Xref: utzoo sci.electronics:7960 sci.physics:9813 Path: utzoo!attcan!utgpu!jarvis.csri.toronto.edu!mailrus!ncar!husc6!m2c!wpi!jhallen From: jhallen@wpi.wpi.edu (Joseph H Allen) Newsgroups: sci.electronics,sci.physics Subject: Re: A violation of the law of conservation of energy Keywords: paradox Message-ID: <4298@wpi.wpi.edu> Date: 27 Sep 89 03:47:43 GMT References: <318@massey.ac.nz> Reply-To: jhallen@wpi.wpi.edu (Joseph H Allen) Organization: Worcester Polytechnic Institute, Worcester, MA. USA Lines: 34 In article <318@massey.ac.nz> ARaman@massey.ac.nz (A.V. Raman) writes: Circuit: +--C--V--R--+ I assume that initially no charge on C when the | | circuit is connected. +-----------+ >which implies E = C * square(V) / 2 >But note that the energy expended in R does not depend on R itself. Yes. >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? An impossible situation. What this means in the ideal case is that some amount of energy is being transfered in 0 time. There is no V which can do this (even if the wires are superconducting). Also, if V is really strong, a superconducting wire would stop being a superconductor since the magnetic field is large. >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? Half the energy transfered ALWAYS becomes heat. In any "clutch" I.E., a device which transfers energy from one side to another until both sides are equal, 1/2 the energy goes into heat. This is true for both electrical systems and mechanical ones. If you have one flywheel which is spinning and one which isn't and you connect them, 1/2 the energy of the system is lost to heat. No matter how you connect them (I.E., no matter what R is).