Norton Equivalent

Discussion in 'Homework Help' started by Tin Whisker, Jul 22, 2009.

  1. Tin Whisker

    Thread Starter New Member

    Dec 3, 2008
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    Can anyone help with the attached circuit. I can't understand how to work it out!

    I thought you had to turn the voltage source into a current source by dividing it by the nearest resistor? But someone else told me you don't........... :confused:

    There's a beer in it for the most helpful person :D



    [​IMG]
     
    Last edited: Jul 22, 2009
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,284
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    Show your attempt.

    Convert the series combination of the 7 volt source and the 7 ohm resistor to a current source in parallel with a suitable resistor (that's what you will calculate).

    Then what do you think you should do next?
     
  3. Tin Whisker

    Thread Starter New Member

    Dec 3, 2008
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    7 Volts / 7 Ohms = 1 Amp

    The 1 Amp is now in parallel with the 7 Ohm resitor

    Add the 1 Amp to the 6 Amps = 7 Amps

    The 7 Ohm and 9 Ohm are now in parallel, so 7X9/7+9 = 63/16 = 3.9 Ohm Resistor

    So the new circuit looks like this:

    [​IMG]

    Now I would turn the current source back to a voltage and put the 3.9 Ohm resistor in series with the voltage source?
     
  4. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    You are right.A Norton's circuit consists of a current source in parallel with a resistor .So you dont need to convert it to a voltage source again
     
  5. Tin Whisker

    Thread Starter New Member

    Dec 3, 2008
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    Do I now need to remove the 3.9 Ohm resistor to get the final answer?
     
  6. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    No you should not remove it.Your answer is right.Just use current division rule and find the current through 3 ohm ( 7 x 3.9/(3.9+3))
     
  7. Tin Whisker

    Thread Starter New Member

    Dec 3, 2008
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    Ok thanks.

    So the answer is 3.95 Amps running through the 3 Ohm resistor (as per the original question)?
     
  8. Tin Whisker

    Thread Starter New Member

    Dec 3, 2008
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    Anyone? Please. :confused:


    Tin Whisker
     
  9. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Yes, that's about right. I get 3.97297 amps through the 3 ohm resistor.
     
  10. Tin Whisker

    Thread Starter New Member

    Dec 3, 2008
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    Thanks for your help everyone.

    I will post another example if you don't mind checking my answer?
     
  11. Tin Whisker

    Thread Starter New Member

    Dec 3, 2008
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    0
    [​IMG]

    Answer:

    6v/6 Ohms = 1 Amp

    1 Amp + 5 Amps = 6 Amps

    6x4/6+4= 24/10 = 2.4 Ohms

    6 x 2.4/2.4 + 2 = 3.2727 Amps
     
  12. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    That's correct.
     
  13. chyadesh

    Member

    Apr 10, 2009
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    please..reply me if my answer is wrong..... I had a thought when we replace the network having 9ohms resistor with short circuit.. in my opinion the current source must be removed since short ciruited and the answer is 1.5 amps...neglecting the current source
     
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