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Superposition Diagram Reduction

  1. Mar 5, 2007 #1
    Please click the image to enlarge it

    http://img237.imageshack.us/img237/3605/untitlednu9.th.jpg [Broken]

    I'm having trouble seeing how this circuit was reduced. Can anyone shed some light on it for me?

    My question pertains to the second solution in the image. I see that the 12k and 4k resistors are in parallel and that leads to the second diagram on the bottom. Why is the 6k resistor drawn up there? When I try to redraw the circuit, I keep it just like the first one- with the exception that the 12k and 4k resistors are combined on the rightmost branch- with the 6k and the current source in the same place. When the 6V source is shorted out why is there a need to redraw that branch?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 6, 2007 #2
    I thought about your problem today while at work and realized that the 4K Ohm resistor and the 12K Ohm resistor on the right are in parallel. Look at them, both resistors connect at both ends (directly and through the short): The definition of parallel. Once they are combined in parallel, you can twist the circuit around and get the final result without much effort.

    (If you still have trouble seeing it, pull the upper ends of the two resistors until they touch and then yank them over to one side. Keep in mind that the 6K resistor is connected to both ends of the current source and the parallel resistors connected on only one end as is the other 12K. Twist it around. You'll see it.)

    (I deleted my original solution because it is too complicated -- this is simplier)
    Last edited: Mar 6, 2007
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