Calculus with Kirkhoff's Equations Help

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

1. sjgallagher2 Thread Starter Active Member

Feb 6, 2013
128
12
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
523
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 Active Member

Feb 6, 2013
128
12
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
523
For the benefit of those who wish to discuss these equations further here they are in the order originally presented in post#1

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

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

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