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Norton Resistance Question

Discussion in 'Electronic Basics' started by [email protected], Nov 18, 2008.

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  1. Guest

    Hi guys,

    I've gotten a lot of repsonses, but no real answers. To clariy, the
    text says to remove RL and short the source. To you guys that think
    this is my homework.... good thing it is not. This is actually on a
    computer aided instruction course at my tech school. It states the
    answer as 21 ohms. I came up with 20.55 ohms. It generally wants two
    decimal places, so I would be shocked if it rounded 20.55 to 21. So, I
    was picking other folks brains.

    Here is the original post:

    Can someone tell me (step by step) how to calculate the Norton's
    resistance for this circuit? (see picture below)

  2. sert

    sert Guest

    wrote in
    There are so many tutorials on the web and in books that you
    shouldn't have a problem with such an easy circuit.

    But, to answer your question, there are several equivalent
    ways to reach your desired conclusion.

    The easiest here is to repeatedly convert your voltage source
    to a current source and vice versa.

    You convert a voltage source E (in series with a resistor R)
    to a current source I by replacing it with a current source
    valued I = E/R. The same resistance is then removed and placed
    parallel to the current source. That leaves you with two
    resistances in parallel which you can combine into one and
    repeat the procedure converting the other way round.

    That means that the first voltage source with a current source
    I = E/R = 50/5 = 10 A and the 5 Ohm resistance will go in
    parallel. You then combine the two 5 Ohm resistances into one
    2.5 Ohm one and replace the 10 A source with a voltage source
    with E = I*R = 10*2.5 = 25 V. You proceed that way until you
    reach the Norton equivalent.
  3. Guest

    Wow.. never mind guys.... I didn't expect this much pent up
    hostility. Honestly.... i will move on. I simply wanted to validate
    if the screen was rounding, I have the computer's answer, but I was
    simply trying to validate it. As I stated in my 2nd post, (and per
    the instruction on the computer) I remove RL and short the load.
    After that I calculate resistance using Ohm's Law. I got 20.55. The
    computer says it is 21 (multiple guess). I was wondering if my Ohm's
    law is flawed, or if the computer is rounding. I spent 3 hours trying
    to get an answer on my own. SO IT IS NOT A LACK OF EFFORT. But C'est
    la vie.... other things need my attention. I will move on. I do
    appreciate the responses.

  4. Guest

    If you think it is homework, then email me on January 1st with your
    answer. (I would suspect that my "homework" would be due before then)
    Let's see if you can do it. And by the way, it's ulterior, not
  5. Guest

    The two 5 ohm resistors in parallel (R1 and R4) give you 2.5 ohms.
    This is added to the 10 ohm (R2). This 12.5 ohms is in parallel with
    an additional 10 ohm (R5). This equates to 5.55555555 ohms added to
    (R3) 15 ohms. 15 + 5.55 is 20.55 (which is what i got). The %^&*
    computer ALWAYS wants two decimal places, but this time the choices
    were 17, 21 and 25... I think they dropped the ball. Or broke their
    own rounding rule. But - THAT - is how I arrived at my answer.
  6. Rich Grise

    Rich Grise Guest

    Did they say two _decimal places_, or two _significant figures_, which
    you're interpreting as decimal places? (the two are not the same.)?

  7. Guest

    Wow... I must say that technical arrogance is not just limited to my
    little corner of the world. Here are the factst. I GRADUATED
    technical school in 1983. I am taking some core classes at the local
    tech school (to get my associates). A friend of mine is an adjunct
    instructor there and we sat down to go through some of these courses
    (It's been 25 years for God's sake) to see what we could remember. We
    were rocking along pretty well until we got here. It piqued my
    curiosity and I wanted to make sure I wasn't totally brain dead.

    I didn't know that a simple question would breed more accusations than
    an episode of the Jerry Springer Show. MY BAD!!! For crying out
    loud, put down the scope probe, turn off the function generator, find
    a girl and do what comes naturally. I know the economy is bad and
    people are a bit uptight.. but JEEEZZZ!!!
  8. Guest

    OK!!! OK!!! I concede. I harken back to the days when the web was a
    friendly place. Like Main Street on a summer's day. I forget
    sometimes that the world has changed around me. But alas, I suppose
    that it is "Change We Need"..... *****SIGH*******
  9. Jasen Betts

    Jasen Betts Guest

    It's good to see evidence of an effort being made.

    the impedance left of R4 is

    0 + 5 = 5

    left of R2

    5 || 5 = 2.5

    left of R5

    2.5 + 10 = 12.5

    left of R3

    12.5 || 10 = 5+(5/9)

    left of RL =

    15 + 5+(5/9) = 20+(5/9) = 20.56

    I get (essentially) the same answer as you.

    my question is do they say "two significant figures" or "two decimal places."

    real resistors typically aren't precision devices a good one will be
    within 2% of the marked value
  10. Guest

    Past lessons were always precise. If you omit one of the 5 ohm
    resistors, then that gives you 15 ohms in parallel with a 10 ohm.
    That becomes a whole 6 ohms added to the 15 ohm, which would give a
    precise 21 ohms. I went back and looked at the lesson criteria and
    question, and there was no verbage stating otherwise. Questions
    before and after included decimals on similar questions. Go figure.
    I just wanted to be sure that I was still trainable.

  11. Rich Grise

    Rich Grise Guest"Norton+Resistance"&btnG=Search
    (people would be significantly less "hostile" if you'd checked that

    Have Fun!
  12. Bob Pownall

    Bob Pownall Guest

    I probably should have posted this link to "How To Ask Questions The
    Smart Way" earlier in this thread...

    It's about asking questions on computer software and hardware problems
    in a way that gets you useful answers, but really - it applies to pretty
    much any topic.

    Bob Pownall
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