Analyzing an RC Circuit with a Sudden Switch Closure

[bassplayer142]

Homework Statement

In a circuit, the switch S has been open for a long time. It is then suddenly closed. Determine the time constant (a) before the switch is closed (b) after the switch is closed. (c) let the switch be closed at t=0. Determine the current in the switch as a function of time.

Homework Equations

for charging cap
q(t) = CE[1-e^(-t/RC)]
I(t) = E/Re^(-t/RC)
c=q/v v=IR

The Attempt at a Solution

I assume that since it has been open for a long time the cap is fully charged to 1E-6F. Using the first equation I can't find Q because if you plug in q=cv in the equation it is ln(1)=0 then the time is 0. I know that the answer to a is t=1.5s from the back of the book. I guess I just don't know how to find Q. Also, since the 50kohm and 100kohm resistors are in series but not next to each other can I used the series law to make it a 150kohm resistor. I guess I don't know since there is either a capacitor or a battery in between them.

 

[kamerling]

Capacitors do not get charged up to a certain capacitance, They get charged until the voltage across them is so high that no more current can flow.
If you have a capacitor, a resistor and a battery in series, that point is reached once the voltage across the capacitor is equal to the voltage across the battery, so the voltage across the resistor is 0.

Unfortunately I can't see the picture yet.

 

[kamerling]

If the switch is open the resistors and the capacitor will all have the same current passing through them. that means it's OK to swap two of them around. You can prove this with kirchhofs loop law. After the switch closes you have two loops with the current in them independent. The sum of those currents goes through the switch

 

[bassplayer142]

So basically I can treat the two resistors as being in series, but when the switch is closed I must use kirchhoff's laws? Sounds right to me.