Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

why is the phas angle the same for voltage and resistance in this series cct

  • 29-06-2005 8:10am
    #1
    Registered Users, Registered Users 2 Posts: 1,073 ✭✭✭


    im just windering why the phas angle the same for voltage and resistance in this series cct

    thanks guyz :D
    heres the question: its question 3.6
    soemoftheballet0139qe.th.jpg


    and heres the soulution (question3.6 just a reminder!)
    soemoftheballet0uh.th.jpg


Comments

  • Registered Users, Registered Users 2 Posts: 717 ✭✭✭Mad Mike


    Do you mean "Why is the angle the same for voltage and impedance (Z) in the line marked (a)"? If so I will try and explain:

    The AC version of Ohms law says I=V/Z where I, V and Z are all complex numbers having a magnitude and phase.

    If we do this sum in polar notation this means that:
    Magnitude (I) = Magnitude(V)/Magnitude(Z)
    and Angle(I) = Angle(V)-Angle(Z)
    These two equations arise from the rules of complex number arithmetic.

    Now remember that in order to assign a phase angle to voltage and currents we need to pick one voltage or current as the reference which we arbitrarily assign a phase of 0. The phase of all other voltages and currents are measured relative to this. You are probably more used to seeing a voltage chosen as the reference. If you choose the voltage as a reference then Angle(V) = 0 and the equation above becomes Angle(I)=-Angle(Z).

    However in this solution the person has chosen to make I the reference this means that Angle(I)=0 and when you put this into the equation you get
    0=Angle(V)-Angle(Z) or Angle(V)=Angle(Z)

    Remember that the choice of reference is entirely arbitrary. It will change the numbers you get for phase angles but only because you are measuring from a different reference. Also the choice of reference doesn't affect the angle of the impedance at all. You could solve thie problem just as easily by choosing voltage as the reference - you may want to try this to test yourself.


  • Registered Users, Registered Users 2 Posts: 1,073 ✭✭✭eurotrotter


    thanks very much for that
    but i am still unclear why the phase angle for the current and the resistance are the same
    Mad Mike wrote:
    Do you mean "Why is the angle the same for voltage and impedance (Z) in the line marked (a)"? If so I will try and explain:

    The AC version of Ohms law says I=V/Z where I, V and Z are all complex numbers having a magnitude and phase.

    If we do this sum in polar notation this means that:
    Magnitude (I) = Magnitude(V)/Magnitude(Z)
    and Angle(I) = Angle(V)-Angle(Z)
    These two equations arise from the rules of complex number arithmetic.

    Now remember that in order to assign a phase angle to voltage and currents we need to pick one voltage or current as the reference which we arbitrarily assign a phase of 0. The phase of all other voltages and currents are measured relative to this. You are probably more used to seeing a voltage chosen as the reference. If you choose the voltage as a reference then Angle(V) = 0 and the equation above becomes Angle(I)=-Angle(Z).

    However in this solution the person has chosen to make I the reference this means that Angle(I)=0 and when you put this into the equation you get
    0=Angle(V)-Angle(Z) or Angle(V)=Angle(Z)

    Remember that the choice of reference is entirely arbitrary. It will change the numbers you get for phase angles but only because you are measuring from a different reference. Also the choice of reference doesn't affect the angle of the impedance at all. You could solve thie problem just as easily by choosing voltage as the reference - you may want to try this to test yourself.


  • Registered Users, Registered Users 2 Posts: 717 ✭✭✭Mad Mike


    I don't quite understand your question - could you be more specific please. What line of the solution are you referring to?

    If you are asking why the voltage across the resistor has the same angle as the current then that is because a resistor causes no phase shift between voltage and current. The voltage across a resistor is always in phase with the current through the resistor.


  • Closed Accounts Posts: 2,256 ✭✭✭Molly


    Out of interest, is this Power Engineering in second year of Elec Eng in U.C.C.


  • Registered Users, Registered Users 2 Posts: 1,073 ✭✭✭eurotrotter


    Molly wrote:
    Out of interest, is this Power Engineering in second year of Elec Eng in U.C.C.
    ya tis!


  • Advertisement
  • Registered Users, Registered Users 2 Posts: 225 ✭✭CathalMc


    ya tis!

    It looked vaguely familiar to me too... small world


  • Registered Users, Registered Users 2 Posts: 363 ✭✭SparkyLarks


    You decided to make the phase angle of the current 0. so you current is a real number.

    V=ZI
    AS ther is no imaginary part (reactive component) the phase angle of V and Z must be the same. Multiply an imaginary number by 5 and the phase angle remains the same remember.

    If you think of it this way. The phase angle of the voltage is determined by the current passing through an impedance. If the impedance is purely active, the Voltage across the impedance will be active.

    If the impedance has a phase angle 90 degrees, purely reactive(inductor). The Voltage across the impedance will lag behind the by 90 degrees.

    If the impedance has a phase angle -90 degrees, purely reactive (capacitor). The Voltage across the impedance will lead behind the by 90 degrees.

    Ti's the same for all the other phase angles( i.e an inpedance with an activve and reactive component


Advertisement