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PhoenixFire
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.
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..
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.
in your second step,,the limit of V is from 10 to 1 and not 1 to 10..
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...
well..seems the only ans right now to me,,i maybe wrong above somewhere..do you have the ans?
for a discharging of capacitor,,the V varies as E(e^ (-t/Rc)) ,,E is constant try this..
well am sorry am a bit confused,,and i even confused you..hmmn..
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.