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Capacitance equation

  1. Apr 26, 2006 #1
    how would i rearrange this equation to find the capacitance, c?

    V=Vo exp(-t/RC)
     
  2. jcsd
  3. Apr 26, 2006 #2
    Do you know what the inverse of the exponential function is?
     
  4. Apr 26, 2006 #3
    would it go to,

    V = Vo ln + (-t/RC)

    then to,

    V = ln Vo - t x 1/RC
     
    Last edited: Apr 26, 2006
  5. Apr 26, 2006 #4
    You're right that you use the natural log function, but your form is incorrect. What would happen if you take the natural log of the exponential function, or ln(e^x) = ?
     
  6. Apr 26, 2006 #5
    ln(e) + x ?????

    not really sure you've lost me a bit sorry
     
  7. Apr 26, 2006 #6

    berkeman

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    Staff: Mentor

    Let's do it first with the more familiar base 10.

    log( 10^2 ) = 2 (right?)

    log( 10^x ) = ?

    ln( e^x ) is analogous to log( 10^x )...
     
  8. Apr 26, 2006 #7
    log(10^x) = x ?

    analogous?? The same as??

    but log e = 1

    does that mean log e^x = 1^x ?

    therefore log e^x = x ?
     
    Last edited: Apr 26, 2006
  9. Apr 26, 2006 #8

    berkeman

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    Staff: Mentor

    No, you need to keep your log() and ln() straight. log() is used with base 10 math, and ln() is used with base e math.

    log( 10^x ) = x

    ln( e^x ) = ?

    Try a few numbers on your calculator to help you keep it straight. There's a reason that most calculators overload the log() key with 10^x and overload the ln() key with e^x.....
     
  10. Apr 27, 2006 #9
    How come you did not take the natural log on the left hand side? :surprised
    If this is your original equation, then:
    [tex]\ln V = \ln V_0 - {t\over RC}[/tex]

    Furthur evaluation is simple.
     
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