The discussion revolves around understanding the behavior of a 40 µF capacitor in a circuit, particularly its role when no voltage source is explicitly provided. Participants are exploring the implications of initial charge, current flow, and energy calculations in capacitor circuits.
There is ongoing exploration of the circuit's behavior, with participants providing insights into the initial conditions and potential current flow. Some guidance has been offered regarding the nature of current flow in relation to the capacitor's charge and the closing of the switch, but no consensus has been reached on all aspects of the problem.
Participants note that the circuit was presented with certain ambiguities, such as the lack of specified initial voltage or charge values for the capacitors. The discussion also touches on the energy loss in real circuits due to resistance and electromagnetic radiation, which are not typically accounted for in basic circuit problems.
fixedmfb said:
Thats how the circuit was given to me in the question lolphinds said:
No doubt the author just re-purposed an existing circuit drawing, thinking it "good enough".superslow991 said:
Yea this was the initial question-gneill said:
No current would flow until the switch closed. Then current would flow very briefly until the new steady-state is reached.superslow991 said:
Ok so where would the current flow from? since i guess no voltage or emf is present, the current would initially flow from the 40µF?gneill said:
Yes, the 40 µF capacitor has an initial charge so it has an initial emf associated with it.superslow991 said:
Ok thanks a lot.gneill said:
Your thinking is correct. Just remember that at steady state capacitor current goes to zero; the circuit reaches an equilibrium where there are no longer potential differences to drive current into or out of the capacitors. In a series circuit that means all current ceases.superslow991 said:
Ok thanks a lotgneill said:
Didn't want to create another thread but for my first question with the picture of the capacitor circuit I calculated the initial energy with q^2/2C and got 0.313micro joules then calculated the final with C=60 then used the same equation for the final energy q^2/2C=0.208 but I want to know how the energy dropped?gneill said:
In a circuit made of real components the energy would be largely dissipated by the small resistance in the wires, the rest lost to a radiating electromagnetic field (which we're not accounting for when we do circuit problems in this way). The bottom line is that when charges are moved around from capacitor to capacitor some potential energy gets lost. After all, the thing that drives the movement of charge is that it is seeking a lower energy state.superslow991 said:
gneill said:
Yes. A mechanical analogy might be water flowing from one reservoir to to another when a path is opened between them. As the water level lowers in one the other rises, but ultimately the center of gravity of the system as a whole has been lowered somewhat so there has been a net loss in potential energy.superslow991 said:
With the switch open, there's no reason why such a circuit could not be assembled using one initially charged capacitor and three uncharged ones. No current would flow until the switch is closed. The question is fine as posed.epenguin said:
gneill said:
OK right, the electrons on the shown negative plate are held there by the opposite positive charge and can't get away till the switch is closed, I was thinking of something else.