## PhoenixFire 3 years ago 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.

1. shubhamsrg

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..

2. PhoenixFire

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.

3. shubhamsrg

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

4. PhoenixFire

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...

5. shubhamsrg

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

6. shubhamsrg

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

7. shubhamsrg

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

8. PhoenixFire

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.