Hi Andy, Jim and all,
This is a great question, because it tells a lot about the
background of the person answering, even if the answer is "correct".
The engineer will say that most of the "missing" 25 uJ will be
dissipated as heat in the resistance in the connecting wire.
The physicist, assuming a resistance-free wire, will talk of energy
radiated from the loop in a damped oscillation.
The mathemetician, assuming a perfect connection except for inductance,
speaks of steady state oscillation.
And the philosopher, assuming an ideal connection, finds the problem
analogous to 'irresistable force vs. immovable object', and concludes
that the stated conditions are inconsistent.
73,
Stewart KK7KA
----- Original Message -----
From: "Andy talbot" <[email protected]>
To: <[email protected]>
Sent: Friday, November 08, 2002 3:57 PM
Subject: RE: LF: To Ponder over the weekend
Can tell you come from a university background Jim, do you ask that of
applicants for courses at herts.ac?
I was asked exactly the same question when applying for Southampton Uni. in
1976.
Andy G4JNT
-----Original Message-----
From: James Moritz [SMTP:[email protected]]
Sent: 2002/11/08 11:52
To: [email protected]
Subject: Re: LF: To Ponder over the weekend
Dear Andy, LF group,
At 09:38 08/11/2002 +0000, you wrote:
>This simple puzzle caused more discussion in my office this morning than it
>had any right to....
Here is another conundrum along similar lines - you have two identical,
loss-free 1uF capacitors, one is charged to 10V, the other is discharged.
You then connect the two in parallel so the total C is now 2uF - charge
will flow from the charged capacitor into the discharged capacitor.
Assuming charge Q is conserved, and that Q=CV, the voltage must now be 5V.
But the stored energy in a capacitor =1/2CV^2, so with the single charged
capacitor, the stored energy is 1/2 x 1u x 100 = 50uJ, while with both
capacitors in parallel it is only 1/2 x 2u x 25 = 25uJ. So where has the
other 25uJ gone?
Cheers, Jim Moritz
73 de M0BMU
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