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Complex power

  1. Oct 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the complex power delivered by the source
    V = 12cos(wt) V

    upload_2016-10-24_21-11-55.png

    2. Relevant equations
    V = IR

    3. The attempt at a solution
    1. I combined 8ohm resistor and 8j ohm inductor in parallel to get 4+4j ohms
    2. I combined that with 4ohm resistor in series to get a Zth of 8+4j ohms
    3. V = IR -> I = V/R = 12/(8+4j) amps
    4. I = 6/5 - 3i/5

    Complex power = V * I conjugate =
    12 * (6/5 + 3i/5)
    = 14.4 + 7.2i

    Did I do this right?
     

    Attached Files:

  2. jcsd
  3. Oct 25, 2016 #2

    cnh1995

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    Hmm..Your calculations are all correct but I am not sure about this one. When we say complex power "absorbed" by the load, we take the conjugate of the current. So for lagging loads, the complex power comes out to be P+jQ, with positive imaginary part, indicating absorption of reactive power. When the load is leading, the complex power "absorbed" by the load becomes P-jQ, where imaginary part is negative, indicating it is supplying reactive power. In your problem, the source is supplying reactive power and the load is absorbing that reactive power. Hence, IMO, complex power supplied by the source should be VI and not VI*. So it should be S=14.4-j7.2.
     
  4. Oct 25, 2016 #3
    The current is I = 6/5 - 3i/5
    Wouldn't the complex conjugate be 6/5 + 3i/5?
     
  5. Oct 25, 2016 #4

    cnh1995

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    Yes but I think we should not take the conjugate here. The expression S=VI* gives the complex power "absorbed" by the load. The question is asking the complex power supplied by the source. Here, reactive power is being supplied by the source, hence, it should be negative.
     
  6. Oct 25, 2016 #5

    gneill

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    I'm pretty sure that the question author intended that the power delivered by the source is to be interpreted as being identical to the power "handled" by the load. I'd use the complex conjugate of the source current to find that power, and leave any interpretations of the imaginary component for later.
     
  7. Oct 25, 2016 #6
    So source delivers 14.4 - 7.2i and load absorbs 14.4 - 7.2i?
     
  8. Oct 25, 2016 #7

    gneill

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    You had it right in your first post.
     
  9. Oct 27, 2016 #8
    How would I know which (conjugate or not) to use on an exam?
     
  10. Oct 27, 2016 #9

    cnh1995

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    I think you should use the formula VI* as gneill said earlier. It holds for both sending and receiving ends. I think the interpretation of reactive power I wrote is incomplete (or wrong, maybe). I'll search and post if I find any explanation for this.
     
  11. Oct 27, 2016 #10

    cnh1995

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    Well, after doing some math, it turns out that the conjugate part has nothing to do with reactive power but it affects the active power P. VI* gives the true active power P dissipated in the circuit. If VI is used, the value of P (real part of complex power) is different from the actual active power dissipated in the circuit. So, always use S=VI*. S=VI is not even a valid expression. My apologies for misguiding you with this equation.
     
  12. Oct 27, 2016 #11
    It's alright, I really appreciate your help!
     
  13. Oct 30, 2016 #12
    Apparently the correct formula was:
    1/2 * S * I*

    Any idea why?
     
  14. Oct 30, 2016 #13

    cnh1995

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    Where did you get this formula?
     
  15. Oct 30, 2016 #14
    Solutions were posted. Solution for this problem is:
    S = 1/2 * Vm * Im* = 7.2 + j3.6 VA

    I looked through my notes and this was in it:
    S = V * I* = 1/2 * Vm * Im<θv-θi = Vrms * Irms <θv-θi

    Also in my notes:
    Vrms = Vm/sqrt(2)
    Irms = Im/sqrt(2)

    I'm assuming the voltage source provided (12cos(wt)) was not in RMS, so in order to use S = V * I*, we have to convert it to RMS first? It's weird because my textbook uses the formula S = V * I* and not S = 1/2 * V * I*

    I will ask the professor in person but do you have any idea when I'm supposed to used S = V * I* and when I'm supposed to use S = 1/2 V * I*?
     
  16. Oct 30, 2016 #15

    cnh1995

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    Ah.. that's right. I missed that the voltage given is peak voltage and not rms. I read only the 12∠0° part and didn't notice V=12coswt written above the diagram, which indicates 12V is actually the peak voltage. So when the voltage equation is given in terms of peak voltage, use
    Or, convert it into rms and then use
    S=VI*.
     
  17. Oct 30, 2016 #16
    So if I were given only the diagram, (12<0 V), should I assume peak or RMS? When the function is given (12cos), should I always assume peak unless it specifically says RMS?
     
  18. Oct 30, 2016 #17

    cnh1995

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    It means the peak voltage is 12V. RMS value is never written in terms of a sinusoidal equation since it's the dc equivalent of the ac waveform. So whenever you see a sinusoidal equation, it gives the peak of the sinusoid. If the voltage is given directly as 12∠0°, it could be rms or peak. It is usually mentioned in the problem whether to read it as rms or peak. But in some of my books, it is assumed to be rms by default.
     
    Last edited: Oct 30, 2016
  19. Oct 30, 2016 #18
    So in other words, if I see the sinusoidal equation, I should always convert to RMS first.
     
  20. Oct 30, 2016 #19

    gneill

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    If you intend to work with power, yes. Otherwise it's not really an issue, particularly if you're expected to supply an answer for currents or voltages in the same form as given (you'd just end up removing the same scaling factor at the end).
     
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