Question by longdecember:
What is the maximum current in the circuit?
A 75 microFarad capacitor is connected in series with a 10 mH inductance and a switch. The capacitor is first charged to a voltage of 220 V. The charging battery is then removed. As soon as the switch is closed, the current begins to oscillate back and forth between one direction and the reversed direction. What is the maximum current in the circuit in units of A?
I am having trouble with this because I know the formula for current for an RL circuit, but this is an LC circuit, and I can't find it anywhere. Any help would be appreciated.
I also know that when current is maximum, the charge has to be zero because current and charge are 90 degrees out of phase, but I don't know where this would fit in.
Best answer:
Answer by Bekki B
When there's no battery in there, the voltage across the capacitor = - voltage across the inductor.
V cap = Q/C = -V ind = -L I' = -LQ''
The solution is a differential equation:
Q = Qmax cos (omega t)
where omega = sqrt (1/LC)
Qmax is the initial charge, which you got when the capacitor voltage Q/C = the battery voltage. So Qmax = VC.
So Q = Qmax cos (omega t)
Current is the time derivative of charge.
So I = Q' = - Qmax (omega) sin (omega t)
The maximum value, therefore, is Qmax * omega. Plug in all the numbers to solve for it.
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