Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. shubhamsrg Group Title
    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 years ago
  2. PhoenixFire Group Title
    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.

    • 2 years ago
  3. shubhamsrg Group Title
    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..

    • 2 years ago
  4. PhoenixFire Group Title
    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...

    • 2 years ago
  5. shubhamsrg Group Title
    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?

    • 2 years ago
  6. shubhamsrg Group Title
    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..

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

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

    • 2 years ago
  8. PhoenixFire Group Title
    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.

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.