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.A capacitor is an electronic component that stores electrical energy in the form of an electric field. It consists of two conductive plates separated by an insulating material, known as a dielectric.
The unit of measurement for capacitance is the farad (F). However, capacitors are commonly measured in smaller units such as microfarads (µF) and picofarads (pF).
A 40µF capacitor circuit works by storing electrical charge on its plates when it is connected to a power source. The amount of charge it can store is determined by its capacitance value and the voltage applied to it. When the circuit is disconnected, the capacitor releases the stored charge, which can be used to power other electronic components.
The purpose of a capacitor in a circuit is to store and release electrical energy as needed. It can be used to filter out unwanted signals, stabilize power supply voltage, and store energy for timing or switching purposes.
The time constant for a 40µF capacitor can be calculated by multiplying the capacitance value (in farads) by the resistance (in ohms) in the circuit. The resulting unit is seconds and represents the time it takes for the capacitor to charge or discharge to 63.2% of its maximum voltage.