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Old May 1st 2017, 01:57 PM   #1
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LC Circuit Problem

At a certain instant, the charge stored by a capacitor in an LC circuit is 60 microCoulombs and the energy stored by the capacitor is nine sixteenths of its maximum value. If the circuit's frequency of oscillation is (40 / pi) Hz, what is the maximum current in the circuit?
I feel like I've tried everything for this problem. I tried to solve it most recently with conservation of energy and I thought I got the answer but I was far off. The answer is 6.4 mA just so you are aware.
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Old May 2nd 2017, 07:28 AM   #2
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OK so you are seeking the peak current in the circuit (I prefer peak to maximum).
So you should start with an equation containing this quantity.
You are right to consider energy, a specific amount of energy is oscillated back and fore between the inductor and the capacitor in an LC circuit.

So


$\displaystyle \frac{1}{2}L{\left( {{I_{\max }}} \right)^2} = \frac{1}{2}C{\left( {{V_{cap\max }}} \right)^2}$


(Ask for the derivation if you have not met this equation or any of the others)

This equation contains the unknown we want and some other quantities that we must obtain from the data given.

Firstly the frequency of oscillation is given by


$\displaystyle f = \frac{1}{{2\pi \sqrt {LC} }}$


We are given f so we can find a connection between L and C replace one in a later equation.

So we can replace L in my first equation.

That leaves the max voltage and the capacitance.

So considering the equations of energy for a capacitor


$\displaystyle {E_{cap\max }} = \frac{1}{2}C{\left( {{V_{cap\max }}} \right)^2} = \frac{1}{2}{Q_{\max }}{V_{cap\max }} = \frac{1}{2}\frac{{{{\left( {{Q_{\max }}} \right)}^2}}}{C}$

also we are given the max voltage in terms of the charge

So using

$\displaystyle Q = CV$

We have

$\displaystyle {Q_{\max }} = C{V_{cap\max }}$

and we are set up for the kill.

Does this help?

Last edited by studiot; May 2nd 2017 at 09:01 AM.
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