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Finding capacitance

  1. May 7, 2013 #1
    1. The problem statement, all variables and given/known data

    This is my data of Irms and angular frequency, and I have to plot the graph and teacher told me that the gradient will be capacitor's capacitance?


    2. Relevant equations


    3. The attempt at a solution

    Did do anything wrong?
     

    Attached Files:

    Last edited by a moderator: May 7, 2013
  2. jcsd
  3. May 7, 2013 #2

    NascentOxygen

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    Hi k31353. [Broken]

    Have you worked out the value of the capacitance? (Can I presume that is the aim of the exercise?)

    How many μF is it?
     
    Last edited by a moderator: May 6, 2017
  4. May 7, 2013 #3

    CWatters

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    The general equation for a straight line is y=mx+c where m is the slope and c is a constant.

    I suggest writing the equivalent equation for your experiment.

    Is the slope C or 1/C or something else?
     
  5. May 13, 2013 #4
    Slope is C but how do i prove that the slope represents the capcitance of capacitor.
    yes and the value is 0.1 uF
     
  6. May 14, 2013 #5

    NascentOxygen

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    The slope is not exactly = C. The voltage you are applying must come into the equation somewhere, since if the voltage you are using is greater, the current must be greater.

    Can you quote an equation relating all of those quantities, C, ω, I and V? That is what you need to examine to see what you are dealing with.

    This information will be useful.
     
  7. May 14, 2013 #6

    CWatters

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    I asked because you plotted ω against I rather than I against ω. Perhaps I was awake too late last night but...

    The imp of a capacitor is

    Z = V/I = 1/jωC

    rearrange to give ω in terms of I...

    I/V = jωC

    ω = I/(VjC)

    ω = (1/VjC) I

    which is now in the form

    y = mx + constant

    So is the slope m = C? 1/C ? or something else?
     
  8. May 14, 2013 #7
    Ok this is my circuit

    [Broken]

    Now and this are the data.

    [Broken]

    Then this is the question :
    From the slope of the graph verify the value of the capacitor ? (Hint: if the system is sinusoidal AC, then I RMS = angular frequency(omega) * C * V RMS). Does it agree with the measure value? If not, why not?

    So this is my answer :

    [Broken]

    So my capacitor answer is way different than it should be.

    So this is the problem
     
    Last edited by a moderator: May 6, 2017
  9. May 14, 2013 #8
    What does it say on your capacitor? Include any and all markings.
     
  10. May 14, 2013 #9
    Nothing actually thats all ive got
     
  11. May 14, 2013 #10
    Do you have a picture of it?
     
  12. May 14, 2013 #11
  13. May 14, 2013 #12
    Ah doh, I missed an error of yours:

    Vrms = Vpp/(2*sqrt(2))

    The peak-to-peak amplitude Vpp is 2*Vp where Vp is the peak amplitude.
     
  14. May 14, 2013 #13
     
  15. May 14, 2013 #14

    NascentOxygen

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    Your table indicates at 1kHz and with a voltage of 7V the unknown capacitor drew 2.28mA.

    Let's compare that with what a known capacitor would do. A capacitor of 0.1μF, at 1kHz and an applied voltage of 7V, would draw 4.4μA.

    Big difference. :eek:

    I wonder was your ammeter scale really μA where you recorded mA?
     
  16. May 14, 2013 #15
    Isnt it if u wanna get rms u just divide by squareroot of 2??
     
  17. May 14, 2013 #16
    I recorded ma.
     
  18. May 14, 2013 #17
  19. May 14, 2013 #18
    His readings are fine, he just didn't calculate the RMS voltage correctly.

    If Vpp = 10 V then:

    Vrms = Vpp/(2*sqrt(2)) = 3.536 V
    C = |Irms|/(2*pi*f*|Vrms|) = 112.5 nF ≈ 0.1 uF

    for f = 100 Hz.
     
  20. May 14, 2013 #19
    Ahhhh gotchya..
    U earn yourself a fortune cookie...
    Thanks
     
  21. May 14, 2013 #20

    NascentOxygen

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    No, I really do make it 4.4mA, so your figures are of the right order, just a factor of 2 difference. That factor of 2 might be accounted by that voltage conversion error milesyoung pointed out.
     
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