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

just simplify this differential eqn..
limit of V would be 10 to 1
and of t would be 0 to 2
R and C are constants..
feed values and solve..

Answer 2

I attempted it but I don't think it is right.
Simplified the differential equation into:
\[{1 \over V} dV = {-dt \over RC}\]
took integrals of each side:
\[\int\limits\limits_{1}^{10}{{{1 \over V} dV}={\int\limits_{0}^{2}{-dt \over RC}}}\]
\[\left[ {\ln V} \right]_{1}^{10} = \left[ {-t \over RC} \right]_{0}^{2}\]
\[2.3 = {-2 \over RC}\]

Solving for R:
\[R = {-2 \over 2.3C}\]
\[C = 3\times10^{-6}\]

Substituting C:
\[R = -289885\]

I somehow don't think this is right.
I've only just recently learnt Differential equations and I still can't grasp the concept and methods of doing things.
Please help.

Answer 3

in your second step,,the limit of V is from 10 to 1 and not 1 to 10..

Answer 4

So that will just make the answer positive instead of negative.
\[\ln 1 = 0\]
\[\ln 10 = 2.3\]
So would that not mean the integral is
\[\ln 1 - \ln 10 = -2.3\]

289885 just seems like a huge resistance...

Answer 5

well..seems the only ans right now to me,,i maybe wrong above somewhere..do you have the ans?

Answer 6

for a discharging of capacitor,,the V varies as E(e^ (-t/Rc)) ,,E is constant
try this..

Answer 7

well am sorry am a bit confused,,and i even confused you..hmmn..

Answer 8

I don't have the answers as it's for an assignmen this is the only information given.

The working looks right. The answer looks wrong. But that could be because I don't know much about electronics. Hmmm thanks for the help.