Norton equivalent

Discussion in 'Homework Help' started by regexp, Dec 27, 2010.

  1. regexp

    Thread Starter New Member

    Nov 20, 2010
    24
    0
    Hi,

    I have the following circuit
    [​IMG]

    And i need to calculate the norton equivalent circuit.

    After i shorted the voltage source and changed the current source into an open circuit, the thevenin resistance R_{th} = 4ohm

    Applying the superposition principle the short circuit current is 2A plus the current generated by the voltage source. This is where i had to stop and think. Are there any current traveling through the 6ohm resistor?
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    Yes the current will be flow through 6ohm resistor.
    And how can 2A is short circuit current?
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    Look at this diagram
     
  4. regexp

    Thread Starter New Member

    Nov 20, 2010
    24
    0
    Thanks, i didn't know you could add them like that.

    Is there any other way besides source transformation?
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    Yes, superposition also work very well
     
  6. regexp

    Thread Starter New Member

    Nov 20, 2010
    24
    0
    Hi,

    Hm, the only way i get 1A from the voltage source is

    if i consider the 6ohm and 3ohm resistor in parallel.

    \frac{6\cdot3}{6+3} = 2

    and then \frac{2\cdot2}{2+2} = 1

    However arent they supposed to be in series when the voltage source appear in between?
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    Hmm, To find Norton equivalent

    First:
    Calculate the output current, IAB, with a short circuit as the load.

    Then find Rth
    Replace all voltage sources with short circuits and all current sources with open circuits.

    [​IMG]

    So first we need to find this short current by using superposition.
    I first remove the current source

    [​IMG]

    And then by inspection I write

    Isc1 = ( [ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω ) / 2Ω = ( (6/(1.2 + 6)) * 1.2 ) / 2 = 0.5A

    And now we replace voltage source with a short circuit

    [​IMG]

    Is2 = 2A * (6Ω||2Ω) / ( 6Ω||2Ω + 2Ω ) = 2A * 2/4 = 1A

    So our final solution is

    Isc = Ics1 + Isc2 = 1.5A
    and Rth= 4Ω


    And of course we can get the same answer with source transformation method, or nodal/mesh analysis .
     
    Last edited: Dec 28, 2010
  8. regexp

    Thread Starter New Member

    Nov 20, 2010
    24
    0
    Hi,

    Thank you very much. There is one part i don't really understand

    Isc1 = ( [ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω ) / 2Ω = ( (6/(1.2 + 6)) * 1.2 ) / 2 = 0.5A

    First, the total current is
    6V/(3Ω||2Ω + 6Ω) = 6/7.2 = 0.83A

    Then we use current division.

    3/(3+2) * 0.83 = 0.498A
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    You use current divider rule, but I use Ohms law.

    First I find total current

    Itot = [ 6V/(3Ω||2Ω + 6Ω) ]

    next I find voltage drop across 3Ω||2Ω

    V = Itot* 3Ω||2Ω =[ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω

    And finally, if I know the voltage across 2Ω resistor I can easily find current that flow through 2Ω resistor.

    I = Isc1 = V/2Ω = ( [ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω ) / 2Ω = ( (6/(1.2 + 6)) * 1.2 ) / 2 = 0.5A

    I hope you understand now.

    And maybe at the end I show how you can use nodal equation to find Isc
    for this circuit:

    [​IMG]

    2A = Va/6Ω + Vb/3Ω + Vb/2Ω (1)

    Vb - Va = 6V ----> Vb = 6V + Va (2)

    2 = Va/6 + 2Vb/6 + 3Vb/6

    2 = (Va + 5Vb)/6

    2= ( Va + 5*( Va+6 ) )/6= ( 6Va + 30) / 6

    2 = (6Va +30) / 6

    12 = 6Va + 30

    -18 = 6Va

    Va = -18/6 =-3V


    Vb = 6V + (-3V) = 3V


    And finaly Isc = Vb/2Ω = 3/2 = 1.5A

    So as you can see you can use any method you want to solve this circuit.
     
    Last edited: Dec 28, 2010
  10. regexp

    Thread Starter New Member

    Nov 20, 2010
    24
    0
    Very helpful, thanks :)
     
Loading...