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Hi, I'm Jacob Block and I'm an ECE Ph.D 
graduate student here at Georgia Tech. 

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Today I will be going over an extra 
problem on RLC circuits. 

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So in this extra problem, we're going to 
be looking at an RLC circuit, except it 

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has a different topology than what you've 
seen before. 

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So, now we have a resistor in series with 
L and a resistor in parallel with C. 

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So let's look at how to derive our common 
expressions that we've been using for RL 

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and C, which will help solve for any 
solution for this problem. 

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so what is VC of t and ILT. 
So our first step is to use k v l and k c 

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l to describe the circuit. 
So our k our first k v l equation is 

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going to be this outer loop. 
So starting from this bottom note here we 

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have And traveling clockwise, we enter 
negative Vs. 

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And then we hit our first resistor and it 
has a current of iL through it, so R1 

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times iL is the voltage drop here, and 
then we needed the, we need the voltage 

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across the inductor. 
So the voltage across the inductor Is 

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equal to L times the current or times the 
derivative of the current going through 

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it. 
So, L dil, dt. 

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And finally the voltage across the 
capacitor or across even R2 is just Vc. 

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So the sum of these voltages is equal to 
0. 

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That's our first equation. 
Our second equation is a KCL at this 

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loo-, at this node here. 
And this will relate again, iL and VC 

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together. 
So what's the current entering through 

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the top branch? 
The top branch has a current of iL. 

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And then, the bottom branch puts the 
voltage through a capacitor. 

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Well, that's just capacitance time the 
derivative of the voltage crossing. 

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Plus. 
So, that's the, the current for the 

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bottom branch. 
And then, we have another current 

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entering R 2. 
And the current entering r 2 is, v c, the 

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voltage across it, across the resistor, 
divided by r 2. 

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So we have two equations and two unknowns 
and both of these are differential 

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equations and they're coupled together. 
So now we want to substitute n the k c l 

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equation into the k v l to get a second 
order differential equation which we can 

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solve. 
So now we're going to combine our two 

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equations together. 
We're going to substitute i l into the 

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first equation so we start off with 
negative v s plus r 1. 

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And we're substituting iL, so that's C 
dvc dt plus vc over R2 plus L times d dt 

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and now, in this case, we are 
substituting in, again, iL And then we'll 

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take the derivative later. 
And finally we have our plus VC, and all 

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of that equals 0. 
We can so now let's kind of expand and 

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see where this takes us. 
So we have negative v s. 

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Plus u, we're going to distribute R1, so 
we have R1C dvc dt plus R1 over R2 VC 

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plus LC, well, here, we have a 
derivative, and it's being distributed 

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across this iL equation. 
So here, we have an L. 

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This C is not varying in time, so the 
derivative so it doesn't, the derivative 

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of C isn't going to change it. 
We can pull it out front, and then we 

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have the derivative of the derivative of 
VC, and so we have a second order here, a 

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secondary derivative of VC, and then we 
have kind of the same process. 

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We can pull out R2 for the second term, 
and take the derivative of VC, which is 

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just a single derivative here. 
Plus VC equal 0. 

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So now we can put we can kind of group 
terms together and put it into a common 

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form. 
So the common form is to have our second 

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derivative in front. 
And then group our first derivative 

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together and then group our constant term 
together and have that equal our second 

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term. 
So, let's fill in these, these spots. 

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So we have our R1 times C, except that's 
being divided by this constant, LC, so we 

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have R1 over L plus we have this dVc over 
here, so dividing by LC, we have 1 over 

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R2 times C. 
And finally for V C we have we have a R 1 

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over R 2, plus 1. 
And then our forcing function is V S over 

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L C. 
So this kind of looks like the 

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second-order differential equation we've 
seen. 

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Where we have d squared y over d t 
squared, plus 2 alpha dy dt. 

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plus omega naught squared y equal to k. 
So, our alpha is 1/2. 

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Of r1 over L plus 1 over r2C, our omega 
knot squared term, or omega knot, is the 

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square root, I'm sorry there's an L over 
C there, is the square root of r1 over r2 

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plus 1 over lc... 
And our K is V s over L C. 

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So now, we can use all of our previous 
solutions for underdamped, overdamped, 

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and critically damped by plugging in our 
respective R 1, L, C, and R 2 into these 

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formulas and use the exact same solutions 
that we've seen previously. 

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And now you can do solve the same or 
create new differential equations for any 

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topology of L's and C's. 
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