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Homework Help: 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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  22. Dec 13, 2012 #21
    So if what i said in post #20 is correct, redoing everything....


    Since Ct=2.42uF
    Q=VC
    Q=25x2.42=60.5uC
    Qt=Q1=Q2 (Series Circuit)

    V2=8.64v

    Finding Voltage across C1, naming it as V1
    Q=VC
    60.5uC=V x 5uF
    V1=12.1V

    V2+V1=8.64+12.1
    = 20.74V


    Et-20.74=4.26V
    4.26V is the voltage across C3 and C4


    Subbing all the values into the energy stored in C formula, the answer i got is 72.59uJ. So can I say post #12 working+concept = wrong while this is the correct one.

    Please help me take a look, thank you.

    PS: Just curious, is part (ii) still doable using the VDR method?
     
    Last edited: Dec 13, 2012
  23. Dec 13, 2012 #22

    gneill

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    Combine two of the capacitors leaving you with two, then apply the voltage divider rule.

    But really, it's probably easier to find the total capacitance first and determine the charge; the same charge will be on each of the capacitors in series, so you can find the voltage on each one quite easily.
     
  24. Dec 13, 2012 #23

    gneill

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    I notice that you are rounding intermediate results as you go along and not keeping enough guard digits to prevent rounding errors from entering your later calculated values. You apply two different (but correct) methods to find a given value and end up with different numbers in the significant figures. This is rounding/truncation errors biting you.

    In red below I've shown the same values that you've calculated but without successive roundings being applied along the way, and with an extra guard digit.
    The method in post #12 was fine, but you made some errors in applying it, such as using '14' for the value of C5 when it should have had units of microfarads included.

    Sure, just combine C1 and C5 into a single capacitor (call it C15) and then use the voltage divider rule.
     
  25. Dec 18, 2012 #24
    Thank you gneill for your explanation so far. So can I say that post #12/23 question has a couple of ways to solve it?

    I would like to ask you a question relating to the picture below:

    "currents and voltages have reached their final values" -> Does it have another meaning as current and voltage = @ max value. ?



    Thank you.
     

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  26. Dec 18, 2012 #25

    gneill

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    Sorry, I don't follow your question, but it is true that most problems can be solved via several approaches.
    It means that the currents and voltages settle into a steady state after which no further changes in their values take place unless the circuit is disturbed in some fashion (like a switch being thrown).

    When switches open or close, sudden changes in potentials or currents can cause transient events like big voltage spikes across inductors, so minimum or maximum magnitudes can occur during the transient state, and then settle down to some constant "end state" over time.

    For the most part, for RL or RC circuits you'll see exponential curves that either decay from a starting value to a minimum, or rise from some starting value to a maximum "plateau".

    Circuits containing both L and C components can be trickier as there will be oscillations that decay over time, so just saying "minimum" or "maximum" value won't work cleanly.

    Anyways, "Assume that currents and voltages have reached their final values" means that the circuit has been left undisturbed for long enough that any transient conditions have died away and operation has reached a steady state (all values remain constant as time goes on).
     
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