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little math help on schematic

Discussion in 'Electronic Basics' started by R.Spinks, Feb 13, 2005.

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  1. R.Spinks

    R.Spinks Guest

    I would like to see the math involved in calculating the current in the 4k
    resistor and the current in the 12k resistor. I have tried to do a node
    current at the top -right of 1k resistor and top of all parallel items)
    which I have called 'node A' (Va) but I keep getting the wrong answer. There
    is 8mA in the 4K and 2.66mA in the 12k (this means the voltage at 'node A'
    must be 32V. (I hand drew the 50mA and 90mA current sources so that's what
    they are supposed to be -- if there was a symbol I missed it). I took the
    current at node A (top) to be:

    ((Va-8)/1k)-50mA - (Va / 12k) - (Va/6k) + 90mA - (Va/4k) = 0 but that does
    not yield 32v. Can someone point an error I've made (as I assume I've set
    something up wrong) or correctly indicate the math for me? Thanks.




    1K
    ___
    |-|___|-----------------------------------|--------
    | | | | | |
    | | | | | |
    | | .-. .-. |90ma .-.
    /+\ | | | | | | | |
    8v ( ) 50ma (\|/) 12K | | | | 6K (/|\) | | 4K
    \-/ | '-' '-' | '-'
    | | | | | |
     
  2. I'm not going to do your homework for you. But I can
    indicate a better approach than you have shown.

    Write a single node equation for the right side of the 1K
    resistor, treating the bottom-most node as 0 V. Use
    superposition to separately calculate the contribution
    of each of the 3 sources and get the net effect by
    adding those independent contributions.
     
  3. John Fields

    John Fields Guest

    ---
    Since the 12K, the 6K, and the 4K resistors are all in parallel, their
    total resistance becomes:


    1
    Rt = -------------------------------- = 2000 ohms
    1 1 1
    ------- + ------- + ------
    12000R 6000R 4000R


    Now, if we redraw your schematic with that in mind, we'll have:


    +8V>--[1K]--+---E2
    |
    [2K]
    |
    0V>---------+


    and we can say:

    8V * 2K
    E2 = --------- = 5.333V
    1K + 2K




    If we now look at the original schematic:


    +8V>--[1K]---+------+------+----5.333V
    | | |
    [12K] [6K] [4K]
    | | |
    0V>----------+------+------+

    we can see that 5.333V will be across all of the parallel resistors,
    so the current in any one of them will be:

    E
    I = ---
    R

    Thus, for the 12000 ohm case,:

    5.333V
    I = -------- ~ 0.000444A ~ 444µA
    12000R

    for the 6000 ohm case,:

    5.333V
    I = -------- ~= 0.000888A ~ 888µA
    6000R


    and for the 4000 ohm case:

    5.333V
    I = -------- ~ 0.00133A ~ 1.33mA
    4000R
     
  4. John Fields

    John Fields Guest

    ---
    Poor, sincere dude's downloaded Andy's program and very quickly
    learned to use it just so he can post a schematic of what he needs
    help with and you chide him because it's homework, and then you hit
    him with nodal analysis and superposition as _the_ solution for his
    problem without even explaining what you mean? From your earlier
    contributions, that seems to be beneath you, so why don't you cut the
    guy some slack and help him out? _seb_, OK?
     
  5. His sincerity was not in doubt and I did not impugn it.
    Hmmm, so "not going to do your homework" becomes
    chiding. Makes me wonder what some real criticism
    or an honest-to-god flame would be called.
    The problem was not likely a real world problem and,
    if it had been, those terms would be well known to the
    OP. Since it looked like a homework problem, and
    because people assigned such problems are exposed
    to both of those concepts, and because learning to use
    them is an important part of learning circuit theory, I
    merely set out to remind him that those methods were
    applicable. The OP was, of course, welcome to ask
    what the terms meant, or do a web search for them,
    if he was not already familiar with them or lacked a
    textbook in which to look them up. There is certainly
    no reason for me to explain them ahead of time.
    I don't grok 'seb', but having tutuored several
    electronics students with good results, I can say
    that the biggest obstacle (at least for those who
    have any business in such a curriculum) is coming
    to the realization that they can apply the abstract
    analytical tools they have (or should have) learned.
    Naming applicable tools was, I thought, the best
    contribution under the circumstances.

    I categorically reject the contention that my reply
    was "beneath" my other contributions. Where
    practical advice is requested, just answering the
    question is a beneficial contribution. Not so for
    homework requests. The OP wanted correction
    of a formula presented without any hint of how it
    was derived. That would be tantamount to doing
    his homework, if it was homework. As everyone
    understands, or can readily figure out, helping any
    student pass courses without doing their homework
    is a bad idea, even for the ostensible beneficiary.

    If the OP was solving a real circuit, and has not
    yet been exposed to superposition and nodal
    analysis, he would clearly benefit from learning
    those techniques. This formum is not the place
    for tutorials on those subjects.
     
  6. R.Spinks

    R.Spinks Guest

    Thanks for your help, John. I am actually not in school, but this problem is
    from a book. The current in the 4k resistor is 8mA not 1.33mA (the solution
    is given). Also, the current in the 12k is 2.66mA (not given but power is
    given at 85.3mW -- works backwards to 2.66mA at 32v which also establishes
    the 8mA in the 4k). In both cases, node A , where you have derived 5.33V is
    actually 32V to establish those voltages. Any ideas?
     

  7. Not counting the 1K resistor, the other resistors are equivalent
    to a single 2K resistor. The nodal impedance at Va is therefor
    1K || 2K == (2/3)K. The voltage divider from the 8V source
    to Va is 2K/(1K+2K) == 2/3. Applying superposition from
    the 3 sources to get 3 terms:
    Va == 8V * (2/3) + 90mA * (2/3)K - 50mA * (2/3)K
    Va == 16/3 V + 80/3 V == 96/3 V == 32 V.

    You can get any of the currents in the parallel collection
    by applying Ohm's law using 32V and the resistance.

    If you are trying to learn this stuff on your own, I have
    to applaud the effort. My father did that and, after a
    long haul and much difficulty, became a professional
    engineer. We still argue about which way current is
    best considered to flow. He keeps talking about
    electrons.
     
  8. R.Spinks

    R.Spinks Guest

    Thanks for your help, John. I am actually not in school, but this problem is
    from a book. The current in the 4k resistor is 8mA not 1.33mA (the solution
    is given). Also, the current in the 12k is 2.66mA (not given but power is
    given at 85.3mW -- works backwards to 2.66mA at 32v which also establishes
    the 8mA in the 4k). In both cases, node A , where you have derived 5.33V is
    actually 32V to establish those voltages. Any ideas?
     
  9. R.Spinks

    R.Spinks Guest

    Thanks to a Mr. Don Taylor for pointing out my math error (sign on the 1K
    resistor contribution).----- Original Message -----
    From: "Don Taylor"

    Sent: Sunday, February 13, 2005 7:23 PM
    Subject: Re: little math help on schematic


    Maybe I'm off base her... by why is ((Va-8)/1k positive
    when -(Va/12k) and -(Va/6k) and -(Va/4k) are all negative?


    Yes, Don. Thanks, that was my error. With the sign correction made the
    proper voltage of 32v at node A is given and the correct resultant currents
    are achieved. Thank you for pointing it out.
     
  10. John Larkin

    John Larkin Guest


    OK, but now you have to add in the current sources. 90-50 = net 40 ma
    flowing into the 5.333 volt node. The node impedance is 1k||2k = 666.6
    ohms, so the additional 40 ma pulls the voltage up from 5.333 to 32.00
    volts.

    John
     
  11. John Fields

    John Fields Guest

    ---
    _Not_ counting the 1k resistor? It's in there isn't it?, so how do
    you propose to do away with it? Oh, I get it... it's inconvenient for
    you to deal with, so you want to want to make it go away so you don't
    have to deal with it. Never mind that the load isn't reactive so you
    can't realize a voltage greater than the DC source voltage feeding it,
    you still want to assume that, somehow, 32VDC is there.
     
  12. You've gone off the deep end here. My language was
    simply a way to refer to the collection of resistors that
    act in a shunt role for the divider fed by the 8V source.
    I was not trying to make the 1K series resistor go away
    and in fact used its value in the next two sentences.

    I urge you to settle down and consider more than the
    first interpretation that occurs to you before your next
    post. Such precaution may keep your foot from any
    further oral insertion.

    You get to 32V, as I did, by summing the contribution
    of the 8V source, as divided by the 1K,2K divider,
    with the product of each current source and the nodal
    impedance. There is no "starting off with 8V" in the
    circuit originally posted. There were current sources
    and a voltage source. Neither implied or was stated
    to have a limit in their effect.

    I hope you are either kidding or having a really bad
    day today because I have found more intelligence in
    most of your other on-topic posts.
    Would a smiley have helped you understand? Or
    do I need to elaborate a trivial aside comment?
     
  13. John Larkin

    John Larkin Guest

    How about the two current sources?

    John
     
  14. John Fields

    John Fields Guest

     
  15. Genome

    Genome Guest

    WHOOPS, I sort of wondered.

    DNA
     
  16. John Fields

    John Fields Guest

     
  17. Genome

    Genome Guest

    That's quite sweet.

    DNA
     
  18. John Larkin

    John Larkin Guest

    It's Valentine's Day!

    John
     
  19. Don Kelly

    Don Kelly Guest

    Taking + currents into node A you should have (8-Va)/1k) rather than
    (Va-8)/1k).
     
  20. Rich Grise

    Rich Grise Guest

    And if we restore the current sources, we have:
    +8V>--[1K]--+-----+-------E2
    | |
    [2K] (/|\) 40 mA
    | |
    0V>---------+-----'

    Solve _this_.
    Yeah, solve the actual circuit, rather than the erroneous simplification.

    Good Luck!
    Rich
     
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