Determining Inductance L in an LC Circuit

[dragonshadowbob]

Homework Statement

In the circuit of the figure below, the battery emf is 40 V, the resistance R is 150 Ω, and the capacitance C is 0.500 µF. The switch S is closed for a long time interval, and zero potential difference is measured across the capacitor. After the switch is opened, the potential difference across the capacitor reaches a maximum value of 150 V. What is the value of the inductance?

Relevant Equations

I=V/R, Omega=Imax/Qmax=1/sqrt(LC)

Here is the circuit I'm working with. So, I was able to get the correct equation by using the Equation Omega=I max/Q max = 1/sqrt(LC). I calculated Qmax by multiplying 150V * .5microfarad, and Imax by doing the 40V/150Ohms. Then, I just solved for L and got the correct answer. My question is this: why does that work? I understand why Qmax is what it is, but why is Imax just equal to the initial voltage through the inductor when the switch is closed?
The way I tried to solve it was with an energy approach, knowing that the final energy in the circuit must be equal to the initial. So, knowing the equation for the potential energy stored in an inductor and a capacitor, I wrote this:
1/2*Qmax^2/C = 1/2*L*I initial^2+1/2*Qinitial/C
and then solved for L. In other words, I figured that since energy is conserved in the circuit, the difference in the initial and final charge, and therefore energy, stored in the capacitor must be equal to the energy initially stored in the magnetic field of the inductor.
I do not understand why the correct approach is to assume that Imax is equal to the initial current resulting from the voltage of the battery. By my way of thinking, the maximum current in the LC circuit would be higher than the initial as once the circuit is open, the capacitor would discharge and further increase the current running through the circuit.

Your help is greatly appreciated, thanks in advance.

 

[TSny]

By my way of thinking, the maximum current in the LC circuit would be higher than the initial as once the circuit is open, the capacitor would discharge and further increase the current running through the circuit.

The problem statement tells you there is zero potential difference across the capacitor at the instant the switch is opened. What is the charge on the capacitor at this instant?

 

[dragonshadowbob]

The problem statement tells you there is zero potential difference across the capacitor at the instant the switch is opened. What is the charge on the capacitor at this instant?

The initial charge carried on the capacitor would be Q=C*Vinitial=2E-5

 

[TSny]

The initial charge carried on the capacitor would be Q=C*Vinitial=2E-5

Just before the switch is opened, it is given that there is no potential difference across the capacitor. So, from Q = CV, there is no charge on the capacitor at the instant the switch is opened. So, immediately after the switch is opened, we have a simple LC circuit with current in the inductor and zero charge on the capacitor. At this instant, all of the energy in the LC circuit is in the inductor and no energy is in the capacitor.

 

[Steve4Physics]

I do not understand why the correct approach is to assume that Imax is equal to the initial current resulting from the voltage of the battery. By my way of thinking, the maximum current in the LC circuit would be higher than the initial as once the circuit is open, the capacitor would discharge and further increase the current running through the circuit.

Hi @dragonshadowbob and welcome to PF. I'll add this to what @TSny has already said.

The voltage across the capacitor before the switch is opened is not 40$\Omega$. It is zero – as stated in the question.

The reason is this. An ideal inductor has zero resistance. The voltage-drop across it ('V=IR') when a constant current flows through it is zero. The separate resistor has the full 40V.

The capacitor is in parallel with the inductor so has the same voltage as the inductor, zero.

As long as there is a constant current flowing through the inductor, the voltage across both it and the capacitor is zero. In this state, the inductor stores energy in its magnetic field; the capacitor’s voltage, charge and energy are zero.

Edits - superfluous stuff removed.

 

[DaveE]

By my way of thinking, the maximum current in the LC circuit would be higher than the initial as once the circuit is open, the capacitor would discharge and further increase the current running through the circuit.

But where would that current flow?
Since the switch is open it must flow through the inductor. Your circuit model after the switch opens only contains L in parallel with C. There is only one current (loop), and only one voltage (branch).
What is the effect of the current flow on the capacitor voltage?
What effect does that voltage have on the inductor current?