# Calculus with Kirkhoff's Equations Help

Discussion in 'Math' started by sjgallagher2, Aug 13, 2013.

1. ### sjgallagher2 Thread Starter Member

Feb 6, 2013
111
7
So I'm fairly new to calculus, and integrals, and I'm working out an RLC circuit with this formula:
$image=http://latex.codecogs.com/gif.latex?V_{0}&space;cos(\omega&space;t)=RI+L\frac{dI_{L}}{dt}+\frac{1}{C}\int&space;Idt&hash=4c9fa5fab53e183d5f30716776df5149$
(if it doesn't show up it's here: http://latex.codecogs.com/gif.latex?V_{0}&space;cos(\omega&space;t)=RI&plus;L\frac{dI_{L}}{dt}&plus;\frac{1}{C}\int&space;Idt)

Where V0cos(ωt) is the AC voltage, ω is the angular velocity (2∏f), I is the total current, and L R and C are the respective component values.
*

I know there are better ways to do this, finding voltages across the components and such, but this way intrigues me. And I don't understand it, and I'd like to. That equation looks all fine and all, so the next step is to simplify the integral. I must be missing something here, because when I simplified it I got:
$image=http://latex.codecogs.com/gif.latex?V_{0}&space;cos(\omega&space;t)=RI+L\frac{dI_{L}}{dt}+\frac{It}{C_{0}}+C&hash=40d14b789c78b799f436785a315525b1$
(If it doesn't show up it's here: http://latex.codecogs.com/gif.latex?V_{0}&space;cos(\omega&space;t)=RI&plus;L\frac{dI_{L}}{dt}&plus;\frac{It}{C_{0}}&plus;C)
*
*Note: In the above schematics, ignore the subscript "L" under the I, it should be simply I.

Where C0 is the capacitor value and C is the constant of integration.
The answer in the book on the other hand, shows the answer as this:
$image=http://latex.codecogs.com/gif.latex?%5Comega%20V_%7B0%7D%20cos%28%5Comega%20t%29%3DL%5Cfrac%7Bd%5E%7B2%7DI%7D%7Bdt%5E%7B2%7D%7D-R%5Cfrac%7BdI%7D%7Bdt%7D+%5Cfrac%7B1%7D%7BC%7DI&hash=8a4fcca5382765472a5699dfdcccfbb8$
(If it doesn't show up it's here: http://latex.codecogs.com/gif.latex?\omega&space;V_{0}&space;cos(\omega&space;t)=L\frac{d^{2}I}{dt^{2}}-R\frac{dI}{dt}&plus;\frac{1}{C}I)
Things that are interesting to me is the extra ω in front and the new stuff being multiplied by R (I thought would still be I but now it's dI/dt).
If anyone can bear with me through all this and help me understand what they did and why, I would be most thankful! Another note by the way, the book says this last function is a "linear second-order nonhomogeneous differential equation" which scares me... Thanks.

2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
That is because the book differentiated the first equation with respect to t to get rid of the intergral in the third term, whereas you tried to integrate it.

Incidentally, this site offers TEX (click on the Ʃ sign at the top right of toolbar in the compose box)

Note also when you write TEX, you should not end on a hanging )

3. ### sjgallagher2 Thread Starter Member

Feb 6, 2013
111
7
Alright thanks. Simple, to the point. I've been looking into differential equations now and realized there were quite a few things I never learned, and so now I'm diving into them much more. I didn't even know what order was, or what a differential equation really was. In any event I'm on the right track now, I've got a long time ahead of me to learn all this math, and a lot to learn, let's do it.
By the way thanks for the tip about TEX haha.

4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
For the benefit of those who wish to discuss these equations further here they are in the order originally presented in post#1

${V_0}\cos (\omega t) = RI + L\frac{{dI}}{{dt}} + \frac{1}{C}\int {Idt}$

${V_0}\cos (\omega t) = RI + L\frac{{dI}}{{dt}} + \frac{{It}}{{{C_0}}} + C$

$\omega {V_0}\cos (\omega ;t) = - R\frac{{dI}}{{dt}} + L\frac{{{d^2}I}}{{d{t^2}}} + \frac{1}{C}I$