Question About AC

Discussion in 'Homework Help' started by zaidqais, Mar 6, 2009.

  1. zaidqais

    Thread Starter Member

    Feb 14, 2009
    16
    0
    Hi every body

    I am stuck on question from the book Introductry circuits Analysis

    [​IMG]


    I tried to do the following

    To find T1 I need to count the time needed v to be 0

    So my solution is :

    v = 200 Sin(2π1000t+60)

    0= 200 Sin(360000+60)

    Sin(360000t+60) = 0

    Sin־1 (0) = 360000t+60

    t = 60/360000 = 0.16666 ms

    But the book I get the question from said that this answer is 1/3 ms ,, can you please tell me where is the mistake in my solution

    Thank you
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Haven't used the TI but I suspect the problem is the mixing of radians and degrees in the equation. If the calculator can be set in radian mode try converting the 60° to radians first and then solve as you have already done. Hope that is the solution!
     
  3. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    zaidqais,

    w*t = 2*pi*f*t= 2*pi*1000*t ===> f = 1000 Hz ===> P = 1 ms

    They want the time of (120/360) period = 1/3 period which is 1/3 ms = 0.333 ms.

    Ratch
     
    Last edited: Mar 7, 2009
  4. zaidqais

    Thread Starter Member

    Feb 14, 2009
    16
    0
    First thank you for your replys

    I have a question
    in the solution i put it in my first post what is the mistake ? I suppose it must calculate T1 or not !!

    and in the equation
    Code ( (Unknown Language)):
    1.  
    2. w*t = 2*pi*f*t= 2*pi*1000*t ===> f = 1000 kHz ===> P = 1 ms
    3.  
    Pi is 180 degree or 3.14 ?
    note f = 1000 Hz and not 1000 kHz

    Thank you
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    OK - take your point - my reply not helpful

    keeping in mind that sin(x) = 0 at x=0 , pi [180°], 2 x pi [360°], etc

    solve
    v = 200 Sin(2π1000t+60) = 0 as you did!
    Consider the cases where
    Sin(2π1000t+60)= 0 , 180°, etc

    case 1:
    200 Sin(2π1000t+60) = 0
    2π1000t+60 = 0 (strictly 2π1000t + π/3 = 0)
    360000t = -60 (using degrees)
    t = - 0.1667 ms - wrong solution as t is less than zero

    case 2:
    2π1000t+60=180 (without being pedantic about radians)
    360000t=180-60
    t = (180-60)/360000
    t = 0.333 ms (1/3 ms exactly) - first positive time at which function is zero - the required solution. t1




     
  6. zaidqais

    Thread Starter Member

    Feb 14, 2009
    16
    0
    You rock :cool:
     
  7. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    t_n_k,

    Why solve for 200 Sin(2π1000t+60) = 0 ? That was 1/6 ms previously. That is not what the problem asked.

    2π1000t + π/3 = 0 , absolutely, no other way. When you introduce time into trig functions, then all frequencies and phase degrees have to be converted to radians. 2n1000t = -n/3 =====> t = -1/6 ms. Isn't that when the function was zero previously?

    You have to use radians with time. 2n1000t + n/3 = n ====> 2n1000t = 2n/3 ====> t = 1/3 ms, which is when the next zero value will occur.

    Arithmetic wrong, 2n1000t = 180-60 = 120 =====> t = 0.019099, which is off base because radians were not used.

    Ratch
     
    Last edited: Mar 7, 2009
  8. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    t_n_k,

    Upon further reflection, I realize you were right and I was wrong. I did not realize that you were converting radians/sec into Hz/sec. Doing it that way you can work exclusively with Hz instead of radians. Sorry for the mixup. At least we both got the same answers.

    Ratch
     
  9. zaidqais

    Thread Starter Member

    Feb 14, 2009
    16
    0
    Really , you helped me so much

    Thank you very much :)
     
Loading...