Xref: utzoo sci.electronics:7941 sci.physics:9802 Path: utzoo!attcan!utgpu!jarvis.csri.toronto.edu!mailrus!cs.utexas.edu!uunet!murtoa.cs.mu.oz.au!munnari.oz.au!comp.vuw.ac.nz!massey!ARaman From: ARaman@massey.ac.nz (A.V. Raman) Newsgroups: sci.electronics,sci.physics Subject: A violation of the law of conservation of energy Keywords: paradox Message-ID: <318@massey.ac.nz> Date: 26 Sep 89 04:21:43 GMT Reply-To: ARaman@massey.ac.nz (A.V. Raman) Organization: Massey University, Palmerston North, New Zealand Lines: 45 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! -- /*----------------------------------------------------------------------*/ Anand Venkataraman - Systems group, Computer Center, Massey University, Palmerston North, New Zealand INTERNET: A.Raman@massey.ac.nz Ph: +64-63-69099 x7943 NZ = GMT + 12