- #1

dragonshadowbob

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- 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)

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.