Capacitor

[MedPR]

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When you close the switch, positive charges begin to accumulate on the top plate of the capacitor and negative charges accumulate on the bottom plate.

Why?

I'm having a hard time remembering which way current flows from a battery..

Also, would the capacitor be charged at the same rate if the resistor was on the top of the circuit instead of the bottom? I know that resistors dissipate energy as heat, meaning less energy can get to the capacitor, but does this matter if the capacitor is the first circuit element to be reached by the current? If that doesn't make sense, this is something I found on mcatquestion..

Showing the above circuit, the question asks.

As the capacitor gets charged, the electrical potential energy it gains:

A. is less than the work done by the battery throughout the charging process
B. equals the work done by the battery throughout the charging process.

Answer is A because:

Current flowing through resistors is dissipated as heat so the work done by the battery is converted into the potential energy of the capacitor plus the heat dissipated by the resistor

So I thought (and have been confirmed by my other thread) that the current will flow to the capacitor first, then the resistor second. So why does it matter that the resistor dissipates heat? None of the energy going to the capacitor has been lost to heat yet, so why wouldn't it equal the work done by the battery?

 

[chiddler]

Positive flow always goes from the larger line of the battery representation. You can do something like "bigger is better". Better is +!

No the order doesn't matter. Think of Kirchhoff's law. Going in a full circle will net 0 V. Starting from the emf source,

V - capacitor - resistor = 0

this is equal to

V - resister - capacitor = 0

The voltage drop will be the same in either case.

 

[MedPR]

Positive flow always goes from the larger line of the battery representation. You can do something like "bigger is better". Better is +!

No the order doesn't matter. Think of Kirchhoff's law. Going in a full circle will net 0 V. Starting from the emf source,

V - capacitor - resistor = 0

this is equal to

V - resister - capacitor = 0

The voltage drop will be the same in either case.

So even though none of the electrical potential energy between the battery and the capacitor has been dissipated as heat, the circuit, in a way, anticipates that the resistor is there knows that the resistor will convert some energy to heat.