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Homework Help: AC Circuit using circuit parameters - HELP!

  1. Apr 16, 2012 #1
    Hey guys, Just need some help with this question, It asks for the answers to be in circuit parameters:
    For some reason I don't think my answers are correct as I'm finding it hard to make sense of my answers to the questions.
    a)Find the voltage Vab (magnitude and angle)
    Vab = Va-Vb
    current going to the first two resistors: I1=(V[itex]\angle0[/itex])/(2R)
    current going through resistor and inductor: I2 = (V[itex]\angle0[/itex])/(R+Xl[itex]\angle90[/itex])
    Itotal = I1+I2 = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex])

    Va=It*2R = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex])*2R = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(R+2RXl[itex]\angle90[/itex])

    Vb = It*(R+Xl[itex]\angle90[/itex]) = (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+RXl[itex]\angle90[/itex])

    Vab = Va-Vb
    Therefore: (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(R+2RXl[itex]\angle90[/itex]) - (V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+RXl[itex]\angle90[/itex]).

    Not sure if that's right.. A little insight would be appreciated!

    b) Calculate the active power dissipated in each reistance.

    Power dissipated in resitor/s (R) = I^2*R = ((V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex]))^2*R

    Power dissipated in inductor = 0.

    c) Calculate the reactive power dissipated in Xl.

    Q = I^2*jXl = I^2*Xl[itex]\angle90[/itex] = ((V[itex]\angle0[/itex](3R+Xl[itex]\angle90[/itex]))/(3R+2RXl[itex]\angle90[/itex]))^2*Xl[itex]\angle90[/itex]

    d) calculate the total current
    Now this is where Im confused it asks me to calculate the total current now even though i thought I already have.. Is this wrong?

    Thanks alot guys.
  2. jcsd
  3. Apr 16, 2012 #2


    User Avatar

    Staff: Mentor

    Is there a schematic to accompany this?
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