How Do You Calculate the Time Constant in a Series Capacitor Circuit?

[Goomba]

Two capacitors in series are charged by a 12.0-V battery that has an internal resistance of 1.00 ohm. There is a 5.00-ohm resistor in series between the capacitor, as shown in this figure:

What is the time constant of the charging circuit?

I know time constant, tau, equals R*C, but I don't know how to apply it to a circuit containing multiple capacitors.

I believe the top plate of the 3.00 microF capacitor is positively charged while the bottom plate is negative. And for the 6.00 microF capacitor, the top plate is negatively charged and its bottom plate is positive.

Also, current flows from the positive terminal of the battery to the negative terminal, so current flow in this circuit would be clockwise.

Does the arrangement of the resistor between the two capacitors matter? Is the time constant only dependent on the capacitor that directly follows the resistor?

 

[Galileo]

How are the total charges on the two capacitors related?

 

[robphy]

Technically speaking, I wouldn't say that these capacitors are arranged in series. They are in a simple 1-loop circuit, yes. It's best to first reason using Kirchhoff's Laws... then possibly equivalent ways of thinking about it.

 

[Hellomynameis]

Bump* I also have no idea how to begin to solve this problem would be great if someone could explain it and figure out an answer.
Thanks a ton

 

[Hellomynameis]

Ah figured it out 1/C + 1/C = 1/Ceq
Ceq*(R1+R2)= time constant
time constant = 1.2*(10^-5)

 

[damionpitt]

for anyone who doesn't completely understand, He is using the capacitors in parallel

(C1*C2/C1+C2) = Ceq and since this is micro f (uf) * (ohms)

you have to multiply Ceq by (1.0*10^-6). giving you your 1.2e-5 in this case.

i kept thinking the Capacitors were in series too.

 

[jomanscool2]

for anyone who doesn't completely understand, He is using the capacitors in parallel

(C1*C2/C1+C2) = Ceq and since this is micro f (uf) * (ohms)

you have to multiply Ceq by (1.0*10^-6). giving you your 1.2e-5 in this case.

i kept thinking the Capacitors were in series too.

I wanted to clear this up a bit - they aren't actually in parallel. The equation does look similar to the 'resistors in parallel' equation, but it is in fact the capacitors in series equation:
http://en.wikipedia.org/wiki/Series_and_parallel_circuits#Capacitors

The problem appears to just be designed to trick you - it is really quite easily once you simply treat it as a simple R-C circuit by combining the resistors and the capacitors. You then just need to use the equation for charging an RC circuit and take what percentage of the total voltage difference is over each capacitor