Maximum Charge in an LC circuit.

AI Thread Summary
In the discussion about calculating the maximum charge (Qmax) in an LC circuit, the circuit consists of a resistor, two inductors, a capacitor, a battery, and a switch. The resistance is 481 Ω, inductances are 281 mH and 163 mH, the capacitance is 85 μF, and the battery voltage is 12 V. The participant is trying to equate the energy stored in the capacitor and the inductor to find Qmax, but their initial approach yielded incorrect results. They note that after the switch is opened, the current from the battery charges the capacitor, and the changing current induces voltage across the inductors, leading to a zero sum of voltages in the closed circuit. Further guidance is sought to correctly formulate the equations needed to solve for Qmax.
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Homework Statement



A circuit is constructed with a resistor, two inductors, one capacitor, one battery and a switch as shown. The value of the resistance is R1 = 481 Ω. The values for the inductances are: L1 = 281 mH and L2 = 163 mH. The capacitance is C = 85 μF and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.

https://www.smartphysics.com/Content/Media/Images/EM/19/h19_LC_energies.png

What is Qmax, the magnitude of the maximum charge on the capacitor after the switch is opened?

Homework Equations



So far I have been trying to find Qmax from these two equations below:

UL = 0.5(L)(I)(I)
UC = 0.5(Q)*(Q)*C
I = V/R

The Attempt at a Solution



So far, I have been assuming that the energy in the capacitor and the energy in the inductor are equal. With that assumption, at first I attempted to set UL = UC, and solve for Q that way, but that answer was wrong.

I'm stumped at the moment. Any suggestions on how to proceed?
 
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Assuming the switch had been closed for a long time before opened the DC current flowing through the inductors is V(battery)/R. When the switch is opened that current flows and charges the capacitor, so its voltage increases. The changing current induces voltage across the inductors. The sum of voltages in the closed circuit is zero.Write up the equation.

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