Question

When a charged capacitor is connected to earth, it discharges and the voltage V across it will decrease according to the equation:
dV/dt=−V/RC
where R (the resistance) and C (the capacitance) are constants. If the voltage V across a certain capacitor with capacitance C = 3×10−6 farads
drops from 10 volts to 1 volt in two seconds, determine the resistance R.

Answer 1

I have this problem too. I get a very strange answer. take a look:
http://openstudy.com/study#/updates/4fbcc124e4b0556534319992

Answer 2

just solve it now t=2sec;v0=10;v=1;c=3e-6

Answer 3

thx so much :)

Answer 4

how'd you end up with ln V/V0??? shouldn't it be lnV - ln(V0)?

Answer 5

log(a)-log(b)=log(a/b)

Answer 6

ahhh, well that explains it. thanks

Answer 7

@PhoenixFire
in other question you ask why it's negative v must be less than v0 because it's discharging

Answer 8

\[R={-t \over {\ln({V \over V_0})*C}}\]
Is that right?

Answer 9

yes

Answer 10

I get: -289529.655
I figured that was the reason for negative, my main worry was the huge resistance value.

Answer 11

no r is positive,you forget -t
ln(1/10)<0

Answer 12

it's the same answ but positive

Answer 13

ah yes... -t.... damnit.
I don't know electronics very well so I thought that resistance value was abnormally large.. but I guess a discharging capacitor is a special case.

Answer 14

look 2 sec is a huge time too

Answer 15

that is true... 2 seconds for electronics is a long time for stuff to happen.
Well, thank you RaphaelFilgueiras!

Answer 16

a flash for example use capacitors,and it discharges,very fast

Answer 17

you are welcome