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Is this the right way to find impedance using complex power and volta

  1. May 18, 2013 #1
    1. The problem statement, all variables and given/known data

    The load (Z) in the circuit below absorbs a complex power of 54+j46 VA when the voltage across it is 3cos(400t+-60)V. Use this information to determine the impedance of the load


    2. Relevant equations

    P = V^2 / Z

    3. The attempt at a solution

    first convert V to polar form
    3 < -60
    convert power to polar form
    70.9<40.4
    Apply the formula
    V^2/P= 9<-120/70.9<-40.4
    =.1268<-160.4
    convert to rectangular

    Z=-0.11947- j0.04248 ohms

    I'm not sure if the answer is correct
     
  2. jcsd
  3. May 18, 2013 #2
    Nope. U cant get such funny angles... take 54 as real power, which is equal to VI cos∅. and take 46 as reactive power which is VI sin∅. You have already found the phase difference (∅) between V and I as 40.4... Now Find RMS value of V from the voltage expression given, and hence the impedance using V and I.
     
    Last edited: May 18, 2013
  4. May 18, 2013 #3
    And BTW, Z=R+jX... R cannot be negative!!!
     
  5. May 18, 2013 #4
    Is this what your trying to say ?

    So Vrms = 3/√2

    Therefore solve for Irms

    which is 54 = 3/√2Icos(40.1)
    I=1.66

    Z=(3/√2) / 1.66

    Then find Z but you don't have an imaginary part ?
     
  6. May 18, 2013 #5
    I will be 39.439 A... You missed a bracket there.... 54 = (3/√2) I cos(40.4)... You have V=3cos(400t+-60)... Current I = 39.439√2 cos(400t+-60+40.4)... Now u can find the imaginary part? This is a complicated question...
     
  7. May 18, 2013 #6
    About that part... V=3cos(400t+-60)... What do u mean by '+-'? Can you please be clear?
     
  8. May 18, 2013 #7
    I just copied the function off the question, i'm assuming it's negative
    Therefore
    V(t)=3<-60
    I(t)=39.439√2<-20.4

    Z= V/I
    =.0537<-39.6

    X=3/29.432sqrt(2)sin(-40.4)

    .0537 -j0.03486

    Is that part correct
     
    Last edited: May 18, 2013
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