capacitor problem

Discussion in 'Homework Help' started by k31453, May 7, 2013.

  1. k31453

    Thread Starter Member

    May 7, 2013
    54
    0
    Hi,

    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?
    [​IMG]


    i run the experiment with unknown capacitor and sinusoidal voltage across the capacitor is Vpeak = 2V and frequency varies.
    then teacher said to plot the graph from following data and he also said that the gradient will be capacitor's capacitance .
    He said to prove your method use known capacitor which is 0.2 micro F and we can do the same experiement on this capacitor as well and the gradient of graph will be capacitance of capacitor but i got wrong answer.. so am i doing correctly?
    and this is my excel result

    [​IMG]
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    Write down the formula for the reactance of the capacitor and solve for the capacitance C.
     
  3. mikeleeson

    New Member

    Aug 22, 2012
    26
    4
    You are varying the frequency and measuring the current. So the frequency is the independent variable (which should be displayed on the X axis) and the current is the dependent variable (which should be displayed on the Y axis).

    Try changing the axes of your chart and then try again. Be careful that you get the units correct as well.

    Then ask yourself, "why does the current vary with the frequency?" See if you can work out the formula.
     
  4. MrChips

    Moderator

    Oct 2, 2009
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    Another way of seeing your error is what are the units of capacitance C?

    C = Q/V

    i.e. charge/voltage = current x time / voltage
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    UNITS! UNITS! UNITS! my friend.

    Capacitance has units of farads, or coulombs/volt.

    Your slope (gradient) has units of Y/X, or frequency/current or, in your case, 1/mAs, which is 1/mC. So you need to fold in voltage somehow -- and don't forget to be consistent and not mix peak and rms values together.
     
  6. k31453

    Thread Starter Member

    May 7, 2013
    54
    0
    I understand what you mean but my lab worksheet says that how i have to put it !!


     
  7. MrChips

    Moderator

    Oct 2, 2009
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    What did your lab worksheet say?
     
  8. mikeleeson

    New Member

    Aug 22, 2012
    26
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    Your lab worksheet is correct if you look at Frequency [Hz] against current [mA]. But when you create the column with ω [radians/sec] you have managed to switch the two axes used for the chart.

    You still need to check the units (amps or milli amps) and convert from peak voltage to rms voltage. Did you work out the equation yet?
     
  9. k31453

    Thread Starter Member

    May 7, 2013
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    0
    My lab sheet says the plot the graph rms current Vs angular frequency
     
  10. MrChips

    Moderator

    Oct 2, 2009
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    Right. That means plot rms current on the y-axis and angular frequency on the x-axis.

    You still have to do the suggestions stated above.

    1) What is the formula for C?

    2) What are the units of C?

    Doing the above will reveal your error.
     
  11. k31453

    Thread Starter Member

    May 7, 2013
    54
    0
    1) c= q/ v

    2) c is a in farade.

    But how gradient is define the value of capacitor?
     
  12. MrChips

    Moderator

    Oct 2, 2009
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    1) Now convert that in terms of current, frequency and voltage.

    2) From (1) write out the units in terms of current, time and voltage.

    The equation of a straight line is y = mx + c

    where

    y = dependent variable
    x = independent variable
    m = slope
    c = intercept, i.e. y at x = 0

    The units of m must be the same as the units of y/x.
     
  13. mikeleeson

    New Member

    Aug 22, 2012
    26
    4
    You need to find an equation that contains the data that you measure (I_{rms} and frequency) and the capacitance that you require. Here is some help with the first few steps, have a go and post your efforts...

    Q = CV we know this equation, but we cannot measure Q (the charge). We can measure the current (rate of change of charge).

    I=\frac{dQ}{dt}=C\frac{dV}{dt}

    Can you find an expression for a voltage that varies with a frequency ω and has a peak amplitude V_{peak} ?

    Can you differentiate this expression to find \frac{dV}{dt}?
     
  14. k31453

    Thread Starter Member

    May 7, 2013
    54
    0
    okay it says i have use this formula but it is not giving me correct answer and i m using data of my spreadsheet
    [​IMG]

    https://www.dropbox.com/s/0thxwsrk3n8wl4w/lab.xlsx

    the worksheet is "2"

    Thanks
     
  15. mikeleeson

    New Member

    Aug 22, 2012
    26
    4
    With the numbers that you've given (worksheet 2), I get C=48.3nF

    But this is not the same capacitor as used in worksheet 3. Look at the value of I_{rms}

    Worksheet 2
    V_{rms}=1.45V f=100Hz I_{rms}=0.044mA

    Worksheet 3
    V_{rms}=1.41V f=100Hz I_{rms}=0.42mA

    For similar values of V and f, the current differs by a factor of 10. Somehow you have got your measured data or your capacitors mixed up - only you can sort that one out!

    In the meantime, here is an explanation of what I can see in your spreadsheets. You have got the formula:

    I_{rms}=\omega CV_{rms}

    Worksheet 2 is asking you to fix the frequency and plot I_{rms} against V_{rms}. The gradient gives you ωC, and you already know ω - so you can calculate C.

    Worksheet 3 is asking you to fix the voltage and plot I_{rms} against ω. The gradient gives you CV_{rms}, and you already know V_{rms} - so you can calculate C.

    Hope that helps.
     
  16. k31453

    Thread Starter Member

    May 7, 2013
    54
    0
    This is my circuit diagram.
    and i have to verify that my capacitor should be 0.1uF
    [​IMG]

    but i got wrong capacitance.


    Check page 2 of this forum.
    http://www.physicsforums.com/showthread.php?p=4382526&posted=1#post4382526
     
    Last edited: May 14, 2013
  17. MrChips

    Moderator

    Oct 2, 2009
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    Are you sure you are determining Vrms correctly?
     
  18. k31453

    Thread Starter Member

    May 7, 2013
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    Yes its v peak -peak / squareroot (2)
     
  19. mikeleeson

    New Member

    Aug 22, 2012
    26
    4
    Looking at your spreadsheet, this may be the problem.

    A sine wave with a peak to peak voltage of 2V has an amplitude of 1V. The rms voltage is therefore \frac{1}{\sqrt{2}} or 0.71 V.
     
  20. k31453

    Thread Starter Member

    May 7, 2013
    54
    0
    but isnt it we wanna get rms value of voltage from peak-peak.. we have to divide by squareroot (2)
     
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