Question

Consider a simple oscillating LC circuit. If the capacitor, C, is charged up then connected to the inductor,L, at t=0, the current will be:

I=I(max)*Sin(2*pi*f*t)

A) show that there is a natural frequency, f , for which the sum of the energy stored in the unductor plus energy stored in the capacitor is constant over time.

B) Consider an instant in time when the current in the circuit is half of its maximum value. How much charge is on the capacitor?

Answer 1

Energy in the inductor at any point in time is
$$
E_L={1\over 2}LI^2\\
E_L={1\over 2}CV^2\\
$$
The voltage across the inductor is the same as the voltage across the capacitor
$$
V_C=V_L=L{dI\over dt}=LI_{max}2\pi f\cos 2\pi f t\\
$$
The total energy is then
$$
E_{total}={1\over 2}LI_{max}^2\sin^2 2\pi t+{1\over 2}CL^2I_{max}^2(2\pi f)^2\cos^2 2\pi f t\\
={1\over 2}LI_{max}^2\left (\sin^22\pi f t+LC(2\pi f )^2\cos^2 2\pi f t\right )
$$
For this energy to be constant, the natural frequency must be
$$
1=LC(2\pi f)^2\\
2\pi f=\sqrt{1\over LC}\\
f={1\over 2\pi}\sqrt{1\over LC}
$$
With this condition
$$
E_{total}={1\over 2}LI_{max}^2
$$

Does this make sense?

Answer 2

For Part b)
$$
C={Q\over V}\\
Q=CV
$$
So we need to know V when \(I={I_{max}\over 2}\)
Use
$$
E_{total}={1\over 2}LI_{max}^2={1\over 2}L{I_{max}\over 2}^2+{1\over 2}CV^2\\
{1\over 2}CV^2={1\over 2}LI_{max}^2-{1\over 2}L{I_{max}\over 2}^2\\
C^2V^2=CLI_{max}^2-CL{I_{max}\over 2}^2\\=CL{I_{max}\over 2}^2\\
Q=CV=I_{max}\sqrt{{CL\over 2}}
$$
This is the charge on the capacitor with the current is half its max.

Make sense?

Answer 3

I think I understand what you did in part A. Did you basically "demanded" the energy to be a constant by saying the scalar multiple on the cos^2 must be 1 so that the equation simplifies to show all the energy being stored in the magnetic field of the inductor.

Answer 4

We know that the solution to this circuit has a natural frequency of \(2\pi f=\sqrt{1/LC}\) so with this knowledge, the trig terms \(cos^2 2\pi f t+\sin^2 2\pi ft =1\), which is a standard identify and we are just left with the constant.

Answer 5

Ah okay, that makes sense to me. Also I followed your reasoning in part B clearly. Thank you for the clear explanations and guidance.

Answer 6

It seems like a "demand" but really we are using knowledge of the solution to this equation

Answer 7

You're welcome