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AC Waveform question

  1. Apr 24, 2009 #1
    1. The problem statement, all variables and given/known data

    v = 60sin(458t + 60º)

    Find the voltage of the waveform when t = 5ms

    It may be a silly question but ive just started studying ac, just wondering how that + 60º affects the equation?

    any help would be appreciated as ive looked everywhere for the method....


    2. Relevant equations

    V = VmSin(2πft)

    3. The attempt at a solution

    v = 60sin(458t + 60º)

    => v=60sin(458(5x10-3) + 60º)

    => v=60(0.039957 + sin60º) ? -----> not sure if this is correct


    I would appreciate any reply :smile:




    ~Newbie
     
  2. jcsd
  3. Apr 24, 2009 #2
    Welcome to PF, newbie991!

    The 60º only sifts the angle, well 60 degrees. You've probably done trigonometry, where sin(∏+∏)=sin(2∏). You just got to be careful, though. Here the voltage is in the form v=60sin(ωt+60º).
     
  4. Apr 24, 2009 #3
    Thanks, very interesting forum indeed! :)

    Thanks for the reply, ive roughly plotted out this function and im getting a different answer to the theoretical value.

    what would you estimate it to be?

    from the forumula ive extracted:

    vmax=60 p-p = 120
    f=72.892Hz
    T=13.71ms

    assuming all that is correct, i found v after 5ms to be between -40 and -20, although im getting -11.66.

    60 Sin (458 (5x10-3) + 1/3 ∏) = -11.66

    I know theres something right in front of me im not seeing!

    would appreciate any further help!

    ~Newbie
     
  5. Apr 24, 2009 #4
    I'm not quite following you here. What does the last part, p-p = 120, mean?

    When I graphically checked the answer, I got the same as plugging in the values to the given signal.
     
  6. Apr 24, 2009 #5
    ah sorry thats kind of irrelevant thats the peak to peak voltage,

    could you tell me what your answer was so i can see if im on the right track?

    or could you tell me where im going wrong here:
    60 Sin (458 (5x10-3) + 1/3 ∏)
     
  7. Apr 24, 2009 #6

    berkeman

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    Staff: Mentor

    The way the problem was originally listed, it looks to all be in degrees. You just made an error with your calculator in multiplying 5ms * 458:

    Re-check 458 * 0.005 = ______
     
  8. Apr 24, 2009 #7
    thanks for the reply :smile:

    458 * 0.005 = 2.29 :tongue:

    ok this is exactly what im entering:

    rad mode:
    60 sin (458(0.005) + 1/3 ∏)

    and im getting: -11.66159....
     
  9. Apr 24, 2009 #8

    berkeman

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    Staff: Mentor

    sin(458 * 0.005) = 0.752

    sin(PI/3) = 0.866

    sin(458 * 0.005 + PI/3) = sin(2.29 + 1.047) = sin(3.337) = sin(3.337 - 3.142) = sin(0.195)

    So the answer cannot be negative. Try doing the calculation in a different order, and also do the individual pieces to check your work.
     
  10. Apr 24, 2009 #9

    berkeman

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    Staff: Mentor

    Maybe your 1/3 PI isn't turning out to be PI/3....
     
  11. Apr 24, 2009 #10

    berkeman

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    Staff: Mentor

    Oops, my bad. I subtracted only one PI inside the sine.... that's not right. Give me a sec...
     
  12. Apr 24, 2009 #11

    berkeman

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    Staff: Mentor

    sin(3.337) = -0.194

    60 * sin(3.337) = -11.65

    Are you sure that's not the right answer? If it's not, then maybe the whole original problem was in degrees after all.

    Is the answer 60sin(2.29 + 60) = 53.1 ?
     
  13. Apr 24, 2009 #12
    I get the same V=-11.66V as well. Is this an online assignment, or how do you know it's not the right answer?

    If the problem was in degrees, I would assume the angular frequency would have the degree symbol added. After all, rad/s is the more commonly used unit.
     
    Last edited: Apr 24, 2009
  14. Apr 24, 2009 #13
    that is the correct answer :wink:

    went over everything again and -11.66 is correct :biggrin:
    was just a silly mistake on my part!

    thanks alot for your help Kruum and berkeman, much appreciated :wink:
     
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