TBR Physics Circuits, Capacitor Charge

[EZR]

So the question states:

After Switch 1 has been closed for a long time (with Switch 2 open), the capacitor becomes fully charged. The charge on the capacitor can now be written as:

C*emf*R3 / (R1+R3)

I thought the answer was C*emf --- but I guess you have to account for the resistor's voltage drop and deduct that from the battery?

So then the capacitor can never have the same potential drop as the battery itself when resistors are in the way?

*I've attached a pic of the circuit.

Thanks in advanced!

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[thais]

Voltage is the same across the resistors only when they are connected in parallel. However, note that R1 is actually in series with the battery. Then R3 and the capacitor are in parallel with EMF - voltage drop over R1.
When I get a circuit like this it usually helps me to write it in another way, so that it reminds me that R1 is in series with the battery. * I took a picture of how I would redraw the circuit.

Except I can't figure out how to attach pictures.. hehe :/

 

[MedPR]

So the question states:

After Switch 1 has been closed for a long time (with Switch 2 open), the capacitor becomes fully charged. The charge on the capacitor can now be written as:

C*emf*R3 / (R1+R3)

I thought the answer was C*emf --- but I guess you have to account for the resistor's voltage drop and deduct that from the battery?

So then the capacitor can never have the same potential drop as the battery itself when resistors are in the way?

*I've attached a pic of the circuit.

Thanks in advanced!

Yes. Everytime the current passes through a resistor, there is a voltage drop (V=IR). So by the time you get to the capacitor, the voltage is not the same, unless the capacitor and resistors are all in parallel. Since R1 is in series with C, the answer won't be C*emf. And since R1 and R3 are in parallel, Req=R1*R3/(R1+R3)

And since C=Q/V and V=IR, C=Q/IR and Q=CIR. So Q=CIR3R1/(R3+R1). I=V/R, so Q=VR3R1/(R3+R1)(R1), so Q=VR3/(R3+R1)

 

[milski]

Yes. Everytime the current passes through a resistor, there is a voltage drop (V=IR). So by the time you get to the capacitor, the voltage is not the same, unless the capacitor and resistors are all in parallel. Since R1 is in series with C, the answer won't be C*emf. And since R1 and R3 are in parallel, Req=R1*R3/(R1+R3)

And since C=Q/V and V=IR, C=Q/IR and Q=CIR. So Q=CIR3R1/(R3+R1). I=V/R, so Q=VR3R1/(R3+R1)(R1), so Q=VR3/(R3+R1)

You cannot do that. V=IR is applicable only for resistors. The current through a capacitor is the exponential decay formula that's being discussed in the other thread.

 

[milski]

If R3 was not there, you are correct, the capacitor would be charged to C*emf. That happens because as the capacitor gets charged, the current through it decreases and that makes the voltage drop over the R1 decrease, which in turn increases the voltage drop over the capacitor. With fully charged capacitor, you have 0 current and 0 voltage drop over the resistor, making the voltage drop over the capacitor emf.

In this problem, you have R3. That means that even when no current is flowing through the capacitor, the current through R1 and R3 is emf/(R1+R3). That makes the voltage drop over R1*emf(R1+R3). The voltage drop over the capacitor then has to be emf-R1*emf/(R1+R3)= emf*(R1+R3-R1)/(R1+R3)=emf*R3/(R1+R3). The charge is C times that.

 

[Lunasly]

I'm having trouble with this problem as well. Can anyone else provide an alternate explanation. I don't quite understand the explanations provided above.

 

[pepocho]

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