Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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.

  • This Question is Closed
  1. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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. PhoenixFire
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  3. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. PhoenixFire
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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...

  5. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  6. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. PhoenixFire
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  9. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy