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Series parallel capactitive circuit

  1. Dec 12, 2012 #1
    Hi guys, I want to ask if I can use the voltage divider rule to solve for part (ii).


    Thanks!
     

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  3. Dec 12, 2012 #2

    gneill

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    You could; do you know the voltage divider rule for capacitors?

    Alternatively, if you've already worked out the total capacitance and the charge on the total capacitance, then take note that C1 and C2 are in series in that total... so what charge must be on them?
     
  4. Dec 12, 2012 #3
    charge is the same, sorry let me sort out something first
     
  5. Dec 12, 2012 #4
    Total capcitance of two capacitors in series:

    Ct=C1xC2/C1+C2 is also the same as 1/Ct=1/C1+1/C2

    Is the any difference between them? How come there are 2 type of ways to find the total C. I have always use the 2nd one "1/Ct" to find the total C. Is there any special for the first one?
     
    Last edited: Dec 12, 2012
  6. Dec 12, 2012 #5

    gneill

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    There's no difference. Simplify algebraically the one on the right and you'll see.
     
  7. Dec 12, 2012 #6
    Ok, thank you. Can I ask if the above diagram capacitors are being replaced by inductor, can I still use the voltage divider rule to solve?


    Thanks.
     
  8. Dec 12, 2012 #7

    gneill

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    There would be a problem replacing all the capacitors with inductors if the power supply remains 25V DC. Do you see why?
     
  9. Dec 12, 2012 #8
    No, why?

    Does the voltage divider rule for inductor exist? Because I can't find the formula in my book but I did some googling and found a formula for it.
     
  10. Dec 12, 2012 #9

    gneill

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    The DC resistance of an ideal inductor is zero. While capacitors behave as open circuits and can charge up to present a potential difference, inductors behave as short circuits and just conduct more and more current. I = (V/L)t.
    It would be more useful for AC circuits where inductors have an impedance (sort of like an "AC resistance"). Voltage divider rules can be applied to any components with impedance.

    In a DC circuit, the eventual voltage across all inductors after any transients die away (some time after power is applied) is zero. It's hard to have a voltage divider mean anything useful if the potentials involved are all zero :smile:
     
  11. Dec 12, 2012 #10
    Is the VDR for capacitor the same as Inductor?
     
  12. Dec 12, 2012 #11

    gneill

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    No, the divider rule for inductors looks like the one for resistors. But again, you would probably only use it for AC circuits (or during transient analysis for DC circuits).
     
  13. Dec 12, 2012 #12
    Ok. I can't get the answer for part (iii.)

    My working:

    V3=CtE/C5 = 2.42x10-6x25/14=4.321uC

    Wc=1/2CV2
    =1/2x8x10-6x(4.321x10-6)2
    =7.468x10-17J

    Ans is:0.733J


    *I combined C3 and C4 together to form C5. From there, i tried to find the voltage across the new C5 reason being voltage for C3 and C4 is the same in parallel. My new V3 is also equal to the V across C3 and C4.

    After getting the new voltage "C5", I sub the values inside the formula including the 8uf (C4) but i cannot get the answer.
     

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  14. Dec 12, 2012 #13

    gneill

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    Shouldn't the result be in volts if it's a voltage? Where are the units for your C5?

    Also, you may be propagating roundoff errors from intermediate results. Don't round intermediate values that are to be used in further calculations -- keep a few extra digits in all intermediate values to prevent previous roundings from creeping into your significant digits in the final values.
    That provided answer looks suspiciously large. 0.733J is a lot of energy for a small capacitor to hold... check the given units.
    Your overall method looks okay. Check the items I pointed out.
     
  15. Dec 12, 2012 #14
    7.468x10-17J is my answer

    0.733J is the book answer.

    So if ignore the rounding off errors am I'm right?
     
  16. Dec 12, 2012 #15

    gneill

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    No, your result has an incorrect power of 10. Check the items I pointed out. Your C5 is not "14"; it needs appropriate units.
     
  17. Dec 12, 2012 #16
    My bad. 74.6uJ. I'm sorry can you help me take a look at another question...

    My working:

    250x10-3=5002x1800x4∏x10-7xA / 200x10-3

    50000x10-6=5002x1800x4∏x10-7xA

    A=50000x10-6[/SUP / 5002x1800x4∏x10-7

    A=0.000008841x101

    A=0.00008841m2

    A=0.08841mm2 My ans

    Book ans is:88.4mm2
     

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  18. Dec 12, 2012 #17

    gneill

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    Yes, that's closer to what I found: 74.32 μJ.

    Your powers of ten seem to have gone awry in that last step. You may find it easier to "normalize" the big (or small) numbers to the form n.nnn x 10n before continuing. That will make slipping decimal places more difficult.
     
  19. Dec 12, 2012 #18
    law of indices division = minus

    -6/-7 = -6 - -7 = 1

    This what i'm thinking..
     
  20. Dec 12, 2012 #19

    gneill

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    Okay, sorry, my eyes slipped over the x101 tacked onto the end of the decimal number. You really should present the figures in normalized form.

    It would appear that your conversion from m2 to mm2 is incorrect. How many square mm are there in a square m?
     
  21. Dec 13, 2012 #20
    Sorry, going back to post #12. Is there something wrong with my concept+answer?

    I missed out C1 earlier on and i use VDR on C2 and C34 instead. I thought VDR only apply for 2 capacitors and how can i use it if i have 3 capacitors at that time?
     

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