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Electrodynamics: AC-circuits and Kirchhoff

  1. Dec 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    Please take a look at the attached circuit. I want to find the current through the capacitor, and of course I will use Kirchhoffs laws for AC-current.

    In this case, ε(t) = ε_0*cos(ωt).

    My question is: If this was an DC-circuit, I would define e.g. counterclockwise to be positive, and then go through the circuit. When I see an EMF which has the terminals the correct way (i.e. it wants to "send" the current through CCW), it gets a "+".

    In this case with the AC-circuit, I don't know which way the EMF wants to send the current, because it changes. Should I always just give ε_0 "+"? Or does the direction in this case even matter?

    Thanks in advance.

    Regards,
    Niles.
     

    Attached Files:

  2. jcsd
  3. Dec 21, 2008 #2

    Hootenanny

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    As for the case of a DC circuit, you may chose your reference direction arbitrarily. Just take care to stick to this direction throughout.
     
  4. Dec 21, 2008 #3
    I have set CW to be positive, and I have solved it two ways:

    1) Saying that the source has polarity as to send currect CCW

    2) Same as #1, but CW.

    When finding the current through the capacitor, the difference I get is in the phase angle. In #1 I get -1.25 and in #2 I get 1.25.

    How is this to be interpreted? If anything, the difference should be 180 degrees?
     
  5. Dec 21, 2008 #4

    Hootenanny

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    Not necessarily. The reference direction that you chose determines the direction of the positive phase angle. So in the first case you chose your reference direction as counter-clockwise and obtained a negative phase angle, i.e. the phase angle is in the opposite direction to your reference direction. In the second case you chose your reference direction to be clockwise and therefore obtained a positive phase angle, i.e. the phase angle is in the same direction as your reference direction.

    Does that make sense?
     
    Last edited: Dec 21, 2008
  6. Dec 21, 2008 #5
    It does, but aren't I allowed to add 2pi, so both phase angles point in the same direction?
     
  7. Dec 21, 2008 #6

    Hootenanny

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    Yes. So as I said in my original post, the choice of reference direction is arbitrary - it doesn't matter as your answers will be the same.
     
  8. Dec 21, 2008 #7
    But the phase-angles only point in the same direction, they don't have the same value; so how can the answers be the same?
     
  9. Dec 21, 2008 #8

    Hootenanny

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    Their value is relative the the initial direction that you chose. In the first case the phase angle is in the opposite direction to your reference direction (hence the negative value). However, in the second case the phase angle is in the same direction as your reference direction (and hence positive).

    Note that both 'values' have the same magnitude and point in the same direction, therefore they are equal.
     
  10. Dec 21, 2008 #9
    Ahh, I see. So I don't even have to add the 2pi to see it.

    But when drawing a phasor diagram, projecting the phasor on the horizontal axis will give you the current. If I project the two phasors in this situation, they will give me the same current (i.e. that why I believed they had to differ by 180 degrees).

    What is the explanation for this?
     
  11. Dec 23, 2008 #10
    I don't mean to be annoying, but can you guys find the error in my reasoning?
     
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