Complex Power

Discussion in 'Homework Help' started by xz4chx, Dec 5, 2012.

  1. xz4chx

    Thread Starter Member

    Sep 17, 2012
    So I have the circuit below and this is the information given:

    Load 1 absorbs 4203.22-2835.11j [VA]
    Voltage sources delivers 7500 [W]
    v_s(t) = 400cos(100[rad/s]t+38°) [V]
    Find all possible values for L_A

    So my idea was to put it into phasor domain.

    So V_s=400<38°
    Since the voltage source only delivers power the angle of voltage and current have to be the same since the power factor is 1 so I came up with

    S_L=.5(V_s)(I_L*) (Complex Power Formula)

    So I_L=(7500/400*2)<38° = 37.5<38°

    Then I could find the voltage of the Load and that comes out to be
    Using S_L=.5(V_L)(I_L*)

    V_L= 2*(4203.2-2835.1j)/(37.5<38°)

    V_L= 269.744+18.86j

    After that i thought I could just do a KVL to get the value for L_A (the inductor)

    So KVL

    So I get
    -400<38°+37.5<38°(100LjΩ+3.3Ω)+ 269.744+18.86j = 0

    But this does not give me a real value for L_A, I am getting a complex value
    For reference the answer is 77.1 [mH]
  2. panic mode

    AAC Fanatic!

    Oct 10, 2011
    you did not post original question (please post it exactly "as is") and reading though your comment gives me a headache ;-)

    7500W is P not S. S is measured in VA not W.

    you have there series circuit so current is the same in L, R and load. note that:
    7500 = Presistor + 4203.22
    Presistor = 3296.78W

    but this is only magnitude of RMS current, we don't know the angle of it yet.

    similarly, Rload=Pload/I^2=4.2074 Ohm

    Re(Z)=R+Rload = 3.3+4.2074=7.5074 Ohm