780 --> 00:12:25,450 So, the Source Tranformation Method. Is to take these two configurations, and 129 00:12:25,450 --> 00:12:29,720 substitute them, interchange them in a circuit. 130 00:12:29,720 --> 00:12:33,807 Whenever I see the, the voltage in series with the resistor, I change it out with a 131 00:12:33,807 --> 00:12:39,240 current source in series with the resistor, whichever is to my benefit. 132 00:12:39,240 --> 00:12:43,940 And by doing that, I can do, I can come up with a graphical. 133 00:12:43,940 --> 00:12:48,738 Way of simplifying my schematic. Now this method differs from the Mesh 134 00:12:48,738 --> 00:12:52,508 node analysis, because that was some, those, those methods were algebraic in 135 00:12:52,508 --> 00:12:57,030 nature. This is one where it's just graphical. 136 00:12:57,030 --> 00:13:04,140 Let's show you an example of that. Looking at this circuit. 137 00:13:13,710 --> 00:13:19,380 I can see this two, this right here a voltage in series with a, with a resistor 138 00:13:19,380 --> 00:13:24,560 and a current. Source in series with the resistor. 139 00:13:24,560 --> 00:13:28,850 This right now is a Norton equivalent circuit and this is a Thevenin equivalent 140 00:13:28,850 --> 00:13:31,901 circuit. What I ultimately want to do is reduce 141 00:13:31,901 --> 00:13:35,510 this down to something everything in series. 142 00:13:35,510 --> 00:13:37,140 So I'd like to reduce to everything in series. 143 00:13:37,140 --> 00:13:40,840 So if I look at this, I can reduce this to something in series. 144 00:13:42,470 --> 00:13:47,480 So let's start with that one. Actually, let's do both of them together. 145 00:13:58,080 --> 00:14:04,940 I've got 1 volt and 2 ohms, the 2 ohms stays the same. 146 00:14:04,940 --> 00:14:10,150 And then the 1 volt is divided by the 2 ohms to give me current of one half. 147 00:14:13,250 --> 00:14:16,469 So I've replaced this Thevenin equivalent with a Norton equivalent. 148 00:14:20,320 --> 00:14:22,600 And that is in series with the 4 ohm resistor. 149 00:14:22,600 --> 00:14:25,980 The reason I did this is because I wanted to combine these resistors, and I know 150 00:14:25,980 --> 00:14:31,130 how to combine resistors in series, in parallel with one another. 151 00:14:31,130 --> 00:14:37,862 Now this top configuration, the Norton configuration, I'm going to replace with 152 00:14:37,862 --> 00:14:43,740 the voltage source in series with the resistor. 153 00:14:43,740 --> 00:14:48,824 And notice my polarity here. The head of the arrow, it's easy maybe to 154 00:14:48,824 --> 00:14:51,800 remember it this way, the head of the arrow is going to be the same direction 155 00:14:51,800 --> 00:14:57,170 and the same polarity as the plus sign. So looking back here, the head of the 156 00:14:57,170 --> 00:15:00,640 arrow, is in the same, polarity or direction as the plus sign. 157 00:15:00,640 --> 00:15:14,336 [BLANK_AUDIO]. And again, this is my a terminal, and my 158 00:15:14,336 --> 00:15:19,248 b node. Now, let me combine these two resistors 159 00:15:19,248 --> 00:15:28,394 in parallel with one another, and that would be equivalent to a resistance. 160 00:15:28,394 --> 00:15:34,408 Let me go ahead and draw that, that would be 8 over 6. 161 00:15:42,980 --> 00:15:46,884 And this one right here, I didn't fill in the value there, the value of this 162 00:15:46,884 --> 00:15:52,200 voltage source would be this current source times that resistance. 163 00:15:52,200 --> 00:16:02,183 So that would be 2 volts. Ok, now I can go back the Thevenin over 164 00:16:02,183 --> 00:16:12,854 here so I've swapped. Here from a, a Thevenin to a Norton, now 165 00:16:12,854 --> 00:16:16,530 I'm going back to a Thevenin. And the whole point is to reduce this 166 00:16:16,530 --> 00:16:19,666 down, step by step, so I get something that's just in series and it's easier to 167 00:16:19,666 --> 00:16:24,176 analyze. The Thevenin here, would be a half this 168 00:16:24,176 --> 00:16:29,612 current times that resistance, which would be 4 over 6, or 2 3rds. 169 00:16:32,840 --> 00:16:35,747 Again, notice that my polarity that I chose on that matches the arrow 170 00:16:35,747 --> 00:16:41,750 direction. The resistance is, recopied there. 171 00:16:54,540 --> 00:16:59,290 So, now I've got. Two resistances in series so, they add. 172 00:16:59,290 --> 00:17:03,700 And then I've got three voltage sources in series and they add. 173 00:17:03,700 --> 00:17:06,472 But notice that two of them are in the same direction, the same polarity, and 174 00:17:06,472 --> 00:17:10,549 one is the opposite. So, this one is going to subtract off. 175 00:17:20,710 --> 00:17:26,681 And this reduces to, I've got a, a plus 2 plus 2 3rds minus 2. 176 00:17:26,681 --> 00:17:35,381 So that ends up to be 2 3rds and then this is 10 plus 8 6ths is going to be 11 177 00:17:35,381 --> 00:17:43,310 and 1 3rd ohms. And that's my simple circuit, my, the 178 00:17:43,310 --> 00:17:48,360 Thevenin equivalent circuit. And it was done by source transformations 179 00:17:48,360 --> 00:17:53,020 where I interchanged the Thevenin and Norton Equivalent circuits. 180 00:17:53,020 --> 00:17:57,178 Those configurations, at will, to reduce down the circuit to something that is 181 00:17:57,178 --> 00:18:00,660 simpler to analyze, like a series circuit. 182 00:18:00,660 --> 00:18:02,800 This was, the final result was a series circuit. 183 00:18:04,080 --> 00:18:07,370 So, in summary, between these two lessons. 184 00:18:07,370 --> 00:18:10,566 We've covered mesh and node analysis. And it's systematic ways to find 185 00:18:10,566 --> 00:18:14,799 independent simultaneous equations. It, it's an algebraic method. 186 00:18:14,799 --> 00:18:18,033 Then, we've applied Thevenin and Norton methods to replace most of the circuit 187 00:18:18,033 --> 00:18:22,114 with a simple equivalent circuit. And we use source transformations to do 188 00:18:22,114 --> 00:18:26,600 this more or less graphically. As I've mentioned before. 189 00:18:26,600 --> 00:18:29,328 As far as this course is concerned, I don't care which of these methods that 190 00:18:29,328 --> 00:18:32,480 you use. They're all systematic methods, and some 191 00:18:32,480 --> 00:18:35,890 students just prefer one method versus another. 192 00:18:35,890 --> 00:18:39,322 There are extra worked problems given on each of these methods, but as you do the 193 00:18:39,322 --> 00:18:42,702 homework, and as you do the quizzes, all I ultimately care about is that you're 194 00:18:42,702 --> 00:18:50,058 able to solve the problems. So, our next lesson, we will cover 195 00:18:50,058 --> 00:18:53,810 maximum power transfer, and I do want to mention that we are going to use Thevenin 196 00:18:53,810 --> 00:18:59,090 equivalent circuits to find the load to maximize the power delivered.