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 2 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.
 2 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.

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shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0just 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..

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0in your second step,,the limit of V is from 10 to 1 and not 1 to 10..

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0So 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...

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0well..seems the only ans right now to me,,i maybe wrong above somewhere..do you have the ans?

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0for a discharging of capacitor,,the V varies as E(e^ (t/Rc)) ,,E is constant try this..

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0well am sorry am a bit confused,,and i even confused you..hmmn..

PhoenixFire
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.
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