Xref: utzoo sci.electronics:7985 sci.physics:9834
Path: utzoo!attcan!utgpu!jarvis.csri.toronto.edu!mailrus!accuvax.nwu.edu!tank!mimsy!fe2o3!michael
From: michael@fe2o3.UUCP (Michael Katzmann)
Newsgroups: sci.electronics,sci.physics
Subject: Re: A violation of the law of conservation of energy
Keywords: paradox
Message-ID: <234@fe2o3.UUCP>
Date: 28 Sep 89 17:05:07 GMT
References: <318@massey.ac.nz>
Reply-To: michael@fe2o3.UUCP (Michael Katzmann)
Followup-To: sci.electronics
Organization: Rusty's BSD machine at home
Lines: 66

In article <318@massey.ac.nz> ARaman@massey.ac.nz (A.V. Raman) writes:
   >This problem has puzzled me for 2 years
   >Can someone please help me out:
   >
   >                       ________________
   >                      |                |
   >                   V ---             ----- C
   >                  ---------          -----
   >                      |                |
   >                       -----RRRRR------
   >
   >Consider the RC circuit above.
   >
   >The aim is to determine an equation for the energy expended in the resistor
   >R at t = infinity.
   >
   >If we denote the energy at t = infinity by E, then
   >E = integral from t = 0 to t = infinity (power (t) dt)
   >where power = R * square (i)
   >since i = f(t) = V/R exp (-t/RC)
   >E = integral from t = 0 to t = infinity (square (V) / R exp (-t/RC))
   >
   >which implies E = C * square(V) / 2
   >which is perfectly logical considering the fact that E = total energy -
   >energy stored in the capacitor.
   >
   >But note that the energy expended in R does not depend on R itself.
   >
   >That won't be quite as astonishing if not for the following reduction
   >of the above circuit.  What if R = 0, V = 1 and C = 1 for example?
   >
   >On creation of such a circuit, the charge that will flow out of V into C
   >is 1 coulomb.
   >By definition then, the energy transferred from V to C is
   >charge * pressure = 1V * 1C = 1 joule
   >But what is the energy stored in the capacitor?
   >It is half C*square(V) = half * 1 * square(1) = half joule.
   >Where has the remaining half joule gone?
   >Beats me!
   >


Oh but we forgot something didn't we!  To charge the capacitor (charge to
flow), we have a current! A flow of current creates a magnetic field, which
stores energy (1/2 L * I * I). Those lines between the plates are allowed
to have zero resistance but as all RF people know even wire has inductance!
Without the resistor we have an LC oscillator. The energy stored in the
in the inductor (wire) will be added to the capacitor and then returned to
the inductor. The voltage source will just look like DC offset and a short 
to the oscillation!


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