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BASIC STUFF... RC Parallel in DC Volt Src

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Newbie level 2
Dec 19, 2012
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So kind of confused with RC circuits in DC right now.

Let me know if these assumptions are correct:

1. Capacitors is resistant to change in voltage so when it is fully charged it acts as an open and technically the capacitor tries to hold the potential across it.

2. if that is the case an RC circuit in parallel with a voltage source. If C becomes fully charged why is the potential changing as Vin increases? I know when you do circuit problems C becomes an open after a certain amount of time and Vin=Vout but if the Cap holds that potential why is the potential increasing across the Cap?


Let me answer your question with a question: What is "fully charged"? A capacitor is not a battery.

The relationship you need to understand is:

V=q/C (q=charge, c=capacitance)

The more charge you add to a cap, the higher the voltage. The cap doesn't limit the voltage across it in any way (theoretically speaking)

If the capacitor did not have this charge soaking up effect, the instant after you turned the voltage on the current flow through the resistor would charge the capacitor up to Vin. Its because the current is charging the capacitor, you get an big voltage drop across the resistor, so only a small volt drop across the capacitor at first. But as the capacitor charges, the current falls (I= {vin -Vcap}/R), so the capacitor takes even longer to get to Vin.

Okay I get that the capacitor is not a battery. So a fully charged capacitor, this time in series with a resistor, would be an open circuit with a potential across the capacitor as v = q/c ? so no current will flow through the fully charged capacitor because of the relationship of i = c dv/dt but will the potential across the capacitor increase as you increase Vin?

Yes, the voltage will increase.

And, again, What is a "fully charged" capacitor? There is no limit to how much charge a cap can store.

To the Ineffable All,

I see a lot of verbiage about a capacitor being "charged" and "fully charged". What is the capacitor being "charged" with? Inquiring minds would like to know, because I think that is a source of confusion.


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