1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
    http://tinypic.com/r/k0k2ed/5
    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

    NascentOxygen

    User Avatar

    Staff: Mentor

    Is there a schematic to accompany this?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook