How Long to Discharge a Capacitor in a Disconnected Circuit?

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thenewbosco
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this seems to be a simple question
for the diagram at http://img.photobucket.com/albums/v11/biggm/ohm.jpg

a)the circuit has been connected a long time...what is the voltage across the capacitor?

this one i calculated and got the right answer of 6V.

b)if the battery is disconnected how long does it take to discharge the capacitor to 1/10 of its initial voltage.

i have the equation t=-RC ln (q/Q)

where R= resistance, C=capacitance, q= final charge and Q=initial charge, i have calculated the initial and final charges using q=CV and final voltage of 3/5V. Capacitance is given...what value do i use for resistance?
i can see given the answer that it should be 18/5...which is the product of the initial and final voltage although this could be coincidental...any help here??
 
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With the voltage source removed and the capacity supplying the voltage, which capacitors are in series?
 
the 1ohm and 4 ohm are in series, and the 8ohm and 2ohm are in series.
do i just add these as in series and then...if i add them together as two then in parallel it does not work out...
 
The 1 and 4 ohm resistors are in series and the 1 and 8 ohm resistors are with respect to the 10V source.

However when 10V source is removed and the capacitor provides the potential difference, resistors 4 and 2 ohm are then in series and resistors 1 and 8 ohm are in series.
 
ah thanks for that tip...it works out now but i fail to see how the 4 and 2 combination and the 1 and 8 ohm combination are in series when the potential is removed.
 
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Some quick photoshopping reveals

http://img299.imageshack.us/img299/1403/untitled14xz.gif
 
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