Final charge on capacitor for an RC circuit

[unseeingdog]

Homework Statement

Homework Equations

$q = CE(1 - e^{-t/RC})$

The Attempt at a Solution

I assumed that, since the problem specifies that sufficient time has passed, it meant to say that enough time passed. thus making the exponential term in the equation go down to 0, and the charge in the capacitor simply $CE$, but the book says the answer is (f), and I don't have any idea how.

P.S: Sorry for the title, I accidentally wrote "Optics", and now it seems I can't change it.

 

[gneill]

There's no problem statement, and your image is essentially empty.

 

[unseeingdog]

There's no problem statement, and your image is essentially empty.

Sorry, I'm trying to fix it. Something went wrong with the attachment

 

[ehild]

Homework Statement

View attachment 204491

Homework Equations

$q = CE(1 - e^{-t/RC})$

The equation is not true if the capacitor is not connected directly to the source.
The capacitor is connected to the resistor R2. How is the capacitor voltage related to the voltage across R2?

 

[gneill]

P.S: Sorry for the title, I accidentally wrote "Optics", and now it seems I can't change it.

I've fixed your title for you

 

[unseeingdog]

I've fixed your title for you

Oh, great! Thanks.

 

[unseeingdog]

The equation is not true if the capacitor is not connected directly to the source.
The capacitor is connected to the resistor R2. How is the capacitor voltage related to the voltage across R2?

Are they equal? Wouldn't the voltage in $R_2$ eventually drop to 0 because of the current going down, or does that equation for the current also not apply?

 

[ehild]

Are they equal? Wouldn't the voltage in $R_2$ eventually drop to 0 because of the current going down, or does that equation for the current also not apply?

Yes, the voltages are equal as the capacitor and R2 are connected parallel. But why do you think that the current is zero through R2? The source, the two resistors and the switch make a closed loop. Current will flow forever through the resistors.

 

[unseeingdog]

Yes, the voltages are equal as the capacitor and R2 are connected parallel. But why do you think that the current is zero through R2? The source, the two resistors and the switch make a closed loop. Current will flow forever through the resistors.

I thought so because of the equation for the current, namely $i =\left(\frac E R \right) (e^{-t/RC})$, but I suppose that one doesn't apply either, since it's derived from the other one I mentioned.

 

[unseeingdog]

Nevermind. It was a million times easier than I thought. I just had to do $iR_2 = \frac q C$, $i$ being $\ \frac {E}{R_1 + R_2}$. Thanks for the help.

 

[ehild]

Nevermind. It was a million times easier than I thought. I just had to do $iR_2 = \frac q C$, $i$ being $\left (\frac {E}{R_1 + R_2} \right)$. Thanks for the help.

Correct, as no current flows through the capacitor in stationary state, after the switch is closed for a long time.

 

[scottdave]

I thought so because of the equation for the current, namely $i =\left(\frac E R \right) (e^{-t/RC})$, but I suppose that one doesn't apply either, since it's derived from the other one I mentioned.

That would be if a capacitor and resistor are in series with a voltage source. Think about this: when a capacitor has no current flowing through it, it essentially acts as an open circuit. So what is the voltage across R2 if the capacitor is just an open circuit?