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Norton's Theorem Question

Discussion in 'General Electronics' started by Mrs. Kerchief, Nov 10, 2003.

  1. Why do you not use Rn to compute the resistance when you're trying to
    arrive at In? (Or do you?) Am I wrong thinking you arrive at In
    using the first resistor closest to the output terminals (as with
    Thevenin's Theorem)?

    My DC instructor spends one class per theorem, obviously not too
    little for other students; and obviously I'm never going to be an
    electrical engineer. I have a B average and would like to get my
    associate's in computer technology. (Plus I would dearly love to
    understand material I have shelled out big bucks to assimilate.)

    Anyway, if Rn=Thn, could someone tell me how you arrive at -48.2mA for
    In in the following problem? The Rn here was given at 56.9. (The 33
    resistor is depicted as being in series with the 3 Volt source; the
    other two are depicted in parallel.)

    |-------------------------------------------------------A
    | | |
    | | |
    | | |
    330 Ohms 100 Ohms 200 Ohms
    | | |
    | | |
    | | |
    + 3 V | |
    | | |
    | | |
    |-------------------------------------------------------B

    Also, as long as we're on Norton's, and if anyone is so inclined,
    could you help with In for the following?
    R3
    |-----15K----|
    R1 | |
    --------22K-----------|Is|----------------------------A
    | | 10mA |
    | | |
    | R2 | |R4
    |12V 8.2K Ohms 10K Ohms
    | | |
    | | |
    | | |
    |-----------------------------------------------------B
     
  2. Ken Taylor

    Ken Taylor Guest

    You need to be more careful with your notation - assuming the answer is
    right, the 200 Ohm resistor is actually 220 Ohm (check the problem) - I only
    got that because it's a standard value and 200 isn't (no prizes to anyone
    who points out precision resistor series! :) You also stated that the
    series resistor was 33 Ohms when it's shown as 330 - you aren't being
    careful enough with the problem. Take a bit more time getting the facts down
    correctly.

    Have another crack at it now and get back to us.

    Ken
     
  3. Sorry for the top-post. Yes, you're correct--my notation was wrong in
    the ways you stated. As for the first schematic, I honestly don't
    know what happened when the post appeared; everything's thrown out of
    whack.

    Anyway, if you or anyone is so inclined, emendations below.

    *************************************************
    (The 330 Ohms resistor is in series with the battery; the others are
    in parallel. I consistently get the right Norton equivalent
    resistance but only once, apparently by fluke, was able to compute
    Norton equivalent current correctly. The answers to the below were
    -48.2mA (In) and 56.9 (Rn).
    _____________________________________________________A
    | | |
    | | |
    | | |
    | | |
    330 Ohms | |
    | | 220 Ohms
    | 100 Ohms |
    | | |
    + 3 V | |
    = | |
    | | |
    | | |
    | | |
    ____________________|___________________|_________________B

    Thanks again.
     
  4. Fraze

    Fraze Guest

    I'm going to have to assume that something is wrong with the values of
    your circuit. Since the RN is correct I would say that the values and
    configuration of you resistors are right. For an IN of -48.2mA the
    supply would have to be -15.9V. Could you recheck the original
    problem in the book and let us know? Anyway, here is Norton in a
    nuttshell. To find the norton current source you would short
    terminals A and B and calculate the current that would flow through
    terminals A and B. RN is the same as RTH but it is now shunted across
    the current source. Lets say that the values on your first circuit
    were correct. If you shorted the terminals you would basically have a
    3 volt supply with 330 ohms in series. This would source 9.091mA, the
    direction would be determined by the polarity of the voltage source
    and whether you were using conventional or electron current flow. If
    you would repost the circuit after checking it in the book I might be
    able to explain it further.
    As for the second circuit, you would need to tell me the polarity of
    the sources and let me know what type current flow you are using in
    your course.
    Hope this helps.
    Fraze
     
  5. Thank you for taking the time to respond. We're using
    conventional current in the course, and in the first problem the
    polarity of the 3 Volts is positive on top, negative on the bottom of
    the schematic. I think the In in this problem *must* be wrong as
    given in the answer section in the back of the book. (I've found
    another wrong question in this text, Principles of Electric Current,
    by Thomas Floyd.)
    The polarity for this second problem is the same (positive on top,
    negative on bottom). As the answer for this question is not given in
    the back of the book, I was simply asking for help because I wanted to
    check my answers. The question was not assigned for homework. Again,
    thank you, Fraze.
     
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