Find the charge q(t) for an LC circuit

[sneakycooky]

Homework Statement

An LC circuit consists of an 82 mH inductor and a 17 microfarad capacitor that initially carries a 180 microC charge. The switch is open for t < 0 and is then closed at t = 0.
a. Find the frequency of the resulting oscillations.
b. At t = 1 ms, find the charge on the capacitor and the current in the circuit.

Homework Equations

f = (omega)/(2pi)
(omega) = 1/sqrt(LC)

q(t) = (q_o)cos([omega][delta t])
I(t) = dq(t)/dt

The Attempt at a Solution

a. I feel pretty comfortable at this part, but just in case, here was my process:

f = (omega)/(2pi)
(omega) = 1/sqrt(LC) = 1/[(82 x 10^-3H)(17 x 10^-6F)] = 846.97 rad/s
then I use f = (omega)/(2pi) and got 134.8 Hz.

b. This is the part I am not so sure about. I have heard that the phase angle can be negated by implementing the change in time, delta t. I think I have the right formula (but it could be wrong):

q(t) = (q_o)cos([omega][delta t])

Now I plug in the numbers.

q(.001) = (180 x 10^-6C)cos([846.97rad/s][.001s]) = 1.192 x 10^-4C = 119.2 microC

Now for the current I use I(t) = dq(t)/dt = -(omega)(q_o)sin([omega][delta t])

Plugging in numbers gives:

I(.001) = -(846.97rad/s)(180 x 10^-6C)sin([846.97rad/s][.001s]) = -0.114 A

I think these are right and that a negative current is fine here, because as far as I know LC circuits are natural AC circuits. Are my calculations and (more importantly) reasonings here correct?​

 

[collinsmark]

Homework Statement

An LC circuit consists of an 82 mH inductor and a 17 microfarad capacitor that initially carries a 180 microC charge. The switch is open for t < 0 and is then closed at t = 0.
a. Find the frequency of the resulting oscillations.
b. At t = 1 ms, find the charge on the capacitor and the current in the circuit.

Homework Equations

f = (omega)/(2pi)
(omega) = 1/sqrt(LC)

q(t) = (q_o)cos([omega][delta t])
I(t) = dq(t)/dt

The Attempt at a Solution

a. I feel pretty comfortable at this part, but just in case, here was my process:

f = (omega)/(2pi)
(omega) = 1/sqrt(LC) = 1/[(82 x 10^-3H)(17 x 10^-6F)] = 846.97 rad/s
then I use f = (omega)/(2pi) and got 134.8 Hz.

b. This is the part I am not so sure about. I have heard that the phase angle can be negated by implementing the change in time, delta t. I think I have the right formula (but it could be wrong):

q(t) = (q_o)cos([omega][delta t])

Now I plug in the numbers.

q(.001) = (180 x 10^-6C)cos([846.97rad/s][.001s]) = 1.192 x 10^-4C = 119.2 microC

Now for the current I use I(t) = dq(t)/dt = -(omega)(q_o)sin([omega][delta t])

Plugging in numbers gives:

I(.001) = -(846.97rad/s)(180 x 10^-6C)sin([846.97rad/s][.001s]) = -0.114 A

I think these are right and that a negative current is fine here, because as far as I know LC circuits are natural AC circuits. Are my calculations and (more importantly) reasonings here correct?​

Hello @sneakycooky,

Welcome to PF!

Your answers look correct to me. Good work!

By the way, as you suggest, the phase of the charge on the capacitor changes depending on how everything is initially set up and how you define when the switch is closed and whatnot. You can sanity-check yourself by applying the given boundary condition. For example, in this problem, you know that $Q = Q_0 \cos \left( \omega t + \theta \right)$ and you already know what $Q_0$ and $\omega$ are. But that's not all you know. You also know that at time $t = 0$, $Q = Q_0$. So go ahead and plug those numbers in and make sure your $\theta = 0$ assumption makes sense.

 

[sneakycooky]

So go ahead and plug those numbers in and make sure your θ=0θ=0 \theta = 0 assumption makes sense.

Ahhh thanks I see it now.

Q_o = Q_ocos(0) so with those conditions theta must be zero.

Thank you :)