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.