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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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shubhamsrgBest 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..
 one year ago

PhoenixFireBest 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.
 one year ago

shubhamsrgBest 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..
 one year ago

PhoenixFireBest 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...
 one year ago

shubhamsrgBest 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?
 one year ago

shubhamsrgBest 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..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
well am sorry am a bit confused,,and i even confused you..hmmn..
 one year ago

PhoenixFireBest 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.
 one year ago
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