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Norton equivalent

  1. Dec 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Having some Difficulty in starting the second question on this sheet. i have managed to find examples and worked my way through the first question, but unfortunately my tutor has decided to move onto pastures new, and we have an assignment to complete but no tutor which has left me in a slight pickle!

    The main problem at the moment is reducing the circuit into a format where i can understand and work with it.
    im assuming the switch is redundant as it states it is closed, but from there onwards, im not sure which are in parallel and which are in series (as im mechanically biased = clueless)
    2. Relevant equations

    i have all the equations necessary and i have worked out equivalent resistances etc before, but it is literally just knowing the order of things to do with this and what each are doing.

    3. The attempt at a solution

    My first attempt was to close the switch, assume that the 180 and 270 ohm resistors are in series, giving me 450ohm in parallel with the 90. and then having the equivalent (75) in series with the 135. giving me a 200ohm resistor, 6A and the load.

    any help would be much appreciated as im a bit aimless at the minute without a tutor to point me in the right direction.

    thanks very much.
     

    Attached Files:

  2. jcsd
  3. Dec 19, 2013 #2

    The Electrician

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    Gold Member

    If the switch is closed, the 135 ohm resistor is shorted and no longer in the circuit.
     
  4. Dec 19, 2013 #3

    gneill

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    Staff: Mentor

    First determine what closing the the switch S1 accomplishes. For example, how is the 135 Ω resistor affected? Once you've sorted that out the network should look easier to solve for the Norton resistance.

    When determining the Norton or Thevenin resistance you want to suppress the sources and then find the resistance of the resulting network as "seen" from the perspective of the load terminals (load removed, of course).

    One suggestion: sometimes it is convenient to solve circuits in stages, converting Norton to Thevenin and back along the way, gobbling up components as you proceed across the circuit. Once you've figured out S1's effect you might find this hint helpful... :wink:
     
    Last edited: Dec 19, 2013
  5. Dec 19, 2013 #4
    Ah, so the switch shorts out the 135 resistor, and because its an open circuit due to removing the current source im just dealing with the top three resistors and the load.
    Right ill have a bash at that tonight and see if i can find an old example somewhere to walk through with it. thank you both for the advice.
     
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