50 Hz waveform

Discussion in 'Homework Help' started by liam, Apr 27, 2008.

  1. liam

    Thread Starter New Member

    Apr 22, 2008
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    could any one help us with this i have no idea where to start Find the instantaneous value of a 50 Hz waveform with a maximum value of 200v. 0.001 sec after it has passed through a voltage zero. many thanks . liam
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    v=Asin(2πft)
    where v is the instantaneous voltage
    A is the peak (maximum) voltage
    f is the frequency
    t is the time

    so v=200sin(2πx50x0.001) =1.097 Volts

    You said when it has passed through zero: so it may be positive like + 1.098 or negative like - 1.097
     
  3. silvrstring

    Active Member

    Mar 27, 2008
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    Shouldn't this be done using radians? v(.001) would then be (+/-)61.8V.
     
  4. mik3

    Senior Member

    Feb 4, 2008
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    No, he said 0.001 seconds not radians
     
  5. silvrstring

    Active Member

    Mar 27, 2008
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    Yes, I know .001 is time, but omega(2pif) is in radians.
     
  6. mik3

    Senior Member

    Feb 4, 2008
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    no its in radians per second and time is in seconds so id you multiply them you get radians
     
  7. silvrstring

    Active Member

    Mar 27, 2008
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    Right. So you get 200sin(pi/10), or 200sin(18degrees). Both are 61.8V, right? Am I doing something wrong?
     
  8. Caveman

    Active Member

    Apr 15, 2008
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    The dimensional analysis is:

    t in seconds.
    2*PI = radians/cycle
    f = cycles/sec.
    2*PI*f*t = radians/cycle * cycles/sec * sec = radians.

    So, take the sine in radians. The 61.8 magnitude is correct.
     
  9. hgmjr

    Moderator

    Jan 28, 2005
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    I got 61.8 as well.

    hgmjr
     
  10. silvrstring

    Active Member

    Mar 27, 2008
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    Ok. Thanks.
     
  11. mik3

    Senior Member

    Feb 4, 2008
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    Oh yes you are right i forgot to make radians in degrees thanks for the correction. Stupid i am !!:)
     
  12. liam

    Thread Starter New Member

    Apr 22, 2008
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    Thanks for the replies guys. but im still none the wiser as were you got the answer from . i understand that the answer is 61.8v. but where did this come from . when i tried working out the equation v=200sin(2πx50x0.001) =1.097 Volts i got - 0.107 ????
     
  13. Caveman

    Active Member

    Apr 15, 2008
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    Okay, step by step
    1. 2*PI radians/cycle * 50 cycles/second * 0.001 seconds = 0.314 radians

    This is just stating that 0.001 seconds into a 50Hz sine wave is 0.314 radians in.

    2. Take the sine of this, but remember to set up your calculator to calculate your trigonometric functions in radians. So, sin(0.314 radians) = 0.309

    Just so you understand, what this says is that at 0.001 seconds into a 50Hz sine wave, the voltage is 0.309 times the peak.

    3. Multiply by the peak voltage, 200V. 200V*0.309 = 61.8V.

    Got it?
     
  14. liam

    Thread Starter New Member

    Apr 22, 2008
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    your a absolute star! thanks for the really easy explanation. ive managed to work it out
     
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