Welcome back to our class on linear 
circuits where today we're going to be 
talking about RLC circuits or circuits 
with resistors, inductors, and capacitors 
altogether. 
We will be using differential equations, 
the second order differential equations 
that you were presented in the last 
lesson to analyse these circuits. 
So as I mentioned in the previous lesson, 
Calvert is solving second-order 
differential equations. 
So in this lesson we'll see how we can 
generate second-order differential 
equations from our our system, and then 
use the techniques from the last lesson 
to solve them. 
And finishing up our RLC Circuits will be 
the concluding portion of this module. 
The lesson objectives are to generate the 
second-order differential equation from 
the circuit. 
Identify the initial and final conditions 
of the circuit. 
Solve the differential equation. 
And then recognize that the system's 
underdamped or overdamped. 
So this is our system. 
We see that we have a resistor. 
Of 20 killiohms, an inductor of 3.3 
millihenrys at a capacitor of 1 100th of 
a microferret/g. 
We want to find vc, the voltage across 
the capacitor. 
Looking at our circuit, we see that the 
switch closes at time t equal 0, but 
before that time, i has to be 0 because 
there's no path for the current to flow. 
We also know that the current cannot 
instantaneously change through an 
inductor, so that means that the current 
at time zero just before the switched is 
closed, which is equal to zero, is also 
equal to the current at time zero just 
after the switch is closed. 
We're going to use this minus here to 
indicate just before time 0 and this just 
after time 0. 
We cannot say just by analyzing the 
circuit or just by inspection of the 
circuit what vc is at time t equals 0, so 
we're going to explicitly state that 
before time t equals 0, before the switch 
closes, that the voltage is 0. 
And because we know that the voltage 
cannot instantaneously change through a 
capacitor, vc at 0 plus is also going to 
be equal to 0. 
We also know that the final condition of 
our system is going to be where this 
inductor starts to look like a wire, and 
this capacitor starts to look like an 
open circuit. 
And if you do the analysis, you'll 
discover that Vc after a long time is 
going to look like 5 volts, the same as 
the input here. 
Now let's see how to get the differential 
equations from the system. 
So the first thing that we're going to 
make use of is Kirchhoff's voltage law, 
which states that vs It's going to be 
equal to the voltage across r plus the 
voltage across the inductor, which will 
call vl, plus vc. 
We also have some expressions for what 
these happen to be, So based on i we know 
vr is equal to i times r. 
Vl is equal to l di dt And then we also 
know that i is equal to cdvcdt. 
So what we want to do is come up with a, 
an equation in terms of these vc's. 
That's a differential equation. 
So, the way we will do that is here in 
the L, we will replace the i's. 
this i here with cdvcdt. 
Which will give us L times C times the 
second derivative of v c with respect to 
d t. 
And here this will be r c d v d t. 
Summing all of those together, we get 
this equation. 
But this does not exactly match up the 
equation that was presented in the 
previous lesson on solving second order 
of differential equations. 
But to get it in that format. 
Which is like this. 
We will divide everything by lc. 
So now we see that a1is r over l. 
A2 Is 1 over lc and k is vs over lc. 
And so now we can take this differential 
equation that we've generated and use the 
techniques from the last lesson to solve 
the system. 
So to do this and solve the system, we'll 
start by looking at the differential 
equation here, where we can then plug in 
the values for R, L and C, to get this. 
Is low as vs, vs was five. 
Now going from this if we want to find 
the transient response, we'll start by 
getting the characteristic polynomial, or 
the characteristic equation, so here 
these derivatives become s squared and 
this one becomes s. 
The coefficients stay, and we set all of 
that equal to zero. 
And then by solving the roots of this 
characteristic equation, using the 
quadratic equation for example, we find 
the roots to be negative 5000 and then 
negative 6.06 times 10 to the 6th. 
Here I'm using e to represent times 10 to 
the power of The 6, just for, brevity in, 
in writing. 
But these is the roots. 
We find that our system is going to have 
a response that looks something like 
this, where this value is our first root 
and this value is our second root and k1 
and k2 are our constants. 
And we don't know what they happen to be. 
They will be dependent upon the initial 
conditions of the system and we will 
settle for them later. 
And so this gives us an idea of the 
transient response of our system. 
Now we need to see what the steady state 
response is. 
To find this, we know from before that 
the Vc, is going to go to K over a2. 
And if K is this, and a2 is this, I find 
out that vc is going to go to vs. 
Which is 5 volts. 
You'll notice that this matches up with 
the behavior that we learned before, 
where this capacitor starts to look like 
an open circuit. 
And so this matches up with our 
intuition. 
To find our complete solution, we need to 
add two responses together. 
So doing that, we get this response. 
But we still need to know what k one and 
k two are. 
So how do we find them? 
Well you're going to use our initial 
conditions. 
We know that Vc at time t equals zero is 
zero. 
So if I plug in zero for our t's and put 
in vc at zero, we know it to be zero, we 
get K1 plus K2 plus 5 is equal to zero. 
We also know that the current before time 
t equals zero Is zero, and we know that 
the current through a capacitor is equal 
to CDVCDT. 
So if I take the derivative of this 
function, multiply it by C and set it 
equal to zero, plugging in zero for T, we 
get this equation. 
So now I have two equations and two 
unknowns. 
So I can just use linear algebra to find 
for k1 and k2 and discover them to be 
-5.00413 and 0.00413 respectively. 
So that's the complete solution. 
Now let's take a look and see, kind of 
what this means Well notice that this 
value is very very small compared to 
this. 
So, this almost looks like the first 
order response with a time constant that 
can be derived from this. 
So that would be negative 5,000 and tau 
is one over that. 
So tau is going to be equal to two times 
ten to the negative fourth. 
So, going to our plot here, that would be 
right about here. 
Moving up we discover that's 
approximately 2 3rds of the way between 
our initial and final states. 
Which is what we observed in first order 
systems, where this is just a decaying 
expodential. 
But this second term is important so that 
we also get the behavior. 
Right here, since we know that the 
initial current is also 0. 
So, there is an effect there that this is 
an important contributor, but it's just 
very difficult to see, because it does 
not take very long before this one 
dominates the behavior. 
But we also notice that it matches all of 
our initial conditions. 
We start at 0, we're going to 5, Looks 
kind of like an exponential decay. 
And because those two roots were real and 
distinct, the system was overdamped, and 
so we're going to be expecting behavior 
that looks kind of like this. 
So to summarize, we've looked at the 
overdamped case and we looked at initial 
and final conditions. 
We've solved the differential equation 
and plotted the final result. 
In the next lesson, we will look at an 
underdamped example and see how it 
differs from the, the overdamped example 
that was presented here.