Whos right? - capacitor to capacitor discharge.

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    Capacitor Discharge
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SUMMARY

The discussion centers on the efficiency loss when discharging one capacitor into another, specifically addressing the "Two Capacitor Problem." The original claim of a 50% energy loss is challenged, with the conclusion that energy loss primarily arises from the equivalent series resistance (ESR) of the capacitors when there is a voltage difference. The mathematical explanation provided demonstrates that when two identical capacitors, C1 and C2, are connected, the energy stored in each after equalization is indeed one-fourth of the original energy, confirming the conservation of energy principle.

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  • Understanding of capacitor basics, including capacitance and voltage.
  • Knowledge of energy storage in capacitors, specifically the formula E = 1/2 C * V^2.
  • Familiarity with equivalent series resistance (ESR) in capacitors.
  • Basic principles of electrical charge conservation.
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  • Study the Two Capacitor Problem in detail, focusing on energy conservation principles.
  • Explore the impact of equivalent series resistance (ESR) on capacitor performance.
  • Learn about energy transfer efficiency in capacitor circuits.
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Electrical engineers, electronics students, and hobbyists interested in capacitor behavior and energy transfer in circuits will benefit from this discussion.

spaceball3000
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I read on another forum, some person says (read below) when you charge one cap, then discharge into another, there is 50% efficiency loss?

I think this is wrong, as the only losses I can think of, are due to the caps ESR when there is a voltage difference between the caps. Can someone help show his\her Math is wrong..


Also you lose 1/2 the energy in this process.

I can illustrate with a little math

Given capacitors C1 and C2 of the same size (C). One charged to V and one uncharged then the energy stored is

1/2 C * V *V

If you hook these two capacitors in parallel then they will settle (after a current pulse) to a voltage in each of 1/2 V. The energy in each will then be

1/2 C * 1/2 V * 1/2 V

or 1/4 of the original energy in each capacitor for 1/2 of the original energy.​
 
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This is the famous Two Capacitor Problem. You have two identical capacitors separated by an open switch. One is charged to V. Then the switch is closed. After the voltage has equalized what is the resultant voltage on both capacitors?

The answer depends on conservation of charge or conservation of energy.

Charge (coulombs), Q = C*V
Energy (joules), E = C*V^2/2

Obviously they can't both apply.
 

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