How Does Closing a Switch Affect the Voltage Across an RC Circuit?

[Sean Reed]

Homework Statement

The switch has been open for a vary long time, and is then at time T=0 closed. Determine the voltage V0(t) for t>0

Homework Equations

voltage from a discharging capacitor is V*e^(-t/RC) Where V is the voltage stored in the capacitor, R is the resistance and C is the capacitance.

The Attempt at a Solution

1) My understanding is that the capacitor will charge up to 12 volts, however I have been told that it only charges up to 6 because the two 2kohm resisters make a voltage divider. I did not think the location in terms of other resisters made a difference for the voltage that a capacitor will charge to but I may be wrong.
2) Once the switch is closed I thought the voltage at the node where the switch is would be 4 volts because it would act as a voltage divider between the first 2kohm resister and the two 2kohm resisters in parallel. My friend says it will be 6 volts because the capacitor acts as an open circuit so it is only a voltage divider between two 2kohm resisters. Now if this is the case then would it matter as the only voltage at V0 is the decaying voltage from the capacitor?
3) For the resistance in the time constant will it just be the 2kohm resister at V0 plus the 2kohm resister in series at the switch, or will it be something else?

Thus my answer is (12e^(-t/.02))-4 While his answer is (6e^(-t/.02))-6 I have no idea who is right or if we are both wrong, so any help is appreciated.

 

[berkeman]

Here's a hint to get you going -- draw a graph of v(t) using the following info...

The voltage on the - side of the cap initially is ground, right? And the voltage on the + side of the cap is initially what? Hint -- there is no voltage divider yet with the switch open. When the switch is closed, what does that instantaneously do to the voltage on the + side of the cap? Remember that the cap looks like a short circuit for instantaneous changes in voltage on one side of it...