mbe@dde.uucp (Martin Berg) (06/01/89)
vaso@mips.COM (Vaso Bovan) writes: >A Paradox of Capacitor Energy Storage >I've heard several competing answers to this paradox. None is entirely >satisfactory: >Consider an ideal 2uF (for computational ease) capacitor charged by a 10 volt >source. Eventually, the energy stored is (1/2)*CV^2=100 joules. ^^^^^^^^^^^^^^^^^^^^^ You forgot that the 'u' in 2 uF stands for 'micro', that is 2*10^-6 F. So the energy stored is 100 uJ (100*10^-6 J). But this does not affect your 'Paradox of Capacitor Energy Storage'. >Consider the capacitor to be isolated from the voltage source, and then >directly shorted across an identical (ideal) capacitor. Eventually, the >voltage across each capacitor will be 5V. Now, there are two equally >charged capacitors, each storing (1/2)*CV^2= 25 joules, for a total of >of 50 joules. What happened to the other 50 joules ? The current in a capacitor is defined as: I = - dV/dt * C (that is: change in voltage pr. sec. times the capacity, forget the sign). This means that you cannot change the voltage across a capacitor instantaneously : this would inply infinite current. It also means that when you short your two capacitors across each other, then the current will rise to an level where internal resistance and wire resistance is not neglectable. When this happens then you loose energy in form of heat. Also: when you start shorting the capacitors, you will get a spark - this is also a waste of enery. If you use an inductor when connecting the capacitors then you would not loose energy (ok, ok a little if the components is not ideal). But then the resulting voltage would not be 5 V - it would be 7.07 V (for energy preservation). By the way: I'm not so sure that the resulting voltage would be 5 V in your example. -- mbe@dde.dk | "The answer is 42" or | D. Adams ..uunet!mcvax!enea!dkuug!dde!mbe |
siegman@sierra.Stanford.EDU (Anthony E. Siegman) (06/02/89)
This problem is sometimes known as the "Case of the Nefarious Joule Thief". Put a small resistor, or a small inductance, or both, in series with the switch; solve the circuit equations for the transient behavior starting when the switch closes; and examine for physical significance, considering particularly the case when the series R or L tend to zero
fransvo@maestro.htsa.aha.nl (Frans van Otten) (06/02/89)
Martin Berg writes: >Vaso Bovan writes: > >>A Paradox of Capacitor Energy Storage > >By the way: I'm not so sure that the resulting voltage would be 5 V >in your example. Of course it would be 5 volt. The following formula applies to capacitors: q = C*u (the charge is the capacitance times the voltage) Math for the problem (both capacitors have capacitance C): Before: q1 = C * 10 V = 10C coulomb q2 = C * 0 V = 0C coulomb Total charge 10 coulomb. after: q1 = C * 5 V = 5C coulomb q2 = C * 5 V = 5C coulomb Total charge 10 coulomb. -- Frans van Otten | fransvo@maestro.htsa.aha.nl or Algemene Hogeschool Amsterdam | fransvo@htsa.uucp or Technische en Maritieme Faculteit | [[...!]backbone!]htsa!fransvo
abali@parts.eng.ohio-state.edu (Bulent Abali) (06/02/89)
In article <147@sierra.stanford.edu> siegman@sierra.UUCP (Anthony E. Siegman) writes: >Put a small resistor, or a small inductance, or both, in series with >the switch; solve the circuit equations for the transient behavior >starting when the switch closes; and examine for physical >significance, considering particularly the case when the series R >or L tend to zero Exactly. A resistor R can be assumed in series with the switch, and the energy dissipated on R can be derived (easy). The resulting equation will show that the energy dissipated is R independent. One half of the original energy of the system will be lost on R, regardless of its value. If L is assumed instead of R, we can arrive at the same conclusion except that the 1/2 energy will not be lost, but will be stored in the inductor regardless of its value. -=- Bulent Abali Ohio State Univ., Dept.of Electrical Eng. 2015 Neil Av. Columbus, Ohio 43210 abali@baloo.eng.ohio-state.edu
siegman@sierra.Stanford.EDU (Anthony E. Siegman) (06/03/89)
There's a mechanical analog to the switched-capacitor "where did half the energy go" problem. A slow conveyor belt rolls past a loading chute which dumps dirt or coal or something on the belt at a fixed rate dm/dt (m=mass of dirt dumped). The dirt is instantaneously accelerated to the belt's steady velocity v as it hits the belt's surface. Force required to keep pulling the belt (no friction) = f = dp/dt where dp = dm v = momentum added to the mass dm. Work done in time dt = f dx = dm v^^2 (using dx = v dt). But kinetic energy added to (or carried by) the dirt is only (1/2) dm v^^2. Where's the other half?