27-03-2013, 01:53 PM
Sorry folks, that's not right!
If you have a charged capacitor, and you connect it in parallel with a discharged capacitor, charge will balance, and energy HAS to be removed. You can't do it 'losslessly' and end up with a new equilibrium.
Charging 1µF to 10V, you have Q = CV = 1µF x 10V = 10µC of charge. Energy = ½CV² = 50µJ
If you connect this to a discharged 1µF capacitor, some of the electrons will flow around. The first capacitor will discharge, the second will charge. It'll balance when the two capacitors share the same voltage. None of the electrons can 'disappear', so charge conservation applies.
The new total capacitance will be 2µF; the new voltage will be 5V. Charge will be Q = CV = 10µC exactly as before. However, energy = ½CV² = 25µJ so half has been lost!
If you parallel-up the capacitors by a resistor and wait for several time constants, equilibrium will have been nearly achieved, and it's easy to see that heat will be generated (any current flowing through a resistor creates heat). It turns out that the total heat generated, if you wait to infinity, is independent of resistor value so even doing the parallelling with a knife switch will give the same energy loss from the capacitors. It's unavoidable.
If you parallel-up using an inductor ("lossless"), and wait a bit, the same applies - the capacitor voltages will equalise at 5V, the energy in the capacitors will be 25µJ, and you'll have lost some. But in this case, there will be significant current flowing in the inductor and the energy 'lost' from the capacitors will exist in the inductor. If you wait a bit longer, the first capacitor voltage will fall below 5V, the second will rise above 5V, and not long after, you'll find that the first has fallen to 0V, the second now is charged to 10V, there is no current in the inductor, charge has been conserved, and all the energy is back in the capacitors only the other way round! At all times during this oscillation, charge will be conserved (C1V1 + C2V2 = constant), but energy in the capacitors alone won't be constant.
If you have a charged capacitor, and you connect it in parallel with a discharged capacitor, charge will balance, and energy HAS to be removed. You can't do it 'losslessly' and end up with a new equilibrium.
Charging 1µF to 10V, you have Q = CV = 1µF x 10V = 10µC of charge. Energy = ½CV² = 50µJ
If you connect this to a discharged 1µF capacitor, some of the electrons will flow around. The first capacitor will discharge, the second will charge. It'll balance when the two capacitors share the same voltage. None of the electrons can 'disappear', so charge conservation applies.
The new total capacitance will be 2µF; the new voltage will be 5V. Charge will be Q = CV = 10µC exactly as before. However, energy = ½CV² = 25µJ so half has been lost!
If you parallel-up the capacitors by a resistor and wait for several time constants, equilibrium will have been nearly achieved, and it's easy to see that heat will be generated (any current flowing through a resistor creates heat). It turns out that the total heat generated, if you wait to infinity, is independent of resistor value so even doing the parallelling with a knife switch will give the same energy loss from the capacitors. It's unavoidable.
If you parallel-up using an inductor ("lossless"), and wait a bit, the same applies - the capacitor voltages will equalise at 5V, the energy in the capacitors will be 25µJ, and you'll have lost some. But in this case, there will be significant current flowing in the inductor and the energy 'lost' from the capacitors will exist in the inductor. If you wait a bit longer, the first capacitor voltage will fall below 5V, the second will rise above 5V, and not long after, you'll find that the first has fallen to 0V, the second now is charged to 10V, there is no current in the inductor, charge has been conserved, and all the energy is back in the capacitors only the other way round! At all times during this oscillation, charge will be conserved (C1V1 + C2V2 = constant), but energy in the capacitors alone won't be constant.
