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Another practice question: nortons theorem

  1. Mar 26, 2007 #1
    hi guys, sorry to be a bothersome newbie but i have another electrical query. this time its nortons theorem. i need to determine the current through RL, the voltage across RL and the power developed in RL using the theorem.

    any pointers on where to start would be really appreciated :!!)

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  2. jcsd
  3. Mar 26, 2007 #2
    R(Norton) = R (Thevenin).
    I(Norton) = [E/(R1+(R2//R3))]*[(R2//R3)/R3]


  4. Mar 27, 2007 #3
    thanks nacer, i'd best get to work :)
  5. Mar 27, 2007 #4
    ok forgive my naivety but what do the double backslashes indicate? // like that? i know the single one is divide. i feel really retarded now.
  6. Mar 27, 2007 #5
    I think he is referring to a parallel combination, i.e R1//R2 = (R1*R2)/(R1+R2)

    Norton equivalent is no different from Thevenin. Norton equi. = Current Source with a resistor in parallel and Thevenin has voltage source and resistor in series. So, basically you can apply your knowledge of Thevenin equivalents from antoher thread :biggrin: and then just replace a thevenin equ that you'll get with Nortons
  7. Mar 28, 2007 #6
    thanks guys :) i appreciate it! can anyone tell me why r4 is not considered when calculating the current (using the equation given by nacer) would i just have to apply that formula to the diagram above to find the norton current?
  8. Mar 28, 2007 #7
    Every resistor excluding Rl&R4 has to be considered, because you're looking for so-called short-circuit current, so when you have a direct connection between the RL terminals, where does the current go? Since the current is looking for the easiest path, the path with least resistance, it will go just via a direct connection, rather then going via R4.

    Check out this link to get more info.

    I suggest that you read up on the Nortons procedure.

    1. Calculate the short-circuit current by applying the direct connection between output terminals
    2. Calculate output resistance by replacing the voltage source with a short-circuit
    3. Norton-source = short-circuit current & Output resistance = resistance from step 2.
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