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SaintJude

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An uncharged capacitor is added to a circuit as follows:

59637.64.h020355.04img01.gif



S4 and S3 are then closed, causing the capacitor to be charged. Which of the following statements is correct?


A.The maximum charge stored in the capacitor would be the same if S3 had remained open.
B. The capacitor would not be charged if S3 had remained open.
C.The maximum charge stored in the capacitor would be lower if S3 had remained open.
D. The maximum charge stored in the capacitor would be higher if S3 had remained open.

Answer: D. UUUUUUmmm???
 

pm1

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An uncharged capacitor is added to a circuit as follows:

59637.64.h020355.04img01.gif



S4 and S3 are then closed, causing the capacitor to be charged. Which of the following statements is correct?


A.The maximum charge stored in the capacitor would be the same if S3 had remained open.
B. The capacitor would not be charged if S3 had remained open.
C.The maximum charge stored in the capacitor would be lower if S3 had remained open.
D. The maximum charge stored in the capacitor would be higher if S3 had remained open.

Answer: D. UUUUUUmmm???

I think I got it! :) How do I load pics here? I think it would be hard to explain without drawing. Thanks!
 

chiddler

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it's because when S3 is closed, voltage splits into two so less max charge possible.

if S3 is open, then half voltage goes to resistor and half goes to capacitor thus less max charge.

edit: wait this is wrong. when circuit is closed, then there is very little/no voltage drop across capacitor so all voltage goes to capacitor. adding resistors should only slow charging, but not low max charge as milski has demonstrated so many times in the past lol

i don't know.
 
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pm1

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An uncharged capacitor is added to a circuit as follows:

59637.64.h020355.04img01.gif



S4 and S3 are then closed, causing the capacitor to be charged. Which of the following statements is correct?


A.The maximum charge stored in the capacitor would be the same if S3 had remained open.
B. The capacitor would not be charged if S3 had remained open.
C.The maximum charge stored in the capacitor would be lower if S3 had remained open.
D. The maximum charge stored in the capacitor would be higher if S3 had remained open.

Answer: D. UUUUUUmmm???

I think it helps a lot to draw it.
Redraw both diagrams: one with S3 open and one with it closed.

http://i.imgur.com/UauD9.jpg

(I hope this image thing works - thanks chiddler)

As you can see when S3 is open the resistors are in series with both the battery and capacitor. The resistors will only cause the capacitor to take longer to be charged and Voltage in series are additive (the voltage drop across the circuit must result the voltage of the battery).
On the other hand, when S3 is closed, there is a parallel component. According to Ohm's law voltage in parallel are the same. Thus, the voltage of the capacitor will be the voltage of the batter minus the voltage drop over R1, R2, and R3. Consequently the voltage will be lower and from C=Q/V since capacitor won't change, Q will also be lower. Thus Q would be higher if S3 remains opened.

Let me know if it makes sense, otherwise I can try to explain in a different way.
 

SaintJude

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pm1, thank you so much. I'm really happy you're here!!

2 questions:

As you can see when S3 is open....Voltage in series are additive (the voltage drop across the circuit must result the voltage of the battery).

Are you saying the voltage of capacitor when S3 is open = voltage of battery ?

On the other hand, when S3 is closed, there is a parallel component. According to Ohm's law voltage in parallel are the same. Thus, the voltage of the capacitor will be the voltage of the batter minus the voltage drop over R1, R2, and R3. Consequently the voltage will be lower and from C=Q/V since capacitor won't change, Q will also be lower.

How did you know capacitor won't change? Is it an intrinsic value of the capacitor?
 

pm1

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pm1, thank you so much. I'm really happy you're here!!

2 questions:



Are you saying the voltage of capacitor when S3 is open = voltage of battery ?



How did you know capacitor won't change? Is it an intrinsic value of the capacitor?

No problem, I'm glad I can help :) You've helped me a lot as well (especially with your recent posts regarding orgo :)

1) Yes.
2) C=kAe/d
 

SaintJude

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Kaplan also says
In the first case (when S3 is open) , under steady state conditions (i.e. in the long run), there will no longer be a current through the circuit. The potential difference across the resistors R1, R2 and R3 is zero from Ohm's law (V = IR = 0(R) = 0.)
What?

That's how they conclude that the potential of battery = that of capacitor.
 

pm1

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Kaplan also says
What?

That's how they conclude that the potential of battery = that of capacitor.

That's true. I'm sorry if something that I said before confused you. Having the resistor with S3 open the resistor will only delay the charging of the capacitor but once the capacitor is charged there is no current going through the resistors. Capacitor stops current at t infinity. After the capacitor is charged all the charges are there and they are kept there.

I guess in regards to what I've said earlier: (the voltage drop across the circuit must result the voltage of the battery). This is true, but once all the charges become stored in the capacitor then no charge goes through the resistor anymore.

Hope it makes sense.
 

chiddler

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edit: i'm wrong. trying to understand why. will add another post later.

I think what pm1 wrote is incorrect.

Lets look at a similar circuit

hdMJZ.png


When switch is closed and current starts flowing, then half voltage goes top side and half goes bottom, right? Ohms law as was mentioned. As the capacitor continues to charge, current in the circuit decreases. Therefore, as current decreases, voltage drop across the resistor decreases. Now that voltage drop across the resistors have decreased, the voltage on the capacitor remains as it would if the resistors were not there because there is no voltage drop in the resistors. It would just take longer to charge.
 
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milski

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edit: i'm wrong. trying to understand why. will add another post later.

I think what pm1 wrote is incorrect.

Lets look at a similar circuit

hdMJZ.png


When switch is closed and current starts flowing, then half voltage goes top side and half goes bottom, right? Ohms law as was mentioned. As the capacitor continues to charge, current in the circuit decreases. Therefore, as current decreases, voltage drop across the resistor decreases. Now that voltage drop across the resistors have decreased, the voltage on the capacitor remains as it would if the resistors were not there because there is no voltage drop in the resistors. It would just take longer to charge.

pm1 has the right explanation. Voltage does not get split - it is the same across both branches of a parallel circuit. Your circuit is not correct though - when S3 is closed, R6 is in parallel with the capacitor only. Thus, whatever the voltage over the capacitor (and its final charge) will be based on the voltage drop across R6 only.
 

chiddler

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pm1 has the right explanation. Voltage does not get split - it is the same across both branches of a parallel circuit. Your circuit is not correct though - when S3 is closed, R6 is in parallel with the capacitor only. Thus, whatever the voltage over the capacitor (and its final charge) will be based on the voltage drop across R6 only.

i wasn't trying to redraw the original circuit. it was just a simpler example circuit.

I don't understand " Thus, the voltage of the capacitor will be the voltage of the batter minus the voltage drop over R1, R2, and R3."
 
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milski

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i wasn't trying to redraw the original circuit. it was just a simpler example circuit.

I don't understand " Thus, the voltage of the capacitor will be the voltage of the batter minus the voltage drop over R1, R2, and R3."

In your sample circuit, the capacitor will charge to the V of the battery.

If you change it by removing the single resistor on top (which does nothing really) and draw another resistor in parallel only to the capacitor, call the new one R4. Now the voltage for the fully charged capacitor will be the same as the voltage over R4, which is the voltage of the battery minus the voltage drops over R1 and R2.
 

chiddler

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In your sample circuit, the capacitor will charge to the V of the battery.

If you change it by removing the single resistor on top (which does nothing really) and draw another resistor in parallel only to the capacitor, call the new one R4. Now the voltage for the fully charged capacitor will be the same as the voltage over R4, which is the voltage of the battery minus the voltage drops over R1 and R2.

why does voltage change in this case?

i know that resistors in series or in parallel to a capacitor (like this) will not alter the max charge for a given voltage. but somehow this is.
 

pm1

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i'd really really like more explanation for this because i don't understand.

please :<

I'll try to help you out.. sorry it has been confusing. I have a good understanding of this but a hard time explaining - which might show not such a strength over this concept after all.. lol
Milsk might be able to help out with his physics, but I'll try.

When S3 is closed you have parallel components, right?
From Ohm's law we know that voltage is the same across each branch. If R1, R2 and R3 were not present the voltage of the battery would be equal to the voltage drop across R6, and consequently, equal to the voltage of the capacitor. They are all branches in a parallel form, hence they all have the same voltage.
However, on this diagram, R1, R2 and R3 are present but the principle holds. The voltage will be the same for each branch in parallel. The way I think of it (Idk how scientifically that it) is that the battery is supplying a certain voltage, but there is a voltage drop (taking away) over R1, R2, and R3. Thus, the total voltage will be the voltage provided by the battery, minus the voltage drops over R1, R2, and R3. Thus, the voltage drop across R6 and Capacitor will be the V from battery minus the voltage drop over R1 R2 and R3. Although the battery gives a higher voltage the resistors besides it will take some "voltage away".

While when S3 is closed, all the components are in series. The resistors will only delay the charging process of the capacitor, but it will not change the capacitor's voltage.

I hope it helps some.. feel free to say you're still confused. it took me a while to figure this out...
 

milski

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why does voltage change in this case?

i know that resistors in series or in parallel to a capacitor (like this) will not alter the max charge for a given voltage. but somehow this is.

One resistors in series or anything in paralel to both the source and the capacitor will not change the max charge. What will change it having at least one resistor in series and then another one in parallel to only the capacitor, effectively creating a circuit allowing the current to flow around the capacitor.

When the capacitor is series with the resistor, the potential across the capacitor will continue to increase until it becomes the same as the potential of the battery - at that point it's fully charged to CV and there is no current flowing through it.

Now, let's add a resistor parallel to it, same resistance as the first one to keep things simple. The two resistors are now in series and form a circuit with the battery - that means that the potential across each of them will be V/2. The capacitor is attached to the ends of only one of them and has to have the same potential as that resistor - V/2. If not, current will flow until the potentials are equalized.
 

chiddler

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I'll try to help you out.. sorry it has been confusing. I have a good understanding of this but a hard time explaining - which might show not such a strength over this concept after all.. lol
Milsk might be able to help out with his physics, but I'll try.

When S3 is closed you have parallel components, right?
From Ohm's law we know that voltage is the same across each branch. If R1, R2 and R3 were not present the voltage of the battery would be equal to the voltage drop across R6, and consequently, equal to the voltage of the capacitor. They are all branches in a parallel form, hence they all have the same voltage.
However, on this diagram, R1, R2 and R3 are present but the principle holds. The voltage will be the same for each branch in parallel. The way I think of it (Idk how scientifically that it) is that the battery is supplying a certain voltage, but there is a voltage drop (taking away) over R1, R2, and R3. Thus, the total voltage will be the voltage provided by the battery, minus the voltage drops over R1, R2, and R3. Thus, the voltage drop across R6 and Capacitor will be the V from battery minus the voltage drop over R1 R2 and R3. Although the battery gives a higher voltage the resistors besides it will take some "voltage away".

While when S3 is closed, all the components are in series. The resistors will only delay the charging process of the capacitor, but it will not change the capacitor's voltage.

I hope it helps some.. feel free to say you're still confused. it took me a while to figure this out...

One resistors in series or anything in paralel to both the source and the capacitor will not change the max charge. What will change it having at least one resistor in series and then another one in parallel to only the capacitor, effectively creating a circuit allowing the current to flow around the capacitor.

When the capacitor is series with the resistor, the potential across the capacitor will continue to increase until it becomes the same as the potential of the battery - at that point it's fully charged to CV and there is no current flowing through it.

Now, let's add a resistor parallel to it, same resistance as the first one to keep things simple. The two resistors are now in series and form a circuit with the battery - that means that the potential across each of them will be V/2. The capacitor is attached to the ends of only one of them and has to have the same potential as that resistor - V/2. If not, current will flow until the potentials are equalized.

Ok!

The major (and shameful :() source of confusion for me was that voltage splits when it goes through two parallel components. So this is, of course, wrong, and it is this realization that helped me understand.

Thanks for the very helpful responses!

i appreciate it :thumbup:
 

johnnydrama

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An uncharged capacitor is added to a circuit as follows:

59637.64.h020355.04img01.gif



S4 and S3 are then closed, causing the capacitor to be charged. Which of the following statements is correct?


A.The maximum charge stored in the capacitor would be the same if S3 had remained open.
B. The capacitor would not be charged if S3 had remained open.
C.The maximum charge stored in the capacitor would be lower if S3 had remained open.
D. The maximum charge stored in the capacitor would be higher if S3 had remained open.

Answer: D. UUUUUUmmm???

The voltage across C is equal to the voltage of the battery minus the voltage drop across R1 + R2 + R3 (let's call that R0).

If both switches are closed, there is current through R0 at steady state, so the voltage across the capacitor drops [ by V * R0 / (R0 + R6) ].

If only the capacitor is connected, the current through R0 is 0, so the voltage drop is R0*0 = 0.

Not sure if that helps.
 

MrNeuro

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So what happens to the voltage across R4 R5 R6 and C if you close all the switches

im asking because wont resistors R1,R2,R3 drop the voltage before reaching the next parallel R or C??? if so the voltages across each parallele won't be the same...???
 

milski

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So what happens to the voltage across R4 R5 R6 and C if you close all the switches

im asking because wont resistors R1,R2,R3 drop the voltage before reaching the next parallel R or C??? if so the voltages across each parallele won't be the same...???

Yes, that's correct. R1 is in series with the rest of the circuit, it will drop the voltage some. Then you have R4 in parallel with everything else - we can ignore it. Then you have R2 in series with the rest of it and so on. The voltage over the capacitor will be the same as the voltage over R6 and will result in the lowest possible charge in this case.
 

MrNeuro

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Yes, that's correct. R1 is in series with the rest of the circuit, it will drop the voltage some. Then you have R4 in parallel with everything else - we can ignore it. Then you have R2 in series with the rest of it and so on. The voltage over the capacitor will be the same as the voltage over R6 and will result in the lowest possible charge in this case.
so for example in this question

the voltage is 120

if R1,R2,R3 are 10 ohms each the voltage drop will be what value X? and that value X will be the voltage over R6 and C1 right?

thanks milski
 

milski

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so for example in this question

the voltage is 120

if R1,R2,R3 are 10 ohms each the voltage drop will be what value X? and that value X will be the voltage over R6 and C1 right?

thanks milski

Depends on R6. :smuggrin:

It will be R6*120/(10+10+10+R6) and yes, that will be the voltage over R6 and C1 (once it's fully charged).
 

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So what is the correct answer? I think it is A because since voltage do not drop amount of charge stored on capacitor should not changed from equation Q=CE(1-e^(t/RC)). E here is electromotive force or maximal V of the battery.
 

johnnydrama

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So what is the correct answer? I think it is A because since voltage do not drop amount of charge stored on capacitor should not changed from equation Q=CE(1-e^(t/RC)). E here is electromotive force or maximal V of the battery.

Wrong. The answer is D.

The voltage across the capacitor is less when S3 is closed due to the current across the first 3 resistors.
 
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