How Long to Discharge a Capacitor in a Disconnected Circuit?

AI Thread Summary
The discussion focuses on calculating the discharge time of a capacitor in a disconnected circuit. After determining the initial voltage across the capacitor as 6V, the user seeks to find the time it takes to discharge to 1/10 of this voltage using the formula t=-RC ln(q/Q). There is confusion regarding the appropriate resistance value to use, with the user attempting to identify which resistors are in series after the battery is removed. Clarifications are made about the series and parallel combinations of resistors, leading to a better understanding of the circuit configuration. Ultimately, the user finds a solution to the problem with assistance from others in the discussion.
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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