Thevenins and Nortons theorem

Discussion in 'Electronics Resources' started by smelly, Nov 1, 2009.

  1. smelly

    Thread Starter New Member

    Oct 19, 2009
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    I am finding Thevenins and Nortons theorem really hard and confusing and i just dont know how to do it, I was hoping some of you would be able to help me.
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Take a look at the material in the AAC ebook on Thevenin and Norton. Here is a link to get you started. There are also worked examples in the worksheets.

    hgmjr
     
  3. bertus

    Administrator

    Apr 5, 2008
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  4. ELECTRONERD

    Senior Member

    May 26, 2009
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    Hi,

    This isn't quite off topic, but is "Source Transformations" pertaining to Thevenin and Norton equivalents?

    Thanks,

    Austin
     
  5. hgmjr

    Moderator

    Jan 28, 2005
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    If wikipedia is to be believed they do equate "source transformation" to circuit manipulation using "Thevenin and Norton equivalencies".

    hgmjr
     
  6. ELECTRONERD

    Senior Member

    May 26, 2009
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    Thanks hgmjr,

    I thought it might be something like that.

    Austin
     
  7. ELECTRONERD

    Senior Member

    May 26, 2009
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    Alright, now that that's cleared up I think I can help smelly out.

    First it's important to distinguish the difference between Thevenin's Theorem and Norton's Theorem.

    A voltage source in series with a resistor, can be converted to a current source in parrallel with a resistor. -Courtesy of http://forum.allaboutcircuits.com/member.php?u=27220.

    That being said, go ahead and take a look at the first schematic in the attachment.

    Notice that the 0.8A current source to the left is in parallel with R1; which complies with Norton's Theorem. But, we can transform that into a Thevenin equivalent in order to simplify it with R2. So using Ohm's Law, we find the voltage transform; 0.8A x 2Ω = 1.6V. Now we have 1.6V in series with the 2Ω resistor which can be added to R2 since it is in series.

    Now, you may notice that we can transform the voltage source back into a Norton equivalent in order to get the new R1 resistor in parallel with the new R2 resistor (see second schematic in attachment). Therefore, using Ohm's Law, I = E/R so I = 1.6V/4 = 0.4A. So we now have a 0.4A current source in parallel with a 4Ω resistor (and another 4Ω resistor-shown in "Equivalent #2").

    Now according to the law of parallel resistors, \frac{1}{R1 + R2...} Then find the reciprocal of that. So two parallel 4Ω resistors makes a single 2Ω resistor. Then we transform that back into a Thevenin equivalent, so 0.4A x 2Ω = 0.8V in series with 2Ω. The final schematic shows the complete simplified Thevenin Equivalent.

    All we have to do now is find the voltage drop across R2 = 2 x 0.8A = 1.6V. So 1.6V - 0.8V = 0.8V. The final voltage with that current is 0.8V if we had a meter on the 0.8A.

    Austin
     
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