# Having trouble rearranging this RC circuit

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1. Nov 30, 2016

### horsedeg

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

2. Relevant equations
P=IV
V=IR
Q=CV

3. 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.

2. Nov 30, 2016

### Staff: Mentor

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?

3. Nov 30, 2016

### 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.

4. Nov 30, 2016

### horsedeg

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.

Last edited: Nov 30, 2016
5. Nov 30, 2016

### Staff: Mentor

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?

6. Nov 30, 2016

### horsedeg

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