What Causes Energy Loss in Capacitor Circuits?

AI Thread Summary
In capacitor circuits, energy loss occurs primarily due to resistance in the wires, which generates heat as current flows between capacitors. When two unequally charged capacitors are connected, the total energy decreases because some energy is dissipated as heat during the voltage equalization process. To understand this energy loss, it is suggested to introduce a real resistance in series and analyze the power lost using differential equations. Adjusting the resistance value can provide insights into how it affects current, time constants, and overall energy loss. Calculating these factors can clarify the mechanisms behind the observed energy loss in the system.
eyehategod
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Circuit 3(connected to a 3v battery,C2 and C3 in series, C1 and C23 in parallel)
C2
C1
C3




c1=c2=c2=1F


Circuit 4(Two unequally charged capacitors are connected to each other,from setup above)


C1 C2


In circuit 4, after the capacitors are reconnected, the total energy of the system is less than before they were connected. what happens to the missing energy?can anyone help me out? is it lost to heat?
 
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eyehategod said:
what happens to the missing energy?can anyone help me out? is it lost to heat?
Yes, exactly.
 
thats were I get stuck b/c I don't really understand how it's lost to heat. Is it b/c of the electrons?
 
All of the wires have some resistance. As current moves from one capacitor to the other the wires heat up.
 
eyehategod said:
Circuit 3(connected to a 3v battery,C2 and C3 in series, C1 and C23 in parallel)
C2
C1
C3




c1=c2=c2=1F


Circuit 4(Two unequally charged capacitors are connected to each other,from setup above)


C1 C2


In circuit 4, after the capacitors are reconnected, the total energy of the system is less than before they were connected. what happens to the missing energy?can anyone help me out? is it lost to heat?


Thread moved to Homework Help forums (even if it is coursework and not homework, it belongs here and you will need to show some work).

This is a classic question in EE and physics. To figure it out, the best thing is to start with some real resistance in series between the two capacitors as you close the switch between them. Calculate the power lost in the resistor as the two voltages equalize (use a differential equation, with appropriate boundary conditions). Now change the value of the resistor, making it smaller. What effects does this have on the currents and time constants? What about the power lost? See any clues here?

Show us those calculations, and we can talk about it more if you aren't seeing the final answer to your original question.
 

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