# LC Circuit max charge problem

reising1

## 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.

Staff Emeritus
Homework Helper

## 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.
The other possibility (the correct one) is that cos(phi)=0.

Try differentiating q(t) and use the other information given in the problem.

reising1
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.

Staff Emeritus
Homework Helper
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.

reising1
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!

Staff Emeritus