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Finding missing caps

  1. Oct 12, 2007 #1
    1. The problem statement, all variables and given/known data

    find c1 & c3

    known values:

    c2= .01uF
    Xc= 2.34k
    c5= .047uF
    c6= .033uF

    given these values:

    Vb 21.91v

    Vc 17.51v

    Vd 14.48v

    It 1.76mA

    Xct 28.34k ohms


    2. Relevant equations

    im not sure which equation to use

    3. The attempt at a solution

    im not sure what to do

    i dont know if im in the right forum

    Last edited: Oct 13, 2007
  2. jcsd
  3. Oct 12, 2007 #2


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    Staff: Mentor

    Capacitive reactance varies with frequency, so you need to know what frequency is being used by the AC source. Otherwise, you cannot mix Xc impedances and C capacitances in the same capacitive voltage divider equation. Do you have some way to determine the frequency? Like, are you given both Xc and C for any of the caps (I don't see that)?

    Welcome to the PF, BTW. And yes, you picked the right subforum for your question.
  4. Oct 12, 2007 #3
    thank you for the reply. the Xc for c4 is 2.34k. so, i need to figure out the frequency first? but how do i figure this out?
  5. Oct 12, 2007 #4


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    Staff: Mentor

    I'm not tracking the whole problem very well, but it looks like they have given you a mixture of capacitances and Xc impedance values, and are asking you to calculate two unknown capacitance values. If you had all the known capacitors as capacitance values, then the frequency of excitation would not matter, and you could just calculate the unknown cap values using the relative impedances. But if you are only given information about a couple of the capacitor positions as Xc values, then you need to know the frequency of excitation that the Xc is given for. You would use the frequency to get back to the capacitor values for those positions.
  6. Oct 13, 2007 #5
    this is more confusing than trying to remember the logic gates. anyone want to try and solve this for me please? :)
  7. Oct 13, 2007 #6


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    Staff: Mentor

    but what is the capacitance of c4?

    In the OP one states c4 = 2.34k. Are the units [itex]\mu[/itex]F, which would be an impressive capacitor, or pF?

    Capacitive reactance, Xc = 1/([itex]\omega[/itex]C).
  8. Oct 13, 2007 #7
    oh sorry, the capacitance of c4 wasnt given. but was given an Xc of 2.34k instead
  9. Oct 13, 2007 #8


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    Staff: Mentor

    Well there are two loops and two unknowns, C1 and C3.

    Take the potenials with respect to ground below the voltage source.

    I take it the Xt is the total reactance, and It is the total current, which must pass through the voltage and those capacitors which are in series with the voltage source.

    What can one say about VA, It and Xt?
  10. Oct 13, 2007 #9
    ...Va, It, Xt = Vs ?
  11. Oct 13, 2007 #10


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    Staff: Mentor

    Well Va = Vs, assuming the ground reference is at the bottom left of the figure, i.e. inlet to source.

    It must go through C1, C5 and C6.

    It*XC6 is related to VE.

    Write some equations.
  12. Oct 15, 2007 #11
    edit: nm
    Last edited: Oct 15, 2007
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