Having trouble rearranging this RC circuit

[horsedeg]

Homework Statement

Switch S shown in the figure below has been closed for a long time, and the electric circuit carries a constant current. Take
C₁ = 3.00 μF

C₂ = 6.00 μF,

R₁ = 4.00 kΩ,
and
R₂ = 7.00 kΩ.
The power delivered to R₂ is 2.30 W.

(a) Find the charge on C1.
(b) Now the switch is opened. After many milliseconds, by how much has the charge on C2 changed?

Homework Equations

P=IV
V=IR
Q=CV

The Attempt at a Solution

To find the charge on C₁ I would use Q=CV, but first I would need to find the voltage or substitute one of the other equations.

I looked at the solution, and it gives me this image (different values):

I have no idea why it can be rearranged like this, since even the capacitors/resistors change places.

My first plan was using the fact that R₂ is given power of 2.30W, so then I would use P=IV and V=IR to get P=I²R. Solve for I and I get the current there. However, according to the solution it determines the current of the whole thing. I don't see how that's possible from either arrangement.

 

[NascentOxygen]

I have no idea why it can be rearranged like this, since even the capacitors/resistors change places.

Parallel components don't care who's on the left or who's on the right. They can be swapped if it's going to make things a little clearer to readers.

Once you have determined the voltage or current for one resistor, you know everything about all the other components here. So what is the voltage across R₂?

 

[amberita]

I don't feel comfortable enough with the math to help there, but I wanted to respond to your comment "I have no idea why it can be rearranged like this, since even the capacitors/resistors change places."

In both diagrams, the block of C1 and R1 is in series with the block of C2 and R2. C1 is in parallel with R1, and C2 is in parallel with R2. In both, current can travel down C1 or R1, then to C2 or R2. It's the same circuit.

 

[horsedeg]

Parallel components don't care who's on the left or who's on the right. They can be swapped if it's going to make things a little clearer to readers.

Once you have determined the voltage or current for one resistor, you know everything about all the other components here. So what is the voltage across R₂?

P=IV, so solving for voltage V=P/I. We don't know what I is, so we plug in V=IR, or I=V/R. Then we get V²=PR₂. Plugging in the values, we get that the voltage drop across R₂ is 126.89V.

So from then we could easily find the current at that place, but that would only be that part of the current, wouldn't it?

EDIT: I'm stupid. I forgot that it's been on for a long time so that the capacitors are fully charged. Right? So then they act as infinite resistors, meaning R1 and R2 are in series. So then I can use V=IR to find the current by plugging in both V and R with what I had before. Then, because that's the total current at that moment, I can use V=IR₁ to find the voltage across the top two devices. Then I can use Q=CV and solve for Q on the capacitor because it's parallel and has the same voltage.

I guess I could also use this info to determine the voltage of the battery with the two resistors in series using V=IR, meaning Voltage of the battery is 199.39V. So then the voltage across each parallel combination of capacitor & resistor would have an overall drop of 199.39V.

 

[NascentOxygen]

it's been on for a long time so that the capacitors are fully charged. Right? So then they act as infinite resistors, meaning R1 and R2 are in series.

Right. The capacitors are charged and no longer drawing current, so in the steady state all current flow is in the resistors.

For (b), can you explain in words what happens when S is later opened?

 

[horsedeg]

Right. The capacitors are charged and no longer drawing current, so in the steady state all current flow is in the resistors.

Okay, I definitely get this one now easily. Thanks.