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How do I find out the capacitor?

  1. Mar 28, 2004 #1
    Hi all

    I've a question to the following and would be greatly appreciated if anyone could lend a hand....

    Question:
    A series combination of a 12 kilo ohm and an unknown capacitor is connected to a 12 V battery. One second after the circuit is completed the voltage across the resistor is 2 V. How do I find out the capacitor?

    pls help me to understand and problem...

    thanks
    cseet
     
  2. jcsd
  3. Mar 28, 2004 #2
    V on capacitor = 12V - 2V = 10V

    now,charging formula for a capacitor is: Vs=Vi(1-[e to the power of t/RC])
    where Vs is the supply voltage
    Vi is the voltage on capacitor
    t is the time
    R is the resistor
    and C is the capacitance

    therefor your formula will becoma 12=10(1- [e to the power of 1/12k * C)

    all you have to do is just subject of the formula
     
  4. Mar 28, 2004 #3
    Also remember that the value of capacitance for a parallel plate capacitor is:

    C= Ae/t where e is the dielectric constant, A is the area of the plates and t is the distance between the plates. Usually values of e are given relative, so multiply them by 8.854E-12 farad/meter to get actual material dielectric (permitivity) constant.
     
  5. Mar 28, 2004 #4
    tigrot made a little mistake, the power of e is negative:
    [tex]V_{(t)} = V_f(1 - e^{-\frac{t}{RC}})[/tex]
    [tex]1 - \frac{V_{(t)}}{V_f} = e^{-\frac{t}{RC}}[/tex]
    And then just use ln()...
     
  6. Mar 29, 2004 #5
    Thank you for correcting me Chen.You are correct.
     
  7. Mar 29, 2004 #6
    re

    Hi there,

    thanks for all your replies. they're really helpful and takes the pain away from learning Physics!

    thanks again!
    Cseet
     
  8. Mar 29, 2004 #7
    yeah ppl are so helpful here
     
  9. Mar 29, 2004 #8
    Hi,
    I came to an answer of 15C.

    the following is how I worked out... correct me if I'm wrong... thanks

    12 = 10 (1-exo(-1/12C))
    12 = 10 * 0.08 * C
    C = 12 / (10 * 0.08)
    C = 15 C

    am I correct?
    thanks
    cseet
     
  10. Mar 29, 2004 #9
    Hi,
    I came to an answer of 15C.

    the following is how I worked out... correct me if I'm wrong... thanks

    12 = 10 (1-exo(-1/12C))
    12 = 10 * 0.08 * C
    C = 12 / (10 * 0.08)
    C = 15 C

    am I correct?
    thanks
    cseet
     
  11. Mar 29, 2004 #10
    hi guys,

    pls ignore my above equation.... I got it all wrong... fond out the answer... thanks anyway...
    cseet
     
  12. Mar 29, 2004 #11
    I don't understand how you got from here:
    [tex]12 = 10(1 - e^{-\frac{1}{12C}})[/tex]
    To here:
    [tex]12 = 10 (0.08C)[/tex]
    I don't think you read the equations above right. The capacity C is part of the power of e, you cannot leave it out and only raise e to the power of 1/12. Do as I said - from this:
    [tex]1 - \frac{V_{(t)}}{V_f} = e^{-\frac{t}{RC}}[/tex]
    Take the ln() of both sides to get:
    [tex]ln(1 - \frac{V_{(t)}}{V_f}) = -\frac{t}{RC}}[/tex]
    [tex]C = -\frac{t}{R ln(1 - \frac{V_{(t)}}{V_f})}}[/tex]

    The answer I get is C = 46.5 micro-farads.
     
  13. Mar 29, 2004 #12
    sorry this may be a very stupid question... but what does In() mean? and where's exp in this equation.

    I do apologise for my ignorance....
    cseet
     
  14. Mar 29, 2004 #13
    ln is Logarithm Natural, i.e log with the natural base e. If:
    [tex]b = a^x[/tex]
    Then:
    [tex]x = \log _ab = \log _aa^x[/tex]
    With ln, the base of the log is simply e. So if:
    [tex]b = e^x[/tex]
    Then:
    [tex]x = \ln b = \ln e^x[/tex]

    So when you have this equation:
    [tex]1 - \frac{V_{(t)}}{V_f} = e^{-\frac{t}{RC}}[/tex]
    You are allowed to take the ln of both sides:
    [tex]\ln (1 - \frac{V_{(t)}}{V_f}) = \ln e^{-\frac{t}{RC}}[/tex]
    The left side stays the same, and the right side just becomes [itex]-\frac{t}{RC}[/itex].
     
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