Finding the Charge and Time in an L-R-C Circuit

[Sparky_]

Homework Statement

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

Homework Equations

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_

 

[tiny-tim]

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

q(0) = 0 yields c1 = 5

Hi Sparky!

Isn't c1 = 0?

 

[Sparky_]

sorry,

my mistake
q(0) = 5

c1 = 5

-Sparky

 

[quantumdude]

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.

 

[Sparky_]

Ahh!

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

Thanks so much!

 

[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_