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## Homework Statement

Given V R C for the figure, determine

a.

The voltage of a with respect to b when the switch is open?

b.

Which point a or b is at a the higher voltage?

c.

What is the final voltage of point b with respect to ground when the switch is closed?

d.

Wow much does the charge on each capacitor change when the switch s is closed?

## Homework Equations

V = IR

V = q/C

charging conductor

[tex] q = CE(1-e^{\frac{-t}{RC}})[/tex]

[tex] i = \frac{E}{R}e^{\frac{-t}{RC}})[/tex]

discharging conductor

[tex] q = Q_0e^{\frac{-t}{RC}}[/tex]

[tex] i = I_0e^{\frac{-t}{RC}}[/tex]

## The Attempt at a Solution

a.

I don't really understand conceptually whats going on, so i'll talk through.. the switch has been open, so the conductor takes all of the charge. getting a is solving the righthalf like a simple series capacitor problem to get voltage at b, voltage at a is just V because 0 current over resister means full voltage reaches a

b.

a would be at higher voltage because it has the maximum possible, b has less because the first capacitor takes some away

c.

At this point I feel like im surely conceptually wrong, but if you open the switch, the conductors are still going to be taking all of the current and nothing should be changing ?

d.

i would think none given above

I have some hitches in the concepts going on I guess. If anyone can clarify what is going on I'm sure I can get this quite easily. I didn't have much a problem when resisters/conductors werent together