# Resistance, Capacitance,

1. Nov 23, 2005

### thenewbosco

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??

2. Nov 23, 2005

### Staff: Mentor

With the voltage source removed and the capacity supplying the voltage, which capacitors are in series?

3. Nov 23, 2005

### thenewbosco

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...

4. Nov 23, 2005

### Staff: Mentor

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.

5. Nov 23, 2005

### thenewbosco

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.

Last edited: Nov 23, 2005
6. Nov 24, 2005

### mezarashi

Some quick photoshopping reveals