RC circuit question

[chiddler]

1. I'm looking at a battery with V1, then a resistor and a capacitor both parallel, R1 and C1.

I know that the potential difference across R1 and C1 do not change over time. My question is regarding current. Both before and after charging the capacitor, how does the current look like in both R1 and C1?



2. Battery, V1, resister and capacitor are in series, R1 and C1.

I know that the charge of the capacitor does not change due to resistors. It would just take longer to fill it with charge. Actually, this sounds wrong. This is wrong, right?

My question was does adding resistance reduce the charge of the capacitor?

C = q/V. A resister must have a voltage drop. V down, q down. Less charge.

helps to visualize.

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[milski]

1: You can consider them independently (as long as the power source is not limited in any way, which it usually is not) - the current through the resistor is constant, V/R, the current through the capacitor decays in the same way as if the resistor was not there.

2: There should be a recent post on it anyway. The capacitor always charges to the same value. The more it gets charged, the higher the voltage drop over it. When it's fully charged there is no current flowing, the voltage drop over the resistor is 0 and the voltage drop over the capacitor is the same as if the resistor was not there - v1. The only thing that the resistor does is limit the maximum current in the beginning which leads to a longer time to fully charge the capacitor.

You are not going to charge the capacitor completely in that case.

 

[chiddler]

all you wrote makes sense, but i'm not understanding one thing:

"the voltage drop over the resistor is 0"

why is that?

 

[milski]

all you wrote makes sense, but i'm not understanding one thing:

"the voltage drop over the resistor is 0"

why is that?

U=IR, if the current is 0, the voltage drop is zero. That is true only for resistors, you cannot say the same thing for the capacitor next to it.

 

[chiddler]

U=IR, if the current is 0, the voltage drop is zero. That is true only for resistors, you cannot say the same thing for the capacitor next to it.

The current is 0 only after it's charged though. Before charging, when current is at its peak, why can't there be a drop?

And since this drop occurs only during charging, this is why I'd expect a lower V on the capacitor.

 

[milski]

The current is 0 only after it's charged though. Before charging, when current is at its peak, why can't there be a drop?

And since this drop occurs only during charging, this is why I'd expect a lower V on the capacitor.

Right, that's only when the capacitor is charged. The voltage drop over the resistor is IR, over the capacitor V-IR. As I->0 V-IR->V.

 

[chiddler]

Right, that's only when the capacitor is charged. The voltage drop over the resistor is IR, over the capacitor V-IR. As I->0 V-IR->V.

huh. interesting. seems paradoxical. because it's low V, then high V because...and the fireworks.

this makes sense now, but loosely. i'm not sure what questions to ask.

i'm going to look at this thread again tomorrow, i think a fresh look will help understanding.

thank you very much.

 

[milski]

huh. interesting. seems paradoxical. because it's low V, then high V because...and the fireworks.

this makes sense now, but loosely. i'm not sure what questions to ask.

i'm going to look at this thread again tomorrow, i think a fresh look will help understanding.

thank you very much.

It's a limit. You can prove that |V-Vc|<&#949; for arbitrary small &#949;.

It starts low and keeps increasing. It will stop changing only when it is exactly V, so it will continue getting closer and closer to V.

 

[chiddler]

It's a limit. You can prove that |V-Vc|<&#949; for arbitrary small &#949;.

It starts low and keeps increasing. It will stop changing only when it is exactly V, so it will continue getting closer and closer to V.

yesh i understand the math very well.

whoah wait. the current eventually goes to 0. The voltage increases because of the current decrease.

Which is why the resistor only provides that initial resistance to slow down but not reduce charge? if this is correct then i understand completely :-3

 

[milski]

Yes, that's correct.

 

[Class1P]

In the first picture there is current flowing to the capacitor based on the resistance of the resistor. The voltage is the same in both the resistor and the capacitor.The current flowing to it will be less than it would in figure 2 since the current in figure 2 is constant throughout the circuit (during charging)

Once the capacitor is charged in Fig. 1 the current will no longer flow to that portion of the circuit and will flow entirely through the resistor

In figure 2 there is a constant current during charging but the resistor causes a voltage drop before the capacitor, causing the charge flow to occur slower to the plate of the capacitor. Once the capacitor is charged, no current will flow through the system.

So do they both charge in about equal time since one is lacking in current and the other is lacking in voltage?

 

[milski]

In the first picture there is current flowing to the capacitor based on the resistance of the resistor. The voltage is the same in both the resistor and the capacitor.The current flowing to it will be less than it would in figure 2 since the current in figure 2 is constant throughout the circuit (during charging)

Once the capacitor is charged in Fig. 1 the current will no longer flow to that portion of the circuit and will flow entirely through the resistor

No. The current through the resistor is constant, U/R. The current through the capacitor follows the standard formula for current through a charging capacitor. For all practical purposes the two elements do not affect each other.

In figure 2 there is a constant current during charging but the resistor causes a voltage drop before the capacitor, causing the charge flow to occur slower to the plate of the capacitor. Once the capacitor is charged, no current will flow through the system.

So do they both charge in about equal time since one is lacking in current and the other is lacking in voltage?

The current is not constant in this case since the capacitor will limit it. The voltage drop over the capacitor will continuously increase so I'm not going to call any formulas for its intermediate values but it starts with whatever you would calculate if the voltage drop over the capacitor is 0 and ends with 0 when the capacitor is charged.

The time will certainly be different.

 

[chiddler]

Hold on milski I think you might be mistaken about my question #1.

click

question is:

Immediately after this circuit is activated (ie, negligible charge on capacitor), what is current through R1?

Answer is I = V/R1

"Initially there is no voltage drop across the capacitor and therefore there is no voltage drop across R2 the instant that the circuit is activated. That is to say that the moment the circuit is activated, all the current goes to the capacitor (following the path of least resistance) rather than through R2."

Specifically, you said that you can consider them independently in both before/after cases. This answer suggests that you have to consider them dependently.

 

[milski]

That does not contradict anything I've said.

In your new circuit you have one resistor in series and one in paralel with the capacitor.

With non-charged capacitor you'll have 0 voltage drop over it and you can calculate both currents for both resistors and from there the maximum current for the capacitor.

For the fully charged case, you won't any current flowing through the capacitor. You can calculate the voltage drops over the two resistors and from there the voltage drop over the capacitor - same as the one over the resistor paralel to it. It will be less than the V of the source in this case.

 

[chiddler]

That does not contradict anything I've said.

In your new circuit you have one resistor in series and one in paralel with the capacitor.

With non-charged capacitor you'll have 0 voltage drop over it and you can calculate both currents for both resistors and from there the maximum current for the capacitor.

For the fully charged case, you won't any current flowing through the capacitor. You can calculate the voltage drops over the two resistors and from there the voltage drop over the capacitor - same as the one over the resistor paralel to it. It will be less than the V of the source in this case.

oh i see.

thanks again.