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Transformers and Active/Reactive power question from HW

  1. Jan 28, 2017 #1
    1. The problem statement, all variables and given/known data
    A 7200-240 V, 60 Hz transformer is connected for step up operation, and a 144∠46° Ω Load is connected to the secondary. Assume the transformer is ideal and the input voltage is 220 V at 60 Hz.
    a)secondary voltage
    b)secondary current
    c)primary current
    d)input impedance
    3)apparent power input to the transformer

    2. Relevant equations
    since its operating as "Step Up" transformer...the turns ratio aka "a' = 30

    1:30 step up

    3. The attempt at a solution
    a)secondary voltage =220*30= 6600 V∠0°
    b)secondary current = 6000V/144Ω ∠46°= 45.8333 A∠-46°
    c)primary current= a x I sec = 45.83333 * 30 = 1375 A∠-46°
    d)Z input= 220 V / 1375A = 0.16∠46°Ω

    Active / reactive and apparent Power
    are all listed as answers in the back of the book
    I am having trouble keeping these numbers straight in my head! Please help me understand my errors on this problem...

    210.1 kW =active power is my true power
    217.6 kVAR = Is my reactive power
    302.5 kVA 1375*220 = 302500 /1000(for the K) = 302.5 kVA

    Can you kindly show me what I am forgetting about active and reactive power guys?

    Thanks Joe
  2. jcsd
  3. Jan 28, 2017 #2
    One-cycle-averaged power is conveniently calculated by using complex representations of voltage and current such as [tex]\tilde S = \frac{{\tilde V{{\tilde I}^*}}}{2} = P + iQ[/tex], where P and Q are an active and reactive power respectively, and absolute value of S is an apparent power.

    Here, I choose that V and I are for the secondary side of the transformer so, [tex]\tilde V = 6600{e^{i0}} = 6600,{\rm{ }}\tilde I = 45.8333{e^{i\left( { - {{46}^ \circ }} \right)}}[/tex]

    Calculation shows that P ≅ 105 kW, Q ≅ 109 kW, and the apparent power is 151 kW.

    Your values for powers are twice than what I calculated. I guess you missed a factor of 1/2 somehow.
  4. Jan 28, 2017 #3
    210.1 kW
    217.6 kVAR

    These are the answers from the book
  5. Jan 29, 2017 #4
    Since you are dealing with effective or rms voltages, the power equation is:

    S = V * I' = P + j Q

    I' should be I-conjugate

    There is no factor of 2 involved.
  6. Jan 30, 2017 #5
    GUYS, I figured it out...My problem was that once I solved for "S"..and I had the angle @ 46°...then I can solve for active and reactive power
    Power triangle means
    s=302.5 theta=46°
    Active=210.13 =302.5cos 46°
    Reactive=217.34 =210.13tan 46

  7. Jan 30, 2017 #6


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  8. Jan 30, 2017 #7
    210.13 tan 46 gets you 217.59 which is what I was looking for..

    Answers from the book are
    210.1 kW Active Power
    217.6 kVAR Reactive Power

    This is per the power triangle
  9. Jan 30, 2017 #8


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    You took Q=PtanΦ and I misread it as Q=StanΦ.
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