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New Battery systems => Super capacitors => Topic started by: pomodoro on January 11, 2017, 01:28:42 AM
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There have been a few threads on tackling the problem of where the energy loss really goes when a charged capacitor is used to charge another uncharged capacitor. The energy is not conserved no matter how its done. Usually the thread ends up in another direction. This is a very important question as OU research often uses capacitors.Have a look at the link below, but do not believe the explanation given instead search for the word 'tweet' and read the comment about charge redistribution. That I believe is the correct answer, not the one mentioning radiation or resistive losses.
http://www.edn.com/electronics-blogs/living-analog/4394290/Capacitor-charge-transfer- (http://www.edn.com/electronics-blogs/living-analog/4394290/Capacitor-charge-transfer-)
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The exact same 'loss' occours when the cap is allowed to double in capacity by allowing the plates to come closer.In both cases C*V is always conserved. Then we can decrease the capacity back to the original and we gain back the loss. Obviously work is involved in this step and it will equal to the 'loss' of energy found in the calcs in the first step, but was the loss converted into heat or in some internal potential energy, like in a spring.
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I have the experiment...! https://www.youtube.com/watch?v=2i4wKmyvpb8&t=95s, i hope it serves for you...!
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I have the experiment...! https://www.youtube.com/watch?v=2i4wKmyvpb8&t=95s (https://www.youtube.com/watch?v=2i4wKmyvpb8&t=95s), i hope it serves for you...!
Thanks but what's going on exactly. The video has no explanations.
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There have been a few threads on tackling the problem of where the energy loss really goes when a charged capacitor is used to charge another uncharged capacitor. The energy is not conserved no matter how its done. Usually the thread ends up in another direction. This is a very important question as OU research often uses capacitors.Have a look at the link below, but do not believe the explanation given instead search for the word 'tweet' and read the comment about charge redistribution. That I believe is the correct answer, not the one mentioning radiation or resistive losses.
http://www.edn.com/electronics-blogs/living-analog/4394290/Capacitor-charge-transfer- (http://www.edn.com/electronics-blogs/living-analog/4394290/Capacitor-charge-transfer-)
for me there is no loss, it just like pressure diminishing.because it is seeking for balance. :)