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Capacitance and Inductance questions

  1. May 20, 2016 #1
    1. The problem statement, all variables and given/known data
    78lfvRG.jpg

    2. Relevant equations


    3. The attempt at a solution
    Here are two questions. I'm stuck on question 1, and need someone to check question 2.

    For the first question I see C2, C3, and C4 in parallel, so Ceq of those 3 are C2 + C3 + C4 = 3C

    but now how do I continue? seems like the 3 remaining capacitors aren't in series or parallel..

    Am I suppose to use Y-delta transformation?

    ______________________________________


    For the 2nd question.

    I see the top 3 inductors are in series and the botton 3 are in series.

    Therefore L1 + L2 + L3 = 3L

    at the bottom row, L4 + L5 + L6 = 3L

    now we have 3L and 3L in parallel connection, 1/Leq = 1/3L + 1/3L

    Leq = 3/2 L
     
  2. jcsd
  3. May 20, 2016 #2

    berkeman

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    What can you replace the voltage source with to keep making progress on combining capacitances here? Hint -- what is the AC impedance looking into a voltage source?
    Correct. :smile:
     
  4. May 20, 2016 #3
    Replace voltage source? AC impedance?

    I'm in a intro class, we only worked with DC so far :(
     
  5. May 20, 2016 #4

    berkeman

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    Well, capacitors and inductors are AC components, so it looks like you've moved on to the next chapter... :smile:
     
  6. May 20, 2016 #5

    berkeman

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    Replace the power supply with an AC short. Then what can you do to keep going...?
     
  7. May 20, 2016 #6
    if the Battery became a short, C1 would be parraell with 3C , and then the Ceq of those 2 would be C + 3C = 4C.

    Then 4C would be in series with C5, then 1/Ceq = 1/4C + 1/C

    Ceq of entire circuit = 4/5 C
     
  8. May 20, 2016 #7

    berkeman

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    Looks good to me. Can you check the answers online?
     
  9. May 20, 2016 #8
    but in what scenario would a battery be replaced by a short? My textbook doesn't seem to cover this, it does say that a capacitor becomes open in a DC circuit, and an inductor becomes closed in a DC circuit.
     
  10. May 20, 2016 #9

    berkeman

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    When you are calculating equivalent impedances, it is common to replace a current source with an open, and a voltage source with a short. That's how the actual circuits for each behave anyway, so it's a reasonable simplification to make when finding equivalent impedances. You will probably be covering it in class soon, since you use it in this question.
     
  11. May 20, 2016 #10

    epenguin

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    Seems to me it is needed to be able to solve this with elementary concepts.

    That C5 might as well not be there – there is no way you can put a charge on it as one end is not connected to anything so there is nowhere a charge can come from to the isolated plate and therefore you cannot maintain a charge on the opposite one.

    Then the potential difference is the same across C2, C3 and C4. Charges on each capacitor would therefore be proportional just to their capacitance.. The total charge is the sum of the three. So the sum the equivalent capacitance of the part 234 will be just the sum of the capacitances, which is in fact the general rule for parallel capacitors

    Replacing then the three capacitors with one equivalent capacitance, we are left with two capacitors in series. Call the second capacitance C234. System there has a conductively isolated 'inside' part with two plates. The charge on one plate is equal but opposite to that on another, because all the electrons on one must have come from the other. So the two capacitors charges are equal to each other, call this charge Q.
    The potential differences in series add up: Vtotal = V1 + V234 = Q1/C1 + Q234/C234. But the charges here, as we previously said, are equal so we just have
    Vtotal = Q(1/C1 + 1/C234)
    Comparing to the general formula V = Q/C you see that the equivalent capacitance C of this circuit is given by
    1/C = (1/C1 + 1/C234)

    So that's practically all the calculation, and all the general formula, from first principles.
     
  12. May 20, 2016 #11
    Maybe it is unclear if it is part of a bigger circuit, or if it is not connected to anything at all. But if it actually is part of a bigger circuit, then the Ceq would be 4/5C then, correct? and If it is actually isolated, then it Ceq would be 3/4 C.
     
  13. May 20, 2016 #12

    gneill

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    That's not the right way to look at it. The open terminals are where you're "looking from" to judge the equivalent capacitance of the network. You have to imagine what the behavior of the circuit would be if some tests were applied at those terminals. Such tests might include applying various voltage or current sources and measuring the results (current injected or voltage that appears across the terminals).

    If open terminals nullified capacitance then an lone capacitor would not have a capacitance...
     
  14. May 20, 2016 #13
    so regardless if the terminals are connected to anything or not, the Ceq of the circuit should be the same/unaffected?
     
  15. May 20, 2016 #14

    gneill

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    Yes.

    An "equivalent value" implies that you are looking at the circuit from some particular port. In this case that's the open terminals of the circuit.
     
  16. May 21, 2016 #15
    is there a way to solve number 1 without using source transformation?
     
  17. May 21, 2016 #16

    cnh1995

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    There is no need of source transformation there. Source transformation means conversion of a voltage source into equivalent current source and vice-versa.
     
  18. May 21, 2016 #17
    I see, is there a way to solve question1 without replacing the battery with a short circuit?
     
  19. May 21, 2016 #18

    cnh1995

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    Equivalent capacitance between two points is the capacitance "seen" by a voltage source connected between those two points. If the equivalent capacitance is asked between the rightmost two points, you need to short the battery. If the equivalent capacitance is asked between the battery terminals, you'll have to open the battery terminals.
     
  20. May 21, 2016 #19
    If I short the terminals on the right, I will have C2 + C3 + C4 + C5 = 4C, since they are in parallel. Then 4C and C1 are in series. Then Ceq would be 4/5 C.
     
  21. May 21, 2016 #20

    cnh1995

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    Why short the terminals on the right? Between which two points is the equivalent capacitance asked?
     
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