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AC Power

  1. Feb 12, 2013 #1
    (i)Find the total W, VAR, VA consumed and the power factor (Fp) in the circuit
    (ii)Find the r.m.s value of the supply current I

    For part I, power factor, I would like to ask for the explanation of the working:
    1-0.65=0.35
    Fp= Cosθ
    =Cos0.35
    =0.99 (Ans)

    Why do you need to minus both Fp (1-0.65=0.35) to get the θ and how do you know it is not the other way round like 0.65-1 = -0.35?

    And for part II, my working:
    P=VICosθ
    1250=(230)(Cos0.35)(I)
    I=5.427A (Ans)

    Book Answer is 6A, I would like to ask where I gone wrong, thank you.
     

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  2. jcsd
  3. Feb 12, 2013 #2

    SteamKing

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    What's the cos(0.35)? What's the cos(-0.35)?
     
  4. Feb 12, 2013 #3
    Hi SteamKing, I am trying to find the power factor..
     
  5. Feb 12, 2013 #4
    Oh Cos 0.35 = Cos-0.35. Then can you explain how come I need to deduct the Fp from load 1 and 2 to get the θ? Is it because θ=θv-θi ?


    Thanks.
     
  6. Feb 12, 2013 #5

    gneill

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    I'm not sure of the context of this 'working' so I might be off target, but it looks a bit odd to me since the power factor 0.65 is already the cosine of an angle. So the operation cos(1 - 0.65) doesn't make a whole lot of sense to me. Also, 0.99 as the overall power factor doesn't look right to me. I'd expect it to be closer to 0.9.
    If the angle is not 0.35, then that would explain the discrepancy.

    Why not start by finding the VA and VAR for both loads (independently). The second load is simple because it has a power factor of 1; what does that tell you about its reactive power (VAR)?
     
  7. Feb 12, 2013 #6
    With reference to the attached picture below, I know how to solve this question. Because I can make use of the information given from the question to get Zt<θ where θ can be plug into the formula to find S,P,Q and pf.

    However honestly speaking, I don't understand post#1 question at all. I cannot make use of the information to get Zt. I believe both question are the same, just different way of asking.

    Would you be kind enough to explain it to me?


    Thanks.
     

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  8. Feb 12, 2013 #7

    gneill

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    This new question involves a single load consisting of a resistor and inductor in series. Given the component values and the frequency of operation you can work out the net load (W, VA, VAR, Fp). The question in post #1 involves two separate loads with different power factors. The idea is to find the net load (W, VA, VAR, Fp) that the voltage supply sees, but since the components comprising the loads are not specified you need to use a slightly different approach than the other question.
     
  9. Feb 12, 2013 #8
    Load 1 VA:769.22
    Load 2 VA: 750

    769.22+750=1519.22 VA.

    Book answer is 1379.7VA

    And how do I go about solving the Fp? Do I add both Fp from Load 1,2 or....?
     
    Last edited: Feb 12, 2013
  10. Feb 12, 2013 #9
    Hi, can I also ask for the 2 question below, I am trying to find the power delivered to the circuit. Formula used is P=i2R.

    For the 2nd question (right pic) my R is 16Ω because that is the only resistor.

    For the 1st question (left pic) how do I determine my R since there is 2 resistor.
     

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  11. Feb 12, 2013 #10

    gneill

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    VA's don't add that way. Think of the VA's as vectors with W and VAR components (look at your power triangle). Add the like components to find the sum of the VA's . Determine the new power factor from the new VA's components.
     
  12. Feb 12, 2013 #11

    gneill

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    How did you arrive at 16Ω? The resistor is said to be 10Ω and everything else is in parallel with it, so how can the resistance be greater than that?
    You'll have to determine the impedance of the circuit. I'd suggest using complex arithmetic.
     
  13. Feb 12, 2013 #12
    Sorry it's 10 type wrong hehe. Can i say for a single resistor, i would use P=i2R
    and for 2 or more resistor i use P=VICos(θv-θi)?



    Thanks.
     
  14. Feb 12, 2013 #13

    gneill

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    I'm not sure that I understand the question.

    In all cases you should find the equivalent impedance. The impedance will have real and imaginary (reactive) components. The real part of that impedance will be the resistance that dissipates energy.
     
  15. Feb 12, 2013 #14
    Va=var+watts?
     
  16. Feb 12, 2013 #15

    gneill

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    Look at the power triangle (the first post in this thread).
     
  17. Feb 12, 2013 #16
    I got it, thanks for your help.
     
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