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Eenglec - alternating current

  1. May 9, 2010 #1
    1. The problem statement, all variables and given/known data

    A 50 cycle alternating current has a maximum value of 42.42A . It crosses the time axis in a positive direction when time is zero. Determine (a) the time when current first reach a value of 30 A. (b) time when current after having gone through its positive maximum value.
     
  2. jcsd
  3. May 9, 2010 #2
    i don't think this problem is phrased well, making it unsolvable. for example, a frequency needs a time unit attached to it. "50 cycles" means nothing. Also, the function "crossing the time axis in a positive direction" does not seem like enough info to restrict the current's function to a single possibility.
     
  4. May 9, 2010 #3

    Zryn

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    I think he possibly means:

    A 50Hz sin wave has 42.42A amplitude.

    a) Show at what time the amplitude is 30A
    b) Show at what time the amplitude is at its maximum

    So translate that into Amplitude*sin(w*t) and solve for the required values.
     
    Last edited: May 9, 2010
  5. May 9, 2010 #4

    vk6kro

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    That is right, except the question was in terms of currents.

    so, you would say

    30 amps = 42.42 amps * sin ( 2 * pi * F * t)
    where F is the frequency in Hz and t is the time you need.
    remembering to calculate the inverse sine in radians.
     
  6. May 9, 2010 #5
    30 amps = 42.42 amps * sin ( 2 * pi * F * t)

    so would it be

    sin-1(2*3.14*50*0)= 42.42 amps - 30 amps?
     
  7. May 9, 2010 #6

    vk6kro

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    30 amps = 42.42 amps * sin ( 2 * pi * F * t)

    so would it be

    sin-1(2*3.14*50*0)= 42.42 amps - 30 amps?


    No, take the 42.42 to the other side and you get 30 / 42.42 = sine of something.

    So, you need to find the something. What angle (in radians) has a sine of 30 /42.42?
    This what the sin^-1 function on your calculator is for.

    So, you would take the inverse sine of 30 /42.42 then put (2 * 3.14159 * 50 * t) equal to that answer. Then solve for t.

    The sine of anything has to be between +1 and -1. You can see that the ratio 30 / 42.42 is in this range.

    Another check is that a 50 Hz sinewave has a period of 0.02 seconds (1 / 50) so a quarter cycle occurs in 0.02 / 4 seconds or 0.005 seconds. So, you must get an answer between zero and 0.005 seconds.
     
  8. May 9, 2010 #7
    I got 0.0025004829.... is that correct for t?
     
  9. May 10, 2010 #8

    vk6kro

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    Yes, but you wouldn't quote it like that. The supplied data has only 4 significant figures, so you can't give an answer with 10 significant figures, even if your calculator produces them like that.

    So 0.0025 seconds would be better.

    These are very important, so try to get good at them. What would you do if the question asked for the current after 0.0015 seconds?

    Got your PM, but I have answered it here so I won't repeat it in a PM.
     
  10. May 10, 2010 #9
    Oh okay, thanks.

    hmm.. for the second one... should the equation look like this..?

    CURRENT = 42.42 amps * sin ( 2 * pi * F * 0.0015) ?

    I am not sure on how to get the answer for B
     
  11. May 10, 2010 #10

    vk6kro

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    Yes, that would be right.

    I gave you a clue to B earlier:

    Another check is that a 50 Hz sinewave has a period of 0.02 seconds (1 / 50) so a quarter cycle occurs in 0.02 / 4 seconds or 0.005 seconds.

    When does a sinewave become a maximum value?
     
  12. May 10, 2010 #11
    is it 1.414 rms?

    I am confuse
     
  13. May 10, 2010 #12

    vk6kro

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    Maybe this will help:


    [PLAIN]http://dl.dropbox.com/u/4222062/sinewaves%202.PNG [Broken]
     
    Last edited by a moderator: May 4, 2017
  14. May 10, 2010 #13
     
    Last edited by a moderator: May 4, 2017
  15. May 10, 2010 #14

    vk6kro

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    Why?

    You can use the same formula if you like:

    42.42 = 42.42 * sin ( 2 * 3.14159 * 50 * t)

    sin ( 2 * 3.14159 * 50 * t) = 1

    ( 2 * 3.14159 * 50 * t) = sin ^-1 (1) = 1.57079 radians

    so t = 1.57079 / 314 = 0.005 seconds

    But you can see this answer from the graphs above.
     
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