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OpenStudy (anonymous):

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.

OpenStudy (phoenixfire):

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

OpenStudy (anonymous):

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thx so much :)

OpenStudy (phoenixfire):

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

OpenStudy (anonymous):

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

OpenStudy (phoenixfire):

ahhh, well that explains it. thanks

OpenStudy (anonymous):

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

OpenStudy (phoenixfire):

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

OpenStudy (anonymous):

yes

OpenStudy (phoenixfire):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

it's the same answ but positive

OpenStudy (phoenixfire):

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.

OpenStudy (anonymous):

look 2 sec is a huge time too

OpenStudy (phoenixfire):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

you are welcome

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