Hi, I'm Jacob Block and I'm an ECE Ph.D graduate student here at Georgia Tech.
Today I will be going over an extra problem on RLC circuits.
So in this extra problem, we're going to be looking at an RLC circuit, except it
has a different topology than what you've seen before.
So, now we have a resistor in series with L and a resistor in parallel with C.
So let's look at how to derive our common expressions that we've been using for RL
and C, which will help solve for any solution for this problem.
so what is VC of t and ILT. So our first step is to use k v l and k c
l to describe the circuit. So our k our first k v l equation is
going to be this outer loop. So starting from this bottom note here we
have And traveling clockwise, we enter negative Vs.
And then we hit our first resistor and it has a current of iL through it, so R1
times iL is the voltage drop here, and then we needed the, we need the voltage
across the inductor. So the voltage across the inductor Is
equal to L times the current or times the derivative of the current going through
it. So, L dil, dt.
And finally the voltage across the capacitor or across even R2 is just Vc.
So the sum of these voltages is equal to 0.
That's our first equation. Our second equation is a KCL at this
loo-, at this node here. And this will relate again, iL and VC
together. So what's the current entering through
the top branch? The top branch has a current of iL.
And then, the bottom branch puts the voltage through a capacitor.
Well, that's just capacitance time the derivative of the voltage crossing.
Plus. So, that's the, the current for the
bottom branch. And then, we have another current
entering R 2. And the current entering r 2 is, v c, the
voltage across it, across the resistor, divided by r 2.
So we have two equations and two unknowns and both of these are differential
equations and they're coupled together. So now we want to substitute n the k c l
equation into the k v l to get a second order differential equation which we can
solve. So now we're going to combine our two
equations together. We're going to substitute i l into the
first equation so we start off with negative v s plus r 1.
And we're substituting iL, so that's C dvc dt plus vc over R2 plus L times d dt
and now, in this case, we are substituting in, again, iL And then we'll
take the derivative later. And finally we have our plus VC, and all
of that equals 0. We can so now let's kind of expand and
see where this takes us. So we have negative v s.
Plus u, we're going to distribute R1, so we have R1C dvc dt plus R1 over R2 VC
plus LC, well, here, we have a derivative, and it's being distributed
across this iL equation. So here, we have an L.
This C is not varying in time, so the derivative so it doesn't, the derivative
of C isn't going to change it. We can pull it out front, and then we
have the derivative of the derivative of VC, and so we have a second order here, a
secondary derivative of VC, and then we have kind of the same process.
We can pull out R2 for the second term, and take the derivative of VC, which is
just a single derivative here. Plus VC equal 0.
So now we can put we can kind of group terms together and put it into a common
form. So the common form is to have our second
derivative in front. And then group our first derivative
together and then group our constant term together and have that equal our second
term. So, let's fill in these, these spots.
So we have our R1 times C, except that's being divided by this constant, LC, so we
have R1 over L plus we have this dVc over here, so dividing by LC, we have 1 over
R2 times C. And finally for V C we have we have a R 1
over R 2, plus 1. And then our forcing function is V S over
L C. So this kind of looks like the
second-order differential equation we've seen.
Where we have d squared y over d t squared, plus 2 alpha dy dt.
plus omega naught squared y equal to k. So, our alpha is 1/2.
Of r1 over L plus 1 over r2C, our omega knot squared term, or omega knot, is the
square root, I'm sorry there's an L over C there, is the square root of r1 over r2
plus 1 over lc... And our K is V s over L C.
So now, we can use all of our previous solutions for underdamped, overdamped,
and critically damped by plugging in our respective R 1, L, C, and R 2 into these
formulas and use the exact same solutions that we've seen previously.
And now you can do solve the same or create new differential equations for any
topology of L's and C's. [BLANK_AUDIO]