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Homework Help.

Discussion in 'Electronic Basics' started by Daniel Pitts, Jan 28, 2013.

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  1. Daniel Pitts

    Daniel Pitts Guest

    Hello Electronics Wizards,

    So, I'm taking a Linear Circuits class at a local JC, but due to
    scheduling constraints I skipped the "AC" electronics course. I have
    one particular homework problem that I think I've gotten correct, but
    I'd like to have someone with more experience confirm. Unfortunately
    most of my classmates finish the homework on the due-date, so I can't
    compare notes.

    The problem as stated:

    Find the peak and average power delivered to Rl in the figure:


    ..--o^-._-^-* -. 2:1 .---->|---o--.
    | Fuse ^ | | Diode^ |
    o S* || *S \
    S || S /
    120V rms S || S \ 220 ohm
    S || S /
    o | | \
    | | | |
    `-------------. .------.---o--.
    |
    -----
    ---
    -


    My approach to solving this is:

    120v rms becomes 60v rms on the secondary of the transformer. Assuming a
    sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

    Using the "ideal" model, the diode only conducts one direction, which
    cuts the Vrms in half.

    Pavg = 30v^2/220ohm ~= 4watts
    Ppeak = 85v^2/220ohm ~= 33watts

    Does my approach look correct, or am I missing something? If I am
    missing something, a pointer to what I'm missing should be enough. I
    know I could get a slightly more accurate result if I account for the
    ..7v drop, but I don't think the instructor cares, as long as I
    understand that I could.

    Thanks,
    Daniel.

    P.S. Sorry if this is a duplicate post, I think my NNTP server flaked
    out on the first attempt (as it does far too often), but I'm not always
    sure.
     
  2. Daniel Pitts

    Daniel Pitts Guest

    I guess I don't fully understand what RMS is then... I thought it was
    the square root of the average of the squares of all the values - a sort
    of Euclidean distance.

    If I understand what you're saying, the Vrms would be divided by 4 when
    the diode is in the circuit. This would lead to a Pavg = 15v^2/220ohm ~=
    1watt.

    Is that the proper approach?

    Thanks,
    Daniel.
     
  3. Daniel Pitts

    Daniel Pitts Guest

    Ah, that would be a factor of sqrt(2). I think I've got it now.

    So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

    That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

    Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
    85v^2/220ohm ~= 33watts?

    Actually, is my analysis of what the transformer does correct?
     
  4. Daniel Pitts

    Daniel Pitts Guest

    I've seen, and sketched, enough half-waves that I'm not that concerned
    about seeing it. I do understand that much.
    I'm on track to be a hobbyist, nothing more ;-). However, I think
    you're correct that it my help me to do that work. Its been a while
    since I've done any real calculus.
    Yes, probably true. Deriving the formulas has always been a useful
    exercise for me. I just didn't really have the time to sit down with a
    pencil and paper to do so. Maybe tonight.

    Was my final answer of 8 watts for the Pavg correct?
     
  5. Grin, Here's another way to get to the answer. (Just to check
    yourself.)

    Without the diode what's the average power?

    Then the diode blocks the signal for 1/2 the time so your answer
    should be 1/2 the above number.... does that check?

    George H.



    Hide quoted text -
     
  6. Daniel Pitts

    Daniel Pitts Guest

    This is exactly the analysis I did originally, but according to Phil
    Hobbs the analysis is wrong.

    So, who is correct?
     
  7. Daniel Pitts

    Daniel Pitts Guest

    Ah, okay. so I think I get it now. That makes sense.
     
  8. Daniel Pitts

    Daniel Pitts Guest

    I see that now. Thanks.
    I think I've got it now. Thanks everyone for your help and suggestions!
     
  9. Now that he's worked out the answer, it would be interesting to know
    what his instructor expects the answer to be. I once took a community ed
    night class at a local college. During the day the room was used for a
    vocational electronics program. They had this same basic problems worked
    out on the blackboard, but had the OP's original wrong answer at the end.
     
  10. Tom Biasi

    Tom Biasi Guest

    Just because the instructor displayed a workout doesn't mean he or she
    said it was correct.
     
  11. Daniel Pitts

    Daniel Pitts Guest

    Some of my classmates were comparing answers, and my answer was the "odd
    man out". Others had calculated the average voltage. When I explained
    how I got my answer (both the "intuitive" way and by integration), they
    seemed very confused. One of them asked the instruct about average
    power and average voltage. I'm not sure the instructor was paying full
    attention but they said yes that would work.

    Mathematically I can see it not working, so I'm sticking with my answer
    and current understanding (Thanks Phil Hobbs), even if the instructor
    does say it is wrong when he finally corrects the homework.

    Anyway, thanks everyone for all your help. I appreciate it very much,
    and I feel like I have a better grasp on this concept than many of my
    classmates.

    Ironic since most of them have taken an "AC" course, and I skipped that
    one.
     
  12. amdx

    amdx Guest

    Mikek
    And don't wait for the next edition of, Art of Electronics. I've been
    waiting 6 or 7 years, maybe 10!
    Mikek

    PS. I found this,

    "UPDATE ( 01 / 24 / 2012 ): Sadly, the schedule for the 3rd edition of
    The Art of Electronics has slipped. This post quotes a mail by Winfield
    Hill in which he explains the current state of the book and says that it
    "should be out by late 2012 or early 2013”. Considering the history of
    delays of the 3rd edition, I think that early 2013 is more probable, and
    I would not be surprised at all if it will be much later. It seems
    obvious that both authors would rather delay the 3rd edition even
    further than to rush their work and deliver a book of lower quality.
    Anyway, as soon as I learn of a new schedule I will add it here."

    Here:
    http://www.wisewarthog.com/electronics/horowitz-hill-the-art-of-electronics.html

    And just for giggles, this from 2004.


    I have found in a german group, that the 3rd editon of the book
    "Horowitz, Hill - art of electronics" shall be released in march 2004.
    This means, next month. Does anybody know something about this?
    Paul

    urban myth

    Thanks,
    - Win


    <Bugs Bunny mode ON>

    Ah... whadda you know, buddy?

    <Bugs Bunny mode OFF>
    Ralph
     
  13. Daniel Pitts

    Daniel Pitts Guest

    Yep, so the instructor has considered this approach and answer wrong...
    His approach was to find the average voltage and use that in v^2/r. Of
    course that isn't the answer I came up with... I used wolframalpha.com
    to calculate my result:

    http://www.wolframalpha.com/input/?...(integrate+(v*sinx)^2/r+from+0+to+(pi))/(2pi)

    {r == 220, v == 60 Sqrt[2], Integrate[(v Sin[x])^2/r, {x, 0, Pi}]/(2 Pi)}

    Unless I don't understand the definition of "Average" (sum over count,
    or integral over range) or "Power" (current times voltage).
     
  14. Guest

    Doesn't require calculus at all, only a counterexample. The average
    of a sine wave is zero. The RMS value is not. QED.
     
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