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How to proceed with nortonization?

  1. Jul 2, 2011 #1
    Hi, :smile:

    Please have a see on the following link: http://img339.imageshack.us/img339/3438/imgmw.jpg

    You also see my question there. There would be potential drop for Vo. I have also read that when the loop is traversed in the direction of loop current, then the IR term is subtracted; if the loop is traversed in the direction opposite to the direction of current, then IR term is added. Please help me. Thanks

    Cheers
     
  2. jcsd
  3. Jul 2, 2011 #2
    PainterGuy,

    When I solved your second circuit and put my values in your equation it was right. Have you noticed that Vo=Vx? this might help simplify your problem. As for the other part of your question I follow a simple convention. When the loop current enters one end of a resistor, that is the positive side. I mark it with polarity + - from the ingoing end (+) to the out going end (-). Then as I write the loop equations I make any transition going from negative up to positive as a positive number and any path that goes from the positive terminal of the resistor (or source) to the negative terminal as a negative number. For example in your second equation you put

    -Vo -2(I2 - I1)=0

    I wrote

    -Vo +Vx=0 then substitute Vx = 2(I1-I2)

    These are the same thing since you reversed the order of I1 and I2 as you'd previously defined, but this is somewhat confusing for me to do it that way. If you always call a voltage drop negative and a voltage rise as positive it may seem less confusing.
     
    Last edited: Jul 2, 2011
  4. Jul 5, 2011 #3
    Thank you, omega minus.

    Hi everyone,

    I have solved the problem. You can have a look if you like:
    Part 1: http://img5.imageshack.us/img5/6241/part1dp.jpg
    Part 2: http://img89.imageshack.us/img89/6678/part2g.jpg [Broken]

    Regards
     
    Last edited by a moderator: May 5, 2017
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