1 00:00:08,530 --> 00:00:13,482 So, let's take a look at this circuit. We want to do a Thevenin equivalent 2 00:00:13,482 --> 00:00:21,615 method of solving for, in this particular case, I'm interested in this current. 3 00:00:21,615 --> 00:00:27,767 The current through that resistor. Now there are different ways of doing 4 00:00:27,767 --> 00:00:30,393 Thevenin Equivalent. There's this source transformation 5 00:00:30,393 --> 00:00:34,313 method. And then there's a method where we're 6 00:00:34,313 --> 00:00:41,350 finding the R thevenin or V thevenin and current directly. 7 00:00:41,350 --> 00:00:47,224 So that, of the two methods, I cannot use the method of solving for V thevenin, or 8 00:00:47,224 --> 00:00:51,772 R thevenin. The reason is I got a dependent source, 9 00:00:51,772 --> 00:00:57,430 this, this box here, this diamond box means that this is a dependent source. 10 00:00:57,430 --> 00:01:01,898 This current depends on this current right here, this i is that current. 11 00:01:01,898 --> 00:01:05,619 But this is, this is and this is also a current source as well, but it's a 12 00:01:05,619 --> 00:01:11,880 dependent current source. So I cannot find R thevenin indirectly. 13 00:01:11,880 --> 00:01:15,240 When we did R thevenin indirectly, we said well let's just 0 out all of our 14 00:01:15,240 --> 00:01:18,448 independent sources. So, I can 0 this one out, but I don't 15 00:01:18,448 --> 00:01:25,946 know what to do with that. So, the only way to do the thevenin. 16 00:01:25,946 --> 00:01:34,325 The only way to do the Thevenin Equivalent method is to use this, this 17 00:01:34,325 --> 00:01:47,762 formula where we find, we know that V thevenin Is equal R thevenin times Isc. 18 00:01:47,762 --> 00:01:59,753 And V thevenin is here, R thevenin is here, this is the equivalent circuit. 19 00:01:59,753 --> 00:02:04,080 And then Isc was the current going through it. 20 00:02:05,530 --> 00:02:12,345 We'll call that a and b, our terminals. So, the only way to find R thevenin, is 21 00:02:12,345 --> 00:02:19,160 to find i short circuit and V thevenin. And since I'm interested in this branch 22 00:02:19,160 --> 00:02:23,390 right here, I'm going to call this a and this b. 23 00:02:24,840 --> 00:02:28,560 And I should mention that we have the dependent source in here. 24 00:02:28,560 --> 00:02:32,844 And I wanted to expose you to dependent sources, because dependent sources often 25 00:02:32,844 --> 00:02:36,649 times come up in models of non-linear circuits. 26 00:02:36,649 --> 00:02:40,679 Like when non-linear circuits like transistor circuits are operating in 27 00:02:40,679 --> 00:02:45,050 their linear range, we oftentimes have dependent sources. 28 00:02:45,050 --> 00:02:47,810 So, it's a good idea to get familiar with how to handle them. 29 00:02:48,980 --> 00:02:53,159 So, going back to solve this, I have to solve for V thevenin and i sub sc. 30 00:02:53,159 --> 00:02:58,320 So if I look at i sub sc, sc stands for shirt, short-circuit. 31 00:02:58,320 --> 00:03:12,150 I short-circuit across the branch a b. Everything else I leave the same. 32 00:03:18,190 --> 00:03:21,862 This relationship between the dependent source and this eye through this branch 33 00:03:21,862 --> 00:03:31,510 stays the same. [SOUND] Voltage sto-, source stays the 34 00:03:31,510 --> 00:03:34,914 same. The only thing is, I short circuit that 35 00:03:34,914 --> 00:03:42,048 branch right there. Now, looking at this, so this is the 36 00:03:42,048 --> 00:03:48,078 circuit I now need to analyze. So, if I look at this current is going up 37 00:03:48,078 --> 00:03:52,720 in this direction into this node. This current is going into that node. 38 00:03:52,720 --> 00:03:56,410 And they're going to add together. So, that's i over 2 plus i is 3i over 2, 39 00:03:56,410 --> 00:04:02,239 has to leave that node. Okay, so that current plus this current 40 00:04:02,239 --> 00:04:08,930 into this node has to equal to the current leaving that node. 41 00:04:11,700 --> 00:04:18,219 I still don't know what i is though. And for doing that I'm going to do 42 00:04:18,219 --> 00:04:23,070 Kirchhoff's Voltage Law around this loop, right here. 43 00:04:23,070 --> 00:04:26,181 So, starting from this point, going in the left direction, the clockwise 44 00:04:26,181 --> 00:04:30,644 direction. I hit the, the plus sign a, this voltage 45 00:04:30,644 --> 00:04:36,560 source first, now I say, a plus 2. And then, I've got the current going 46 00:04:36,560 --> 00:04:39,030 through here times 5, is that voltage drop. 47 00:04:43,230 --> 00:04:49,926 And then this current is 3i over 2 times 2 is a voltage drop here and then that 48 00:04:49,926 --> 00:04:58,614 has to equal 0. If I solve for i, I would get i is equal 49 00:04:58,614 --> 00:05:08,304 to minus 1, 4th and if I is equal to minus 1, 4th I sub sc is 3 halves times 50 00:05:08,304 --> 00:05:14,495 that. 51 00:05:14,495 --> 00:05:17,512 [SOUND] 52 00:05:17,512 --> 00:05:28,780 Or minus 3, 8ths. So, that's this current right there. 53 00:05:30,740 --> 00:05:39,683 Let's look at the Thevenin Voltage. And I'm going to now redraw this circuit 54 00:05:39,683 --> 00:05:44,682 where I open circuit, this branch and I solve for the voltage drop there. 55 00:05:44,682 --> 00:06:22,640 Let me just redraw the circuit. I need to solve for this, this V 56 00:06:22,640 --> 00:06:26,498 thevenin. Well, if we look at this, looking again 57 00:06:26,498 --> 00:06:30,458 at these nodes, this node here, and looking at the current that goes into 58 00:06:30,458 --> 00:06:34,550 this, well, I have a current of i going into this, a value of i going into this 59 00:06:34,550 --> 00:06:41,566 node from this direction. And from this branch, I have a value of i 60 00:06:41,566 --> 00:06:45,858 over 2. So, I've got 3i over 2 going in this 61 00:06:45,858 --> 00:06:51,808 direction here. That's much the, what we found up here, 62 00:06:51,808 --> 00:06:56,510 3i over 2. But you see, there's a problem. 63 00:06:56,510 --> 00:06:58,620 The problem is that this is open-circuited. 64 00:06:58,620 --> 00:07:04,480 So, if this is open-circuited, that means there's 0 current flowing through here. 65 00:07:04,480 --> 00:07:06,450 That means that this current i has to equal 0. 66 00:07:06,450 --> 00:07:12,561 Well, if the current i equals 0, that means there's no voltage drop here, and 67 00:07:12,561 --> 00:07:21,258 there's no voltage drop here. So, if I look at the voltage around this 68 00:07:21,258 --> 00:07:25,162 loop right here that means that the V thevenin has to balance this voltage 69 00:07:25,162 --> 00:07:31,848 source right here. And since they're both really in the same 70 00:07:31,848 --> 00:07:36,002 direction, as I go around the voltage, the loop in this direction, and did a 71 00:07:36,002 --> 00:07:41,098 KVL, I would get a plus 2 and then a plus V thevenin. 72 00:07:41,098 --> 00:07:46,468 And that means that they're going to have to have negative signs of one another. 73 00:07:46,468 --> 00:07:58,435 So, V thevinin minus 2 volts. So, if I want to redraw the thevanin 74 00:07:58,435 --> 00:08:03,891 equivalent circuit from up here, I just redraw it with my values in there. 75 00:08:03,891 --> 00:08:17,967 Plus minus, minus 2 volts, R thevenin is found from taking V thevenin minus i sub 76 00:08:17,967 --> 00:08:28,448 sc. And I can actually write that here. 77 00:08:28,448 --> 00:08:40,958 And that is equal to 16 over 3 ohms. Two volts, 16 over 3 ohms, that's a, 78 00:08:40,958 --> 00:08:48,350 that's b. So, this is this part of the circuit. 79 00:08:48,350 --> 00:08:52,658 Let me show it in red right here. I've taken this part of the circuit to 80 00:08:52,658 --> 00:08:57,050 the left and simplified it to this right here. 81 00:08:57,050 --> 00:08:59,716 Then if I want to connect it up with the rest of the circuit which is this final 82 00:08:59,716 --> 00:09:05,509 branch. Which is 1 0, and now I'm interested in 83 00:09:05,509 --> 00:09:11,640 solving for I sub a. I sub 0, which is the current through 84 00:09:11,640 --> 00:09:14,925 that, that branch. So, then I just look at the, this is a 85 00:09:14,925 --> 00:09:18,984 simple circuit now. I've got 2 resistors in series and the 86 00:09:18,984 --> 00:09:27,190 voltage. So, I have i0 is equal to the voltage 87 00:09:27,190 --> 00:09:34,678 which is minus 2, divided by this resistance. 88 00:09:34,678 --> 00:09:42,229 Which is, 19, if I add these together I have 19 over 3. 89 00:09:47,650 --> 00:09:52,036 In other words, it's minus 6 over 19 amps. 90 00:09:52,036 --> 00:09:57,463 And that's that current through here. So, just to summarize this problem, I 91 00:09:57,463 --> 00:10:01,884 wanted to do Thevenin Equivalent. So, I have to identify what branch I want 92 00:10:01,884 --> 00:10:08,100 to isolate and keep that branch as is. And take the rest of the circuit, and 93 00:10:08,100 --> 00:10:12,680 simplify it to a Thevenin or Norton circuit. 94 00:10:12,680 --> 00:10:16,450 In this particular case, because I had a dependent source in there, I had to use 95 00:10:16,450 --> 00:10:21,846 this equation, here, to find R thevenin. So, I had to find i sub sc, which is 96 00:10:21,846 --> 00:10:28,290 finding the current if I short circuit across those terminals. 97 00:10:28,290 --> 00:10:31,755 And I had to find V sub thevenin, which is the voltage if i open circuit across 98 00:10:31,755 --> 00:10:36,453 those terminals. And then I plug those, and then I find R 99 00:10:36,453 --> 00:10:40,722 from those two values. If I have V and I, I find R, and I just 100 00:10:40,722 --> 00:10:47,311 redraw my thevenin equivalent circuit. And then add in that final branch, and 101 00:10:47,311 --> 00:10:51,180 then solve for whatever parameter I was interested in solving for.