# Capacitor energy loss

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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Dale
Mentor
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 dont really understand how it's lost to heat. Is it b/c of the electrons?

Dale
Mentor
All of the wires have some resistance. As current moves from one capacitor to the other the wires heat up.

berkeman
Mentor
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.