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!

Calculating Current Supplied by Voltage Source (Complex algebra)

  1. Apr 29, 2014 #1
    I've attached the question that I am referring to.

    I believe I'm heading in the right direction with this one by stating that:

    1/Rt = 1/(6+j8)Ω + 1/(9-j12)Ω

    But I am confusing myself with my algebra.

    Any help is appreciated
     

    Attached Files:

  2. jcsd
  3. Apr 29, 2014 #2

    berkeman

    User Avatar

    Staff: Mentor

    Welcome to the PF.

    [STRIKE]Looks good so far[/STRIKE]. (Oops, I missed the typo that jambaugh caught below) Go ahead and keep working though the algebra, and post your work so we can check it.

    BTW, it's best if you format the equations with LaTeX, instead of having to carry a bunch of parenthesis along with plain text. There is an introduction to LaTeX in the PF FAQ thread in the Feedback Forum here:

    https://www.physicsforums.com/showthread.php?t=617567

    :smile:
     
    Last edited: Apr 29, 2014
  4. Apr 29, 2014 #3

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    If your 6 is a typo and should be an 8, then fine but I'd call it 1/Z not 1/Rt. 1/Z is the Admittance = 1/Imped.
    Then apply impedance to Ohm's law (ac form)
    V = IZ

    where [itex]V = V_0 exp(j \omega t), I = I_0 exp(j \omega t + j \phi )[/itex].

    You get I as a complex multiple of V and so can get its magnitude to find peak current. For power dissipation you will need to take into account the phase shift [itex] \phi [/itex] since current and voltage won't peak at the same time.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted