- #1

horsedeg

- 39

- 1

## 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

C

_{1 }= 3.00 μF

C

_{2 }= 6.00 μF,

R

_{1}= 4.00 kΩ,

and

R

_{2}= 7.00 kΩ.

The power delivered to

*R*

_{2}is 2.30 W.

(a) Find the charge on

*C*1.

(b) Now the switch is opened. After many milliseconds, by how much has the charge on

*C*2 changed?

## Homework Equations

P=IV

V=IR

Q=CV

## The Attempt at a Solution

To find the charge on C

_{1}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 R

_{2}is given power of 2.30W, so then I would use P=IV and V=IR to get P=I

^{2}R. 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.