Node/Super/Norton

Discussion in 'Homework Help' started by ihaveaquestion, May 5, 2009.

  1. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    Read my other two posts also, I made mistakes in this first post in my writings shown in the links

    The original circuit is drawn at the top
    http://img18.imageshack.us/img18/6787/73694572.jpg
    I'm supposed to find Vout using the Node method and Superposition, then I'm supposed to find the Norton Equivalent...

    As you can see in my Node method attempt, I got stuck at the point where I'm not sure how the current source will fit into my equation for e2...

    Here's my superposition, which I think I got right:
    http://img18.imageshack.us/img18/739/28731840.jpg

    And my Norton Equivalent:
    http://img18.imageshack.us/img18/5115/32868024.jpg

    Weird numbers for my answer for Norton..
     
    Last edited: May 5, 2009
  2. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    Just noticed some things I did wrong:

    On my superposition link:

    1) I noticed I miscalculated -> so i = 6/7 A not 7/8 A, therefore Vout due to current source is 36/7 V


    2) Needed to combine resistors before doing first voltage division:

    V1 = (8(8/3)) / ((8/3) + 2) = (64/3) / (14/3) = 64/14 = 32/7 V

    Vout(for only the voltage source) = ((32/7)*6) / 8 = (192/7) / 8 = 192/56

    Vo-total = 36/7 + 192/56 = 480/56 V

    3) I think I'm just not going to include the current source into my e2 node equation, so it would just be

    e2/6 + (e2-e1)/2 = 0
     
  3. ihaveaquestion

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    May 1, 2009
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    Adjusting my Norton evaluation:

    Voltage at node 'N' = 32/9 V, so i due to voltage source only is i = V/R = (32/9) / 2 = 16/9 A

    i(norton) = 6/7 A + 16/9 A = 166/63 A
    R(norton) = 15/7 Ohms

    correct?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Just had a quick look.....

    It's getting difficult keeping up with all these calculations!

    Anyway, I have (provisionally)

    Inorton = 4A
    Rnorton = 2.25Ω

    Will re-check unless confirmed by others.
     
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You found the mistakes I was about to tell you about! Good going.
     
  6. The Electrician

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    i(norton) isn't correct, but I see in a later post you did get it correct. It's 4 amps. EDIT I see it's TNK that got it correct.

    You have R(norton) correct here, but in that later post you changed it to 2.25 Ω. If you convert 15/7 (which is the correct value) to decimal, it's 2.134286--did you make a slight mistake in this conversion? EDIT I also see this was TNK

    Finally, in your nodal method, your first equation is a little off; it should be:

    (e1 - 8)/2 + e1/4 + (e1 - e2)/2 -6 = 0

    Watch out for those arithmetic mistakes on your final!
     
    Last edited: May 5, 2009
  7. t_n_k

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    Ignore my values - I'm on the wrong planet today.
     
  8. t_n_k

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    Hopefully ....

    Inorton = 4A
    Rnorton = 15/7Ω
     
  9. The Electrician

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    You can't just calculate i due to the voltage source only; you also have to take into account the current source's contribution.

    Anyway, you already have enough info to get i(norton). Vth is 60/7 volts, and Rth (which is the same as r(norton)) is 15/7, so I(norton) is (60/7)/(15/7) = 4 amps.
     
  10. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    Haha I think you're confusing my post with tnk's Electrician..

    Ya my Rnorton is correct @ 15.7, but my current isn't...

    I acutally didn't get 4 A yet..

    The following is still what I would have intuitively gone with:

    Voltage at node 'N' = 32/9 V, so i due to voltage source only is i = V/R = (32/9) / 2 = 16/9 A

    i(norton) = 6/7 A + 16/9 A = 166/63 A


    Not sure why this isn't right.
     
  11. ihaveaquestion

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    May 1, 2009
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    I thought I did take the current source's contribution into account by adding 6/7A?
     
  12. ihaveaquestion

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    May 1, 2009
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  13. The Electrician

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    As I said in the previous post "You can't just calculate i due to the voltage source only; you also have to take into account the current source's contribution."

    To calculate the Norton current, you have to short the output. That means the 6 Ω resistor is replaced with a short and then you can use superposition to calculate the current in the right-hand 2 Ω resistor.

    To calculate the contribution due to the voltage source, you have the right-hand 2 Ω resistor in parallel with the 4 Ω resistor, which gives you 16/5 volts at node 'N'. This gives 16/10 amp in the right-hand 2 Ω resistor.

    With the voltage source shorted, you have 2 Ω, 4 Ω, and 2 Ω in parallel from node 'N' to ground, driven by a 6 amp current source. This gives 24/5 volts at node 'N', and 24/10 amps in the right-hand 2 Ω resistor. The total is 16/10 + 24/10 = 4 amps.
     
  14. The Electrician

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    The 6/7 amp was derived when the 6 Ω resistor wasn't shorted. That was when you were calculating the open circuit output voltage.

    Now you're calculating the short circuit output current. You can't use the 6/7 amp from the open circuit calculation.
     
  15. ihaveaquestion

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    May 1, 2009
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    Ok so I found where the misunderstanding was...

    I didn't think you had to get rid of the 6 ohm resistor... according to the books definition (which of course has no examples like this to work off of), you short the designated terminal pair of the original network...

    So I would think to short that open circuit to the right of the original network, linked here for convenience:

    http://img18.imageshack.us/img18/6787/73694572.jpg
    In other words, have a short in parallel to the right of the 6 ohm resistor... Could you explain why this isn't the case etc, thanks
     
  16. The Electrician

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    It is the case that there is a short to the right of the 6 Ω resistor. Remember this image you posted?: http://img18.imageshack.us/img18/5115/32868024.jpg

    And, that short is directly across the 6 Ω resistor, which means that you might as well delete the 6 Ω resistor.

    Calculate the effective resistance of a parallel combination of a zero Ω resistor (which is what a short is) and a 6 Ω resistor. The resistance of the parallel combination is just zero Ω. There will be no voltage across any resistor which is in parallel with a short, and no current will flow in such a resistor. This means that you can delete that resistor and there will be no effect on circuit performance.
     
  17. ihaveaquestion

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    May 1, 2009
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    Ahh... makes sense, got it - cool.
     
  18. The Electrician

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    I'm glad this point was cleared up before your final!

    Just remember, any resistor in parallel with a short is effectively not there. Delete it in any further calculations.

    This is for a perfect short. If the short isn't exactly zero ohms (for example, it might be 1/1000 ohm), then you will have to calculate the effective parallel resistance of the imperfect short and the parallel resistor to get exact currents and voltages in the circuit.
     
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