circuit/capicator/resistor problem!

[physics213]

1. The problem statement, all variables and given/known data

IMAGE ATTACHED!

In the figure below, suppose that the switch has been closed for a length of time sufficiently long enough for the capacitor to become fully charged. (R = 11.0 k, R2 = 15.0 k, R3 = 3.00 k, ΔV = 9.30 V)

Find the steady-state current in each resistor.

Find the charge on the capacitor.


2. Relevant equations

V=IR
Series/Parallel resistor equations.
C=Q/V

3. The attempt at a solution

I believe you have to use the loop rule but I can't figure out how to simplify the circuit, or even what to do after that. Please help!

[mikelepore]

When the capacitor is fully charged, current stops flowing into it, so you can treat its branch as an open circuit. See if that gives you a hint about the next step. How does that affect the resistor that's in series with it? What simplifictions will then become possible?

[physics213]

So the current in the resistor that is in series with the capacitor would be zero. Which would mean that the current in the other two resistors should be equal to each other, correct? Which would also mean the other two resistors are in series if I'm not mistaken.

[physics213]

ok I got the correct answer for the current through the resistors problem! Thanks a lot.

[physics213]

Wouldn't the second part of this question (What is the charge on the capacitor?) just be Q=CV? I can't get the correct answer from that though...

[mikelepore]

Correct, R and R2 will be in series after the current in the R3 branch decays to zero. After R3 has no current, next step, what is the voltage across R3? Using that fact, how does the voltage across the capacitor compare to the voltage across R2: would it be smaller or greater or equal? Finally, will the full voltage source V be applied across R, or will the full voltage source be applied across R2, or will V be divided somehow between R and R2? You have to get the capacitor voltage right before you can apply Q=CV.