One capacitor charging another capacitor

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    Capacitor Charging
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In the discussion about charging one capacitor (C1) with another (C2), the voltage on C2 is calculated to be 8V after charge redistribution. The conversation highlights the complexities of ideal versus real-world scenarios, emphasizing that energy losses must be considered, even in theoretical models. Participants debate the implications of zero resistance and inductance, arguing that without these factors, oscillations and energy redistribution would be infinite and undefined. The conclusion stresses that while the initial question is theoretically flawed, understanding charge conservation and capacitance relationships is crucial for solving such problems. Ultimately, the discussion reveals the intricacies of capacitor behavior in idealized conditions versus practical applications.
  • #31
NascentOxygen said:
Alas... I stayed on to give you the benefit of the doubt, in case you had something insightful to say.

I am absolutely dumbfounded that you do not find my conclusion above fascinating.

I suppose we live in an age where everyone wants a soundbite, spoon fed to them.

The conclusion, that it doesn't matter how the energy is lost, it is the same amount of energy, is fascinating. In fact, I suppose it is a form of 'Gibbs free energy' type calculation, in which we look at the 'before' and 'after' state and recongise that both are defined, so the route to the end-point is immaterial.

The example given above with the spring is similarly so. Imagine you drop a mass on a spring, through lossy fluids, magnetic fields, whathaveyou that might impede its decent, yet at the end, once equilibrium is achieved, exactly the same amount of potential energy would have been expended.

I think that is fascinating. You don't? Time for you to move on to graze for your next sound-bite...
 
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  • #32
I would like to go right back to your first post and do what I think the question is asking.
I am going to assume that when C1 is connected to C2 they are in parallel... that is a sensible way to use one capacitor to put charge on another.
First of all C1 is 1μF charged to 10V so energy stored is 0.5 x 1 x 10^-6 x 10^2
This gives 5.0 x 10^-5J
also the charge on C1 = C x V = 1 x 10^-6 x 10 = 1 x 10^-5 C
When C1 is connected (in parallel) to C2 the combined capacitance = C1 + C2 = 1.25x10^-6 F
The charge in the combination remains the same (1x10^-5C)
therefore the Voltage across this parallel combination is V =Q/C = 1x10^-5/1.25x 10^-6 = 8V
The energy stored on the combination is 0.5xCxV^2 = 0.5 x 1.25x10^6 x 8^2 = 4 x 10^-5J
So there has been an energy loss of 5x10^-5J - 4x10^-5J = 1x10^-5J
In an earlier post I explained how energy is lost when charge flows on to or from a capacitor.
Resistance of any connecting wire, sparking at the switch and in the last resort electromagnetic radiation (probably radio waves).
This radiation can be detected on a radio in the vicinity, it can be shown on an oscilloscope with a pickup coil of a few hundred turns connected to the input of the CRO
 

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