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  #1  
Old 11-20-2017, 04:48 AM
ricards ricards is offline
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Join Date: Mar 2015
Posts: 137
Energy In Capacitor

The Scenario is this:
A charged capacitor of let say 12v 40000uf was connected to another uncharged capacitor of the same value capacitance.

after a while they would balance the energy in both capacitor
the actual voltage on both capacitor is somehow close to 6.09v (which make sense)..

the question now is how do you determine the efficiency of the energy transfer:
considering the calculation of Energy stored in a capacitor
w=0.5*v^2*C , w= 2.88 Joules

do you use the formula above to calculate the energy in 2 capacitor which will result in total of 1.48 Joules (0.74 Joules each capacitor) lose like half to entropy e.g. heat. EM wave etc.. approx. 50% efficiency

or do you use the energy (2.88 Joules) and calculate the voltage by changing the value of capacitance to 80000 uf and come with 8.48 volts and lost x amount of Joules in entropy e.g. heat. EM wave etc.. (1.39 volts worth of charge).. approx 72% efficiency
v=sqrt ( w / (0.5 * 0.08 F))

or are these math only to determine the supposed energy a charged capacitor has... and is it more sensible to think of voltage just as pressure, and the pressure (voltage) dropped by half because the space doubled (capacitance)..
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Last edited by ricards; 11-20-2017 at 05:09 AM.
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  #2  
Old 11-20-2017, 06:30 AM
bistander bistander is online now
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Join Date: Apr 2015
Posts: 941
Capacitor energy

Quote:
Originally Posted by ricards View Post
The Scenario is this:
A charged capacitor of let say 12v 40000uf was connected to another uncharged capacitor of the same value capacitance.

after a while they would balance the energy in both capacitor
the actual voltage on both capacitor is somehow close to 6.09v (which make sense)..

the question now is how do you determine the efficiency of the energy transfer:
considering the calculation of Energy stored in a capacitor
w=0.5*v^2*C , w= 2.88 Joules

do you use the formula above to calculate the energy in 2 capacitor which will result in total of 1.48 Joules (0.74 Joules each capacitor) lose like half to entropy e.g. heat. EM wave etc.. approx. 50% efficiency

or do you use the energy (2.88 Joules) and calculate the voltage by changing the value of capacitance to 80000 uf and come with 8.48 volts and lost x amount of Joules in entropy e.g. heat. EM wave etc.. (1.39 volts worth of charge).. approx 72% efficiency
v=sqrt ( w / (0.5 * 0.08 F))

or are these math only to determine the supposed energy a charged capacitor has... and is it more sensible to think of voltage just as pressure, and the pressure (voltage) dropped by half because the space doubled (capacitance)..
Hi ricards,

Quote:
after a while they would balance the energy in both capacitor
You got that right. So each cap will have 1.44J. And voltage is 8.48V. That assumes ideal capacitors and circuit, meaning no resistance internal to the caps or in the connections, and no switching loss like an arc. Those are the only mechanisms for loss. So any difference in measured energy (using a voltmeter and the formula) and the ideal calculation of energy is lost as heat through the resistance (I^2*R).

Regards,

bi
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Old 11-20-2017, 09:39 AM
ricards ricards is offline
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Join Date: Mar 2015
Posts: 137
Hi Bistander,

thanks for the quick reply, I was actually trying to calculate the overall efficiency of the energy transfer including losses from entropy and gain from magnetic field collapse.. It got up from 72% eff. to 88%..

I'd like to try now and see if increasing the amount of copper coils would give a further gain from magnetic field collapse..
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