# Differential Equation Help

1. Mar 13, 2008

### Sparky_

1. The problem statement, all variables and given/known data
Find the charge on the capacitor in an L-R-C circuit at time t = 0.001

L = 0.05H, R = 2 ohms, C = 0.01F

q(0) = 5
i(0) = 0
E(t) = 0

2. Relevant equations

3. The attempt at a solution

$$L \frac {di(t)}{dt} + R \frac {dq(t)}{dt} + \frac {q}{C} = 0$$

$$\frac {dq^2(t)}{dt^2} + \frac {R} {L} \frac {dq(t)}{dt} + \frac {q}{LC} = 0$$

$$\frac {dq^2(t)}{dt^2} + 40 \frac {dq(t)}{dt} + 2000q = 0$$

$$m^2 = 40m + 2000 = 0$$

$$q(t) = e^{-20t} (c1 * cos(40t) + c2 * sin(40t))$$

q(0) = 0 yields c1 = 5

$$q'(t) = i(t) = -20e^{-20t} (c1 * cos(40t) + c2 * sin(40t) ) + e^{-20t}(-200 sin (40t) + 40*c2*cos(40t))$$

c1 = 5
c2 = 5/2

I get q(0.01) = 4.11 coulombs

The book has q(0.01) = 4.568 coulombs

Can you help resolve my error?

Thanks
-Sparky_

Last edited: Mar 13, 2008
2. Mar 13, 2008

### tiny-tim

Hi Sparky!

Isn't c1 = 0?

3. Mar 13, 2008

### Sparky_

sorry,

my mistake
q(0) = 5

c1 = 5

-Sparky

4. Mar 13, 2008

### Tom Mattson

Staff Emeritus
Sparky:

You got the wrong answer because your calculator was in degree mode. Put it in radian mode (like it should be), and you get the book's answer.

5. Mar 14, 2008

### Sparky_

Ahh!!

I was using a calculator I wasn't familar with and didn't check the trig settings.

Thanks so much!

6. Mar 18, 2008

### Sparky_

Do you agree with my q(t) =

$$q(t) = e^{-20t} (5 cos(40t) + \frac{5} {2} sin(40t))$$

??

There is a second part to this problem -

Find the first time q is equal to 0.

Thanks to Kreizhn I have

$$0 = e^{-20t} (5cos(40t) + \frac{5} {2}sin(40t))$$

$$0 = (5cos(40t) + \frac{5} {2}sin(40t))$$

$$cos(40t) = -\frac{1} {2}sin(40t))$$

$$40t = -1.1.07$$

$$t = -0.0276$$

$$40t = -1.1.07 + pi$$

$$40t = 2.0345$$

$$t = 0.0508$$

The book gets t = 0.0669.

Suggestions?

Thanks
-Sparky_