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PhoenixFire
Group Title
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
 2 years ago
PhoenixFire Group Title
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
 2 years ago

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shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
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 years ago

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

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

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
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...
 2 years ago

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

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

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

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago
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