Anyone can check if these homework questions are correct

Discussion in 'Recreational Math' started by BIGEYE, Dec 6, 2004.

  1. BIGEYE

    BIGEYE Guest

    I have a couple of homework questions, appreciate if someone can take the
    time to check my answers.
    Q1 Given that voltages v1 = 12 sin 100 pi t and v2 = 20 sin ( 100 pi t +
    (pi / 3) ). State the minimum value and the phase angle (relative to v1) of
    the resultant voltage v1 + v2 by writing the sum as a single sinusoid.

    My answer 28 sin ( 100 pi t + 0.667 )



    Q2 In an ac circuit i = 100 sin 20 pi t amperes and v = 50 sin (20 pi t -
    ( pi / 6 )) volts

    Instantaneous power p, is given by p = vi watts

    Use "products-to-sums" formulae to express p in a form involving only one
    sinusoid and hence state the maximum power.

    My answer 2165 - 2500 cos ( 20 pi t - (pi / 6) )



    TIA
     
    BIGEYE, Dec 6, 2004
    #1
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  2. BIGEYE

    Guess who Guest

    It is always possible to get a right answer by wrong methods. Can you
    show your work on how you came by those answers?
     
    Guess who, Dec 6, 2004
    #2
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  3. BIGEYE

    BIGEYE Guest

    OK Q1
    sin(A+B) = sinAcosB+cosAsinB
    to make it easier to write out I made 100pi = w
    20sin(wt + (pi/3))
    20 sin wt cos (pi/3) + 20 cos wt sin (pi/3)
    = 20 sin wt * 0.5 + 20 cos wt * 0.866
    = 10 sin wt + 18.32 cos wt
    => 12 sin wt + 10 sin wt + 18.32 cos wt
    = 22 sin wt + 17.32 cos wt
    I then put into R = wt + a (where a =alpha)
    R^2 - SQRT 22^2 + 17.32^2
    R = 28
    tan a = 17.32/22
    a = 0.667 radians
    therefore v1 + v2 = 28 sin (wt + 0.667)
    substituting 100pi back
    v1 + v2 = 28 sin (100 pi t + 0.667)

    Q2
    w = 20 pi
    50 sin (wt - (pi/6)) * 100 sin wt
    = 5000 sin (wt - (pi/6)) sin wt
    sinA sinB = 0.5[cos (A - B) - cos (A + B)]
    => 5000 * 0.5[ cos (wt - (pi/6) - wt) - cos (wt - (pi/6) + wt)]
    = 2500[cos ( -pi/6) - cos (2wt - (pi/6))]
    = 2500[0.866 - cos (2wt - (pi/6))]
    = 2165 - 2500 cos (2wt - (pi/6))
    Substituting 20 pi back for w
    2165 - 2500 cos (20 pi t - (pi/6))
     
    BIGEYE, Dec 6, 2004
    #3
  4. BIGEYE

    BIGEYE Guest

    In Q1
    R^2 - SQRT 22^2 + 17.32^2
    should read
    R^2 = SQRT 22^2 + 17.32^2

     
    BIGEYE, Dec 6, 2004
    #4
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