LC Circuit max charge problem

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  • #1
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Homework Statement



In an oscillating LC circuit, L = 2.93 mH and C = 3.21 μF. At t = 0 the charge on the capacitor is zero and the current is 2.14 A. What is the maximum charge (in C) that will appear on the capacitor?

Homework Equations



I know that for an LC Circuit, the charge q at a given time t is:
q = Q cos(wt + phi)

The Attempt at a Solution



At t = 0, q = 0;
Thus,
0 = Q cos(phi),
which implies Q = 0.

So the maximum charge is 0.

However this is not right.
Any help please?
 

Answers and Replies

  • #2
vela
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Homework Statement



In an oscillating LC circuit, L = 2.93 mH and C = 3.21 μF. At t = 0 the charge on the capacitor is zero and the current is 2.14 A. What is the maximum charge (in C) that will appear on the capacitor?

Homework Equations



I know that for an LC Circuit, the charge q at a given time t is:
q = Q cos(wt + phi)

The Attempt at a Solution



At t = 0, q = 0;
Thus,
0 = Q cos(phi),
which implies Q = 0.

So the maximum charge is 0.

However this is not right.
Any help please?
The other possibility (the correct one) is that cos(phi)=0.

Try differentiating q(t) and use the other information given in the problem.
 
  • #3
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Okay, so

i = dq/dt = -wq sin(wt + phi)

Plugging in,
since w = 1 / sqrt(LC)
2.14 A = Q (-1 / sqrt(2.93 mH * 3.21 microF)) sin(phi)

However I'm not sure what phi is. In fact, I'm not quite sure specifically what phi is. I know it is the phase difference. But in this case, the phase difference between what?

Thanks so much.
 
  • #4
vela
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The phase phi simply tells you where in the cycle you're starting.

With the information you're given, you can solve for phi. First, you know cos(phi)=0, so phi=pi/2 or -pi/2. From your equation for the current, you can figure out which of the two possibilities is the correct one.
 
  • #5
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Okay. I understand that one. Now here's the next part;

(b) At what earliest time t > 0 is the rate at which energy is stored in the capacitor greatest?

So, I know that U = q^2 / 2C
So U = (Q^2)(sin^2(wt)) / (2C)

deriving this (to find the "rate at which energy is stored"), we get

dU/dt = (Q^2)(w)(sin(wt))(cos(wt)) / C

The maximum occurs when sin(wt)cos(wt) is at a max, which is when wt = pi/4

So, wt = pi/4 implies that the time when the rate at which energy is stored in the capacitor is greatest is
t = (pi/4)(1/w)
plugging in values, I get
t = 7.61686 E-5 seconds

However, this is incorrect.
See my flaw anywhere?
Thanks!
 
  • #6
vela
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Other than perhaps the number of significant figures, it looks right to me.
 
  • #7
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Going back to before, why is it that we know cos(phi) = 0?
 

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