1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

AC Circuit Sinusiodal

  1. Dec 8, 2012 #1
    From the 1st screenshot part ii of the question:

    My working:
    Vm x √2 = 28.284 x √2 = 40v (Answer Correct)

    I would like to ask a question from the 2nd screenshot. "Determine the effective (r.m.s) value of this voltage waveform over one complete cycle.

    I don't know how to solve this part. Is the 1st screenshot question which I got the answer the same method as the 2nd ?


    I would appreciate if someone can give some guidance to me, thanks.
     

    Attached Files:

    • a.jpg
      a.jpg
      File size:
      73.6 KB
      Views:
      49
    • b.jpg
      b.jpg
      File size:
      19.9 KB
      Views:
      48
  2. jcsd
  3. Dec 8, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Things are not quite so simple when the waveform is not a sinusoid. Apply the definition of rms and integrate over the period. Here you can "do it manually" since the waveform is composed of simple segments of constant voltage.
     
  4. Dec 8, 2012 #3
    Ok, just wanna check is the 1st question answer correct 40v?

    Effective RMS= Vm x √2

    Rms=Vm/√2

    Am I correct?
     
  5. Dec 8, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    The RMS value of a sinusoid should be less than the peak value. If Vm is your peak, then divide by root 2, not multiply.
     
  6. Dec 8, 2012 #5
    So the answer is actually 20v, the book answer is wrong?

    And how do you determine if the Vm is at peak or not? Since you say if Vm is at peak, x 2. So I need to determine if the Vm is at peak or not

    Thanks gneill.
     
  7. Dec 8, 2012 #6

    gneill

    User Avatar

    Staff: Mentor

    Yup.
    You are given a waveform:

    v(t) = 28.284 sin(....) V

    The peak value of the sine function is 1, so the peak value of the voltage function is 28.284 V.
     
  8. Dec 8, 2012 #7
    I would like to clarify one thing:

    -Calculate the RMS values for voltage
    -Determine the effective RMS value of v(t)

    Are both question asking the thing? RMS is also effective RMS?
     
  9. Dec 8, 2012 #8

    gneill

    User Avatar

    Staff: Mentor

    Yes, they're the same thing.
     
  10. Dec 8, 2012 #9
    Ok. so for the 2nd question, since I'm not given an expression and the waveform is not sinusoidal, do i need to work out an expression for myself and from there mutiply by √2?


    Omg how do i write an expression that is not sinusoidal o_0
     
  11. Dec 8, 2012 #10

    gneill

    User Avatar

    Staff: Mentor

    Note that the constant factor √2 applies to the relationship between peak and rms for sinusoids. It does not so apply for arbitrary waveforms.

    You should apply the definition of RMS to the given waveform.
     
  12. Dec 8, 2012 #11
    Do you mean that the concept of x √2 or / √2 does not apply to this question because it is not a sin wave form?

    If yes how do i find the rms value for square wave form?
     
  13. Dec 8, 2012 #12

    gneill

    User Avatar

    Staff: Mentor

    Yes, that's right.
    Apply the definition of RMS. The name gives a hint on how to remember what to do; it's the square Root of the Mean of the Square value of the function over its period.

    $$f_{rms} = \sqrt{\frac{1}{T}\int_0^T f(t)^2 dt}$$

    Since your voltage signal is comprised of several constant values over the period of 10ms, you can break the sum of the squares into a few pieces that are easy to do by inspection.
     
  14. Dec 8, 2012 #13
    I can also use the 2nd formula from the picture? It's the same as the 1st one and also the one from you right?

    I managed to find the area which is 45, please correct me if I'm wrong.

    But i don't know the meaning of [i2(t)], can you please explain to me?


    Thanks.
     

    Attached Files:

  15. Dec 8, 2012 #14

    gneill

    User Avatar

    Staff: Mentor

    i(t)2 is the square of the current function. Say the current is 4A at some instant in time. Then its square is 16A2 at that instant.

    In your attachment showing the voltage waveform over a period the first portion of the period shows a value of 6V for a time of 5ms. Dropping the units for now, integrating the square over that portion of the curve yields (6)2x5 = 180. Do the same for the rest of the segments and sum the results. Take the mean by dividing the sum by the whole period (10).
     
  16. Dec 8, 2012 #15
    Before I go on to find the rms value, can you help me check the answer for part vii.

    My working for area:
    (6x5)+(9x3)+(-6x2)=45

    Average value:
    45/10=4.5v (Ans)
     
  17. Dec 8, 2012 #16

    gneill

    User Avatar

    Staff: Mentor

    Sure, looks good.
     
  18. Dec 8, 2012 #17
    Ok thanks, back to finding the value of rms.

    I don't understand why (6)2x5 = 180. this working.

    i(t)2 is the square of the current function. Say the current is 4A at some instant in time. Then its square is 16A2 at that instant.

    I would put 45 in the area, since i found it earlier and T = 10s. So for v(t)2 is 36+81+36=153

    √45(153)/10 = 26.23V (Ans Wrong)
     
    Last edited: Dec 8, 2012
  19. Dec 8, 2012 #18

    gneill

    User Avatar

    Staff: Mentor

    That's 6 squared multiplied by 5. 6V is the value of the voltage over a period of 5 milliseconds. So v2t becomes (6)2 x 5 = 180.
    You are no longer working with area. It's no longer just voltage x time (which corresponds to an area under the curve), but v2 x time.

    If you wish, replace each of the unique segments of the curve with a function and do the integration piecewise. So for example, for the second segment it would be:

    $$\int_5^8 9^2 dt$$

    Sum up the values for each segment and then divide by the total period, yielding the MEAN of the SUM of the SQUARE VALUES.
     
  20. Dec 8, 2012 #19
    Why is it not voltage x time which is the area but v2 x time?

    Didn't the formula put area[v2(t)
     
  21. Dec 8, 2012 #20

    gneill

    User Avatar

    Staff: Mentor

    The definition of RMS squares the voltages and multiplies by time. It's not area for RMS.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: AC Circuit Sinusiodal
  1. Ac circuit (Replies: 7)

  2. AC circuit (Replies: 3)

  3. Ac circuits (Replies: 18)

  4. AC Circuits (RL circuit) (Replies: 30)

  5. AC circuit problem (Replies: 4)

Loading...