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Average power to ZL coupling coefficent and mutual inductance M?

  1. Jun 12, 2013 #1
    1. The problem statement, all variables and given/known data
    2roktc7.jpg

    In the circuit of Figure 4-2, let N1=2000 turns, N2=1000 turns, R=0, Z1=0 and ZL=300+j400 (Ω). If V1(max)=1500 V, what is the average power delivered to ZL?
    A. 5400 W
    B. 2700 W
    C. 337.5 W
    D. 675 W


    2. Relevant equations
    V2=(N2/N1)V1


    3. The attempt at a solution
    since turn factor is 2:1 then V2= (1/2)V1=750V. and I2=750/(300+j400)= 0.9-1.2i=1.5A (magnitude of polar form)

    here is where i get confused. my book for some examples uses (1/2)I2R for power, however i also know the general power equation is I2R. in this case i decided to go with the book's example so to find average power to ZL I used (1/2)I2R and only used resistive part of ZL in calculating power:
    1/2 (1.5)2*(300) = 337.5. so i put C as my answer.
     
    Last edited: Jun 12, 2013
  2. jcsd
  3. Jun 12, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    The 1/2 comes from taking the RMS value of the given peak value. So ##I_{RMS} = I_{peak}/\sqrt{2}##, and squaring it yields a factor of (1/√2)2 = 1/2.
     
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