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An interesting LC circuit problem

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Marko

Tue Apr 28 2009, 08:21PM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

Hi guys,

I've ran into this in my physics competition preparations and it has caused much of controversy, despite it looked simple to me at glance.

In an oscillating electrical circuit that consists of a plate capacitor, inductor and negligible ohmic resistance, oscillates an amount of energy W. Capacitor plates are slowly separated until the frequency doubles. What amount of work was done in this action? (Hint: Use the average value of the force. Average value of the function (cos^2 x) is 1/2.)

The result should only relate to the initial energy W.
I've solved it over conservation of charge and energy, without using the coulomb force, finding a result which completely makes sense to me but disagrees with their solution. I'm really interested to see how would you solve it?

Cheers,

Marko

Mattski

Tue Apr 28 2009, 11:47PM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

I get W*ln(4)/2.
Edit: Incidentally, that's the same as W*ln(2) = 0.7*W (approximately)

I attached the work I used to get the answer, hopefully it makes sense, I just did some quick and sloppy work I added the 1/2 factor after the attached work because the voltage will be some Vmax*cos(wt), where Vmax changes with plate separation, and force is proportional to voltage squared, because F_coulomb is q*q/dist where q = C*V. Thus the average force is 1/2 of the peak force, and that is then integrated over.

What answer did you get, and what answer is the "correct" one?
]cap_work_prob_4hv.pdf[/file]

klugesmith

Wed Apr 29 2009, 02:50PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

Here is a quick answer based on conservation of charge, and reason why it's wrong.
From the initial to final state, C is reduced by a factor of 4.
IF the peak charge on capacitor plates were invariant then Vpeak would increase
by a factor of 4. Then tank circuit energy has quadrupled, so the work input is 3W.

But there is no reason to expect charge invariance.
A tank circuit with any finite initial energy can be pumped up to arbitrary
intensity by cyclically varying C or L.

Respectfully,
Rich

Oh, and I looked at Mattski's notes. Not quite following the "assume W const" explanation.
Moving the plates apart requires mechanical work, which adds directly to the tank circuit energy.
We agree that if there were no inductor (DC charge on capacitor) then force has constant value
F = Wo/Xo, and stored electrical energy would go up in direct proportion to plate spacing.
In the oscillating tank circuit, initial attractive force (averaged over 1 cycle) is only half as great.
So the first 1% increase in X causes only a 0.5% increase in stored energy.
(specifically, -1% change in C and +0.75% change in Vpeak). That's as far as this old mind could figure it tonight, sorry!

Marko

Thu Apr 30 2009, 09:09PM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

Hi guys - thanks for your replies!

The ''correct'' answer is, indeed, 3W, and I myself also have another simple objection against it:

The given energy is equal to W = C*U^2/2, where U is the peak voltage on the capacitor.

If I imagine the case of capacitor alone simply being charged to DC voltage equal to this peak voltage, I can reason like Klugesmith did:

With charge being conserved in this case, reducing the capacitance by 1/4 will increase voltage 4 times, and as energy is proportional to square of voltage, it will increase 4 times - resulting in E = 3W...

My argument is, that AC circuit like described can't possibly develop this same amount of work, because the capacitor isn't charged whole time! It must be less than 3W, to my reasoning. Still, I couldn't find anything explicitly wrong in the original solution done using force. I'll post this solution tomorrow here.

My own solution simply considered that force is dependent on the RMS voltage across the capacitor. And since this voltage is 1/sqrt2 of peak voltage and the force is dependent on square of the voltage, it should be 1/2 of the average force in previously described DC case, from which I could already assume that the work done is 3/2W.

I then continue to prove it, using one controversial assumption though: I pondered that slowly separating the plates over a long period of time is equal to separating them nearly instantly at the moment when the momentary voltage hits the RMS value.
And for this brief moment, I could assume charge to be conserved;

After doing the same math for the DC case, except using voltage of U/sqrt2, I again end with 3/2W as the result.

Nice to have several conflicting solutions for one problem!

Cheers,

Marko

Marko

Sat May 02 2009, 08:00PM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

Hello,

Here is the ''correct'' answer that I just can't agree with, translated best I could from the original paper:

Charge on the capacitor varies in the following way:

q = Qm cosÏ‰t

Where Ï‰^2 = 1/LC , and C is the momentary capacitance of the capacitor (S is the plate surface)

C = e0S/y

y is distance between the plates. Considering that the frequency has increased Î· times, value Ï‰^2 changes Î·^2 times. Considering that

Ï‰^2 = y/(e0*S*L), distance between the plates changes from y0 to y0*Î·^2 (wheel reinvented.)

Voltage on the capacitor is V = (Qm/C)cosÏ‰t = (y*Qm/e0*S)cosÏ‰t

Electric field between capacitor plates is E = (Qm/e0*S)cosÏ‰t

Then, the force between the plates is F = ((Qm^2)/e0*S)cosÏ‰t

Considering that the force is always positive, and plates are separating slowly, we can use the average force:

F = F = ((Qm^2)/e0*S*2)

The work that was done is A = F(y0*Î·^2 - y0) = (Î·^2 - 1)*((Qm^2)*y0/e0*S*2)

((Qm^2)*y0/e0*S*2) = (Qm^2)/2C0 = W - which is the starting energy,

It's obvious that A = (Î·^2 - 1)W = 3W

I'm about to disappear now, but I'll probably reappear by the end of the following week.

Cheers,

Marko

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