Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Phasor representation of AC voltage and current

  1. Nov 10, 2007 #1


    User Avatar
    Staff Emeritus
    Science Advisor

    Phasor representation of AC voltage and current.



    in general

    [tex]V\,=\,A\angle{\theta^o}\,=\,A cos{\theta}\,+\,jA sin{\theta}\,V[/tex]

    and similarly for I

    It is assumed that the angular frequency [itex]\omega[/itex] is the same throughout the system, and it is assumed that the Voltage and Current are RMS values.

    For the above phasor values, the voltage and current are:

    v(t) = 141.4 cos ([itex]\omega[/itex]t + 30°)


    i(t) = 7.07 cos [itex]\omega[/itex]t
  2. jcsd
  3. Nov 14, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    AC Power in Phasor Form

    [tex]p(t)\,=\,[V_{max}\,cos(\omega{t}+\theta)] \times [I_{max}\,cos(\omega{t}+\phi)][/tex]



    The average power is

    [tex] P\,=\,V_{rms}I_{rms}\,cos(\theta-\phi)[/tex]

    In phasor notation,










    The real part of power is given by


    and the reactive or imaginary part of power is


    and the quantity [itex]cos(\theta-\phi)[/itex] is known as the power factor.

    The apparent power, S, expressed as volt-amperes (VA) is given by

    S (volt-amps) = P (Watts) + jQ (volt-amps-reactive) = VI*

    |S|2 = |P|2 + |Q|2 = Vrms2 Irms2

    PF = |P|/|S|

    VAR is commonly used as a unit for "volt-amperes-reactive"

    Some useful background on AC power and phasors.


    http://www.physclips.unsw.edu.au/jw/AC.html [Broken]

    http://www.walter-fendt.de/ph11e/accircuit.htm [Broken]
    Last edited by a moderator: May 3, 2017
  4. Jun 11, 2008 #3
    phasor representation

    so Phasor representation of an AC voltage is what magnitude? RMS
  5. Jun 11, 2008 #4


    User Avatar

    you might want to explicitly relate Vmax to Vrms and similar for the currents. in fact, Astronuc, i might define the sinusoids as

    [tex] v(t) \triangleq V_{max} cos(\omega t + \theta) = \sqrt{2} V_{rms} cos(\omega t + \theta) [/tex]


    [tex] i(t) \triangleq I_{max} cos(\omega t + \phi) = \sqrt{2} I_{rms} cos(\omega t + \phi) [/tex]

    and then crank out the instantaneous and mean power as you did.

    i dunno. just a suggestion.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook