1 00:00:03,360 --> 00:00:06,170 Welcome back to our continuing series in linear circuits. 2 00:00:06,170 --> 00:00:09,594 Again, I'm Nathan Parrish. And today we're talking about our first 3 00:00:09,594 --> 00:00:13,064 real circuit elements, resistors. In this class we're going to introduce 4 00:00:13,064 --> 00:00:15,040 resistors and how to analyze them in circuits. 5 00:00:17,230 --> 00:00:19,982 Putting it in context, we've already talked about a little bit of how within a 6 00:00:19,982 --> 00:00:23,830 device we see the voltage and the current relate to each other. 7 00:00:23,830 --> 00:00:26,854 Then we talk about Kirchhoff's laws, which allows us to see how devices work 8 00:00:26,854 --> 00:00:29,712 together. Now when we have real resistors, we can 9 00:00:29,712 --> 00:00:33,070 put these two ideas together and start doing some basic simple analysis. 10 00:00:34,290 --> 00:00:37,692 So, again, from our previous class, we covered Ohm's law and then we covered 11 00:00:37,692 --> 00:00:41,148 Kirchhoff's laws, which are these single device, as well as multiple device 12 00:00:41,148 --> 00:00:46,605 interacting properties. The objective to this lesson is to apply 13 00:00:46,605 --> 00:00:50,345 those two laws, Ohm's law and Kirchhoff's law to simple resistive circuits, then 14 00:00:50,345 --> 00:00:54,030 calculate an equivalent resistance of resistors when their placed in Parallel 15 00:00:54,030 --> 00:00:57,495 or series and then we are going to do an example, where we can see that, you can 16 00:00:57,495 --> 00:01:01,290 apply these laws, sucessfully to solve a complete solution and find an equivalent 17 00:01:01,290 --> 00:01:09,770 resistance. So to review, first of all, Ohm's law. 18 00:01:09,770 --> 00:01:12,812 They are going from the plus to the minus V equals i times R. 19 00:01:12,812 --> 00:01:16,469 If you flip your references, multiply by negative 1, Kirchhoff's Current Law, you 20 00:01:16,469 --> 00:01:20,020 sum all of the currents coming in to have to equal the currents going out, and then 21 00:01:20,020 --> 00:01:23,465 Kichhoff's Voltage Law, if you make any closed loop and sum up all the voltages 22 00:01:23,465 --> 00:01:29,140 along that closed loop, Then the sum has to be equal to zero. 23 00:01:29,140 --> 00:01:34,393 You end up back where you started. If I stick resistors in series, we can 24 00:01:34,393 --> 00:01:38,480 find that, our equivalent resistance by simply adding the 2 resistors. 25 00:01:38,480 --> 00:01:41,897 And to show why this happens, there's a derivation here that shows that, because 26 00:01:41,897 --> 00:01:46,008 they're in series. They have to have the same current, and 27 00:01:46,008 --> 00:01:49,870 using that, you can then factor out to discover that R is equal to R1 plus R2. 28 00:01:49,870 --> 00:01:55,600 This brings us to the idea of a voltage divider. 29 00:01:55,600 --> 00:01:59,188 What I've placed two resistors in series, in this configuration, and I've put some 30 00:01:59,188 --> 00:02:02,464 voltage across them, we'll see some voltage dropping in the first resistor 31 00:02:02,464 --> 00:02:06,450 and some voltage dropping in the second resistor. 32 00:02:06,450 --> 00:02:10,162 And to see exactly how that interaction works and to see what the voltage is on 33 00:02:10,162 --> 00:02:13,874 that center node in between them, we can use this analysis called the voltage 34 00:02:13,874 --> 00:02:18,829 divider. And essentially what it gives us is if i 35 00:02:18,829 --> 00:02:24,369 want to know the voltage v2 Which is across this resistor. 36 00:02:24,369 --> 00:02:26,100 I find the resistance of this resistor, r2, here. 37 00:02:26,100 --> 00:02:28,676 And then I take the sum of the 2 resistances, r1 plus r2. 38 00:02:28,676 --> 00:02:30,929 So, if r2, were, for example, 3 ohms. And r1 is 1 ohm. 39 00:02:30,929 --> 00:02:41,565 3/4 of the voltage drop would occur across r2. 40 00:02:41,565 --> 00:02:45,250 1/4 of the voltage drop would occur across r1. 41 00:02:45,250 --> 00:02:53,040 And similarly, you can calculate v1. But just using an r1 instead of an r2. 42 00:02:53,040 --> 00:02:54,790 When I place them in a parallel configuration. 43 00:02:54,790 --> 00:02:58,790 Remember that, now, the voltages have to be equal, due to Kirchoff's voltage law. 44 00:02:58,790 --> 00:03:02,600 Doing a little bit of analysis, we discover that the equivalent resistance. 45 00:03:02,600 --> 00:03:07,002 Of resistors placed in a parallel configuration, is the sum of the inverse 46 00:03:07,002 --> 00:03:12,665 of the resistances, and then you take that quantity and invert it. 47 00:03:12,665 --> 00:03:15,980 And when you're doing the analysis for this, it sometimes ends up being easier 48 00:03:15,980 --> 00:03:19,499 to analyze using the conductance rather than the resistance, and then change your 49 00:03:19,499 --> 00:03:24,108 answer when you get to the end. And so that's the way that the derivation 50 00:03:24,108 --> 00:03:26,956 is presented here. And you can take a closer look at that if 51 00:03:26,956 --> 00:03:31,200 you have questions on, on where this calculation came from. 52 00:03:32,940 --> 00:03:36,510 Like we had the voltage divider, there's also a current divider. 53 00:03:36,510 --> 00:03:40,538 This comes from Kirchoff's current law. Because the current coming into this set 54 00:03:40,538 --> 00:03:44,654 of resistors. Has to be split between all of the 55 00:03:44,654 --> 00:03:48,430 different paths or I is equal to i1 plus i2. 56 00:03:48,430 --> 00:03:53,734 When we use that, we can use the, the previous calculation of the currents that 57 00:03:53,734 --> 00:04:00,170 are going i1 and i2 from before. And plug them into to discover that i1 is 58 00:04:00,170 --> 00:04:06,170 going to be equal to r2. Divided by R1 plus R2 plus i. 59 00:04:06,170 --> 00:04:11,537 Be careful, because this is i1 and here we're using R2 in the numerator. 60 00:04:11,537 --> 00:04:12,587 This is i2, now we're using R1 in the numerator. 61 00:04:12,587 --> 00:04:23,500 If you have more than 2 elements, these equations change a little bit. 62 00:04:23,500 --> 00:04:26,338 Normally, we're only going to do two at a time, but if you're interested, you can 63 00:04:26,338 --> 00:04:29,262 do the same derivation and discover what these equations would be if you had three 64 00:04:29,262 --> 00:04:37,715 or more resistors. Finally, we're going to do an example to 65 00:04:37,715 --> 00:04:43,460 apply the things that we've learned. Here we have a 7-ohm resistor and a 5-ohm 66 00:04:43,460 --> 00:04:47,022 resistor. And we see that these are now in a series 67 00:04:47,022 --> 00:04:51,445 configuration right now. The six ohm resistor is then in parallel 68 00:04:51,445 --> 00:04:55,332 with that combination. And then this eight ohm resistor is in 69 00:04:55,332 --> 00:04:59,256 series with the whole thing. So to be able to solve this, what we're 70 00:04:59,256 --> 00:05:02,470 going to do is, we're going to do them each piece at a time. 71 00:05:02,470 --> 00:05:05,362 So first of all, this series is very easy to calculate. 72 00:05:05,362 --> 00:05:11,743 The equivalent resistance is going to be just 7 plus 5 is 12 ohms. 73 00:05:11,743 --> 00:05:18,800 Because, R is going to be equal to R1 plus R2. 74 00:05:18,800 --> 00:05:22,896 Now, that combination is in parallel with the 6 ohm resistor. 75 00:05:22,896 --> 00:05:31,340 So, to do that will do one over 6. Plus 1 over 12. 76 00:05:31,340 --> 00:05:36,244 Then we will invert that whole result. So that's 2 12ths plus 1 12th is 3 12ths, 77 00:05:36,244 --> 00:05:45,332 so 1 4th, and invert, giving us 4 Ohms. because in this case, R is equal to 1 78 00:05:45,332 --> 00:05:53,019 over R1 plus 1 over r2 inverse. That gives us the equal resistance of 79 00:05:53,019 --> 00:05:57,670 that whole set. Finally we have this 8 ohm resistor which 80 00:05:57,670 --> 00:06:01,830 is now in series with this combination, so the final result is going to be the 8 81 00:06:01,830 --> 00:06:09,830 ohm resistor plus our 4 ohm resistor. Giving 12 ohms. 82 00:06:09,830 --> 00:06:14,686 So what's the significance of this? Well if I have all of these resistors, I 83 00:06:14,686 --> 00:06:20,173 could replace the whole network with a single resistor that has a 12 ohm 84 00:06:20,173 --> 00:06:26,690 resistance. So this could be applied both ways. 85 00:06:26,690 --> 00:06:29,480 If you have a 12 Ohm resistor, you could replace these four resistors with a 86 00:06:29,480 --> 00:06:32,135 single resistor, or if you don't have the resistor you want, you can use 87 00:06:32,135 --> 00:06:36,950 combinations of resistors and parallel series to get the desired quantity. 88 00:06:41,240 --> 00:06:44,828 To summarize, we introduced resistors as circuit elements, and then we showed how 89 00:06:44,828 --> 00:06:48,260 they can be combined when they're in a series Or in a parallel configuration and 90 00:06:48,260 --> 00:06:51,796 we found those equivalent resistances by using these laws and we found that we can 91 00:06:51,796 --> 00:06:57,816 use them successively. applying them again and again until we 92 00:06:57,816 --> 00:07:01,388 get the desired result. In the next class there will actually be 93 00:07:01,388 --> 00:07:04,703 a demo showing that all of things that we've been talking about actually works 94 00:07:04,703 --> 00:07:08,069 in real life and we'll be able to see the way that the voltage is dropping it would 95 00:07:08,069 --> 00:07:13,233 cross through this resistance. And that will be demonstrated. 96 00:07:13,233 --> 00:07:16,382 If you have any questions on the stuff that was covered in this in this lesson, 97 00:07:16,382 --> 00:07:22,069 as always go to the forums. And until then, see you later.