1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem with Capacitance

  1. Jan 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Imagine that you have a light-bulb that has a resistance of about 10 ohms and that can tolerate a maximum voltage of 3 volts. Imagine that you want to connect this to a charged capacitor large enough to keep the bulb glowing reasonably brightly for more than 10 seconds. Roughly what should the capacitor's capacitance be?

    2. Relevant equations
    VC(t) = V0e(-t/RC)

    3. The attempt at a solution
    I feel as if I need another equation for this problem, because I don't have 4 variables to plug in! The starting potential of the capacitor should be 3 volts, right?

    So

    VC(t) = (3 volts)e(-10 seconds/(10 ohms)*C)

    So now I have unknowns of C and VC, and I need to solve for C. Clearly I'm missing something, but I'm not sure what... Can somebody point me in the approximately right direction? Thank you!!
     
  2. jcsd
  3. Jan 17, 2015 #2

    berkeman

    User Avatar

    Staff: Mentor

    Good start. :-)

    I would probably start with the equation [tex]I = C \frac{dV}{dt}[/tex]
     
  4. Jan 17, 2015 #3

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The light-bulb will get dimmer as the voltage goes down, but it should not get too dim. You can define what exactly that means by fixing Vc(10s) to some reasonable value.
     
  5. Jan 17, 2015 #4
    So I can arbitrarily say that Vc(10 s) is, say, 1.5 volts? And then solve for C, which would give 1.4 farads. Isn't that considered a really really high capacitance?
     
  6. Jan 17, 2015 #5
    Would dV/dt just be the derivative of the right side of my equation above? That is, dV/dt = -(V0/RC)*e-t/RC?

    Then, since I = C*dV/dt,

    I = -3 volts/10 ohms * e(10 s)/((10 ohms)*C)

    I feel like that doesn't help, because then I just have another unknown (I).
     
  7. Jan 17, 2015 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Right.
    Maybe 20 years ago, but now those are cheap elements available everywhere.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Problem with Capacitance
  1. Capacitance problem (Replies: 1)

  2. Capacitance Problem (Replies: 6)

  3. Capacitance Problem (Replies: 2)

  4. Capacitance Problem (Replies: 5)

  5. Capacitance Problem (Replies: 4)

Loading...