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**1. 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

**2. Homework Equations**

**3. The Attempt at a Solution**

[tex] L \frac {di(t)}{dt} + R \frac {dq(t)}{dt} + \frac {q}{C} = 0 [/tex]

[tex] \frac {dq^2(t)}{dt^2} + 40 \frac {dq(t)}{dt} + 2000q = 0 [/tex]

[tex] m^2 = 40m + 2000 = 0 [/tex]

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

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

Do you agree with my q(t) =

[tex] q(t) = e^{-20t} (5 cos(40t) + \frac{5} {2} sin(40t)) [/tex]

??

There is a second part to this problem -

Find the first time q is equal to 0.

Thanks to Kreizhn I have

[tex] 0 = e^{-20t} (5cos(40t) + \frac{5} {2}sin(40t)) [/tex]

[tex] 0 = (5cos(40t) + \frac{5} {2}sin(40t)) [/tex]

[tex] cos(40t) = -\frac{1} {2}sin(40t)) [/tex]

[tex] 40t = -1.1.07 [/tex]

[tex] t = -0.0276 [/tex]

[tex] 40t = -1.1.07 + pi [/tex]

[tex] t = 0.0508 [/tex]

The book gets t = 0.0669.

Suggestions?

Thanks

-Sparky_

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