1
00:00:02,580 --> 00:00:05,610
Welcome back to our class on linear 
circuits where today we're going to be 

2
00:00:05,610 --> 00:00:10,540
talking about RLC circuits or circuits 
with resistors, inductors, and capacitors 

3
00:00:10,540 --> 00:00:13,400
altogether. 
We will be using differential equations, 

4
00:00:13,400 --> 00:00:16,610
the second order differential equations 
that you were presented in the last 

5
00:00:16,610 --> 00:00:21,940
lesson to analyse these circuits. 
So as I mentioned in the previous lesson, 

6
00:00:21,940 --> 00:00:24,760
Calvert is solving second-order 
differential equations. 

7
00:00:24,760 --> 00:00:28,589
So in this lesson we'll see how we can 
generate second-order differential 

8
00:00:28,589 --> 00:00:33,210
equations from our our system, and then 
use the techniques from the last lesson 

9
00:00:33,210 --> 00:00:37,070
to solve them. 
And finishing up our RLC Circuits will be 

10
00:00:37,070 --> 00:00:42,635
the concluding portion of this module. 
The lesson objectives are to generate the 

11
00:00:42,635 --> 00:00:45,470
second-order differential equation from 
the circuit. 

12
00:00:45,470 --> 00:00:48,730
Identify the initial and final conditions 
of the circuit. 

13
00:00:48,730 --> 00:00:52,056
Solve the differential equation. 
And then recognize that the system's 

14
00:00:52,056 --> 00:00:56,680
underdamped or overdamped. 
So this is our system. 

15
00:00:56,680 --> 00:01:06,380
We see that we have a resistor. 
Of 20 killiohms, an inductor of 3.3 

16
00:01:06,380 --> 00:01:07,260
millihenrys at a capacitor of 1 100th of 
a microferret/g. 

17
00:01:07,260 --> 00:01:11,780
We want to find vc, the voltage across 
the capacitor. 

18
00:01:11,780 --> 00:01:15,780
Looking at our circuit, we see that the 
switch closes at time t equal 0, but 

19
00:01:15,780 --> 00:01:21,710
before that time, i has to be 0 because 
there's no path for the current to flow. 

20
00:01:21,710 --> 00:01:24,320
We also know that the current cannot 
instantaneously change through an 

21
00:01:24,320 --> 00:01:30,880
inductor, so that means that the current 
at time zero just before the switched is 

22
00:01:30,880 --> 00:01:38,050
closed, which is equal to zero, is also 
equal to the current at time zero just 

23
00:01:38,050 --> 00:01:41,320
after the switch is closed. 
We're going to use this minus here to 

24
00:01:41,320 --> 00:01:47,730
indicate just before time 0 and this just 
after time 0. 

25
00:01:47,730 --> 00:01:51,850
We cannot say just by analyzing the 
circuit or just by inspection of the 

26
00:01:51,850 --> 00:01:56,990
circuit what vc is at time t equals 0, so 
we're going to explicitly state that 

27
00:01:56,990 --> 00:02:01,940
before time t equals 0, before the switch 
closes, that the voltage is 0. 

28
00:02:01,940 --> 00:02:04,590
And because we know that the voltage 
cannot instantaneously change through a 

29
00:02:04,590 --> 00:02:11,120
capacitor, vc at 0 plus is also going to 
be equal to 0. 

30
00:02:11,120 --> 00:02:15,020
We also know that the final condition of 
our system is going to be where this 

31
00:02:15,020 --> 00:02:18,610
inductor starts to look like a wire, and 
this capacitor starts to look like an 

32
00:02:18,610 --> 00:02:21,530
open circuit. 
And if you do the analysis, you'll 

33
00:02:21,530 --> 00:02:26,260
discover that Vc after a long time is 
going to look like 5 volts, the same as 

34
00:02:26,260 --> 00:02:31,680
the input here. 
Now let's see how to get the differential 

35
00:02:31,680 --> 00:02:34,440
equations from the system. 
So the first thing that we're going to 

36
00:02:34,440 --> 00:02:40,010
make use of is Kirchhoff's voltage law, 
which states that vs It's going to be 

37
00:02:40,010 --> 00:02:46,720
equal to the voltage across r plus the 
voltage across the inductor, which will 

38
00:02:46,720 --> 00:02:49,330
call vl, plus vc. 
We also have some expressions for what 

39
00:02:49,330 --> 00:02:52,417
these happen to be, So based on i we know 
vr is equal to i times r. 

40
00:02:52,417 --> 00:03:05,080
Vl is equal to l di dt And then we also 
know that i is equal to cdvcdt. 

41
00:03:05,080 --> 00:03:10,662
So what we want to do is come up with a, 
an equation in terms of these vc's. 

42
00:03:10,662 --> 00:03:15,590
That's a differential equation. 
So, the way we will do that is here in 

43
00:03:15,590 --> 00:03:21,440
the L, we will replace the i's. 
this i here with cdvcdt. 

44
00:03:21,440 --> 00:03:25,440
Which will give us L times C times the 
second derivative of v c with respect to 

45
00:03:25,440 --> 00:03:29,760
d t. 
And here this will be r c d v d t. 

46
00:03:29,760 --> 00:03:33,680
Summing all of those together, we get 
this equation. 

47
00:03:37,130 --> 00:03:41,120
But this does not exactly match up the 
equation that was presented in the 

48
00:03:41,120 --> 00:03:43,910
previous lesson on solving second order 
of differential equations. 

49
00:03:43,910 --> 00:03:47,880
But to get it in that format. 
Which is like this. 

50
00:03:47,880 --> 00:03:52,502
We will divide everything by lc. 
So now we see that a1is r over l. 

51
00:03:52,502 --> 00:04:05,720
A2 Is 1 over lc and k is vs over lc. 
And so now we can take this differential 

52
00:04:05,720 --> 00:04:09,980
equation that we've generated and use the 
techniques from the last lesson to solve 

53
00:04:09,980 --> 00:04:14,430
the system. 
So to do this and solve the system, we'll 

54
00:04:14,430 --> 00:04:19,260
start by looking at the differential 
equation here, where we can then plug in 

55
00:04:19,260 --> 00:04:25,055
the values for R, L and C, to get this. 
Is low as vs, vs was five. 

56
00:04:25,055 --> 00:04:31,059
Now going from this if we want to find 
the transient response, we'll start by 

57
00:04:31,059 --> 00:04:35,810
getting the characteristic polynomial, or 
the characteristic equation, so here 

58
00:04:35,810 --> 00:04:40,930
these derivatives become s squared and 
this one becomes s. 

59
00:04:40,930 --> 00:04:44,430
The coefficients stay, and we set all of 
that equal to zero. 

60
00:04:44,430 --> 00:04:48,535
And then by solving the roots of this 
characteristic equation, using the 

61
00:04:48,535 --> 00:04:54,100
quadratic equation for example, we find 
the roots to be negative 5000 and then 

62
00:04:54,100 --> 00:04:59,840
negative 6.06 times 10 to the 6th. 
Here I'm using e to represent times 10 to 

63
00:04:59,840 --> 00:05:04,720
the power of The 6, just for, brevity in, 
in writing. 

64
00:05:06,510 --> 00:05:10,170
But these is the roots. 
We find that our system is going to have 

65
00:05:10,170 --> 00:05:17,910
a response that looks something like 
this, where this value is our first root 

66
00:05:17,910 --> 00:05:22,880
and this value is our second root and k1 
and k2 are our constants. 

67
00:05:22,880 --> 00:05:25,860
And we don't know what they happen to be. 
They will be dependent upon the initial 

68
00:05:25,860 --> 00:05:28,165
conditions of the system and we will 
settle for them later. 

69
00:05:28,165 --> 00:05:32,600
And so this gives us an idea of the 
transient response of our system. 

70
00:05:32,600 --> 00:05:34,400
Now we need to see what the steady state 
response is. 

71
00:05:36,110 --> 00:05:44,742
To find this, we know from before that 
the Vc, is going to go to K over a2. 

72
00:05:44,742 --> 00:05:51,585
And if K is this, and a2 is this, I find 
out that vc is going to go to vs. 

73
00:05:51,585 --> 00:05:58,480
Which is 5 volts. 
You'll notice that this matches up with 

74
00:05:58,480 --> 00:06:02,220
the behavior that we learned before, 
where this capacitor starts to look like 

75
00:06:02,220 --> 00:06:05,560
an open circuit. 
And so this matches up with our 

76
00:06:05,560 --> 00:06:09,610
intuition. 
To find our complete solution, we need to 

77
00:06:09,610 --> 00:06:17,080
add two responses together. 
So doing that, we get this response. 

78
00:06:17,080 --> 00:06:19,090
But we still need to know what k one and 
k two are. 

79
00:06:20,400 --> 00:06:22,590
So how do we find them? 
Well you're going to use our initial 

80
00:06:22,590 --> 00:06:26,380
conditions. 
We know that Vc at time t equals zero is 

81
00:06:26,380 --> 00:06:31,879
zero. 
So if I plug in zero for our t's and put 

82
00:06:31,879 --> 00:06:39,860
in vc at zero, we know it to be zero, we 
get K1 plus K2 plus 5 is equal to zero. 

83
00:06:42,370 --> 00:06:47,640
We also know that the current before time 
t equals zero Is zero, and we know that 

84
00:06:47,640 --> 00:06:53,400
the current through a capacitor is equal 
to CDVCDT. 

85
00:06:53,400 --> 00:06:57,750
So if I take the derivative of this 
function, multiply it by C and set it 

86
00:06:57,750 --> 00:07:03,000
equal to zero, plugging in zero for T, we 
get this equation. 

87
00:07:03,000 --> 00:07:05,180
So now I have two equations and two 
unknowns. 

88
00:07:05,180 --> 00:07:10,482
So I can just use linear algebra to find 
for k1 and k2 and discover them to be 

89
00:07:10,482 --> 00:07:20,870
-5.00413 and 0.00413 respectively. 
So that's the complete solution. 

90
00:07:20,870 --> 00:07:25,680
Now let's take a look and see, kind of 
what this means Well notice that this 

91
00:07:25,680 --> 00:07:29,755
value is very very small compared to 
this. 

92
00:07:29,755 --> 00:07:37,430
So, this almost looks like the first 
order response with a time constant that 

93
00:07:37,430 --> 00:07:46,990
can be derived from this. 
So that would be negative 5,000 and tau 

94
00:07:46,990 --> 00:07:54,570
is one over that. 
So tau is going to be equal to two times 

95
00:07:54,570 --> 00:07:58,260
ten to the negative fourth. 
So, going to our plot here, that would be 

96
00:07:58,260 --> 00:08:07,160
right about here. 
Moving up we discover that's 

97
00:08:07,160 --> 00:08:12,150
approximately 2 3rds of the way between 
our initial and final states. 

98
00:08:12,150 --> 00:08:16,305
Which is what we observed in first order 
systems, where this is just a decaying 

99
00:08:16,305 --> 00:08:21,260
expodential. 
But this second term is important so that 

100
00:08:21,260 --> 00:08:24,740
we also get the behavior. 
Right here, since we know that the 

101
00:08:24,740 --> 00:08:29,410
initial current is also 0. 
So, there is an effect there that this is 

102
00:08:29,410 --> 00:08:33,690
an important contributor, but it's just 
very difficult to see, because it does 

103
00:08:33,690 --> 00:08:36,170
not take very long before this one 
dominates the behavior. 

104
00:08:36,170 --> 00:08:38,910
But we also notice that it matches all of 
our initial conditions. 

105
00:08:38,910 --> 00:08:42,740
We start at 0, we're going to 5, Looks 
kind of like an exponential decay. 

106
00:08:42,740 --> 00:08:48,830
And because those two roots were real and 
distinct, the system was overdamped, and 

107
00:08:48,830 --> 00:08:51,310
so we're going to be expecting behavior 
that looks kind of like this. 

108
00:08:53,450 --> 00:08:56,450
So to summarize, we've looked at the 
overdamped case and we looked at initial 

109
00:08:56,450 --> 00:08:59,260
and final conditions. 
We've solved the differential equation 

110
00:08:59,260 --> 00:09:02,815
and plotted the final result. 
In the next lesson, we will look at an 

111
00:09:02,815 --> 00:09:07,002
underdamped example and see how it 
differs from the, the overdamped example 

112
00:09:07,002 --> 00:09:08,691
that was presented here.