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. 
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