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Nortons and Thevenins Circuits

  1. Feb 13, 2014 #1
    Hello, im dario and have a problem that deals with electrical circuits
    I have a practice problem for Nortons and Thevenins equivalents, and don't really know how to approach it. I'm confused when it comes to choosing a load, and reducing it into the final simple circuit.

    1. The problem statement, all variables and given/known data
    The problem is i the attached picture provided, it is a drawing of a circuit.

    2. Relevant equations
    V = IR

    Current divider: IRx = ( Rtotal / Rx ) Itotal

    Voltage Divider: VRx = (Rx / Rtotal ) * Vtotal

    Thevenins Resistance = Nortons Resistance

    Nortons Current = Thevenins Voltage / Thevenins or Nortons Resistance

    3. The attempt at a solution
    I do not have any progress really because I do not really know how to approach this circuit. Would i neglect the output on the right because no current will flow there and remove the load R1 to determine the circuit? I don't see how else the circuit could be reduced to a Nortons circuit.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Feb 13, 2014 #2

    Simon Bridge

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    In neither approach would you have to choose a load.

    In such circuit diagrams, the load would normally be attached to the terminals A and B.
    You need to pay close attention to what the question says: what are you supposed to find the equivalent for?
    Put that part of the circuit inside a box.

    There is a standard procedure to finding the Norton or Thevenin equivalent to a circuit.
    You will have this procedure in your course notes.
    Start at step one and work your way through.
    Show me where you get stuck.

    If you are having trouble with your notes, then you can look up the procedures online.
    i.e. for thevenin:
    ... there are two methods. The one I use is the second one.

    For norton:
    ... again, two methods: I use the second one.

    Just follow them step-by-step.
  4. Feb 13, 2014 #3
    Here is my attempt at a solution for Nortons equivalent. Im confused for one thing though. For the nortons current, the 30 A current will divide through R1 and R2 and 30 A will then flow through R3. I want to say that the norton's current will end up being 30 A as well but feel as if that is wrong.

    For the norton's resistance I found it to be 18.33 ohms. I open circuited the current generator and simplified the circuit to find the total resistance.

    Attached Files:

  5. Feb 13, 2014 #4

    Simon Bridge

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    Um OK: You did -

    step 1. find the Norton resistance.
    open circuit all current generators: there's just one.
    find the resistance between A and B ... which you did from:

    $$R_N= \left(\frac{1}{R_1}+\frac{1}{R_2}\right)^{-1} + R_3$$
    ... notice that this is just the equivalent resistance in the circuit?

    You have a linear network with an ideal current source of 30A in parallel with an equivalent resistance RN ... the Norton equivalent to this is an ideal current source of IN in parallel with resistance RN ... sketch the two networks next to each other.

    So what value does IN have to be for the Norton network to be the same as yours?
  6. Mar 5, 2014 #5
    Sorry for the really late response, I totally forgot about my new account on here. I better start remembering so I can put it to use more. Thank you for the help Simon, I actually got my professor to help me with this the following day so this problem is finished.
  7. Mar 7, 2014 #6

    Simon Bridge

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    That's great - can I twist your arm to post the answer you worked out.
    That way, anyone else who got stuck the same way you did and googled to this thread can get the answer they need?

    One way to avoid forgetting about these threads is to go into your profile and enable e-mail notifications for your threads. Then you'll get a note when someone replies.
    Cheers :)
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