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Homework Help: Tricky norton equilivent

  1. Sep 10, 2008 #1

    Well that's the problem. I tried to do this problem several ways but it seems to be impervious to several ways or maybe it's just 3:07 am that's the problem :P

    Anyway i got the Rth pretty easy because shorting the Vs also shorts the R1 so R2 = Rth = 56k ohms.

    Now i did get close with getting Voc = 5,600,000*I which with my Rth gives 100*I as the norton current. but according to some software it's supposed to be I^2 instead of I for both Voc and Isc.

    Here this is what i have so far:

    (v2 - v1)/r_1 = i and v_3/r_2 = i*100

    so v1 -> Vs and v2 is node over R_1 and v3 = Voc = node over R2
    Last edited: Sep 10, 2008
  2. jcsd
  3. Sep 10, 2008 #2


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    Hi, I don't think you can find Rth that way because there's a dependent current source in the circuit. Instead find Rth by calculating Voc and Isc.
  4. Sep 10, 2008 #3
    Ya i think your right about the Rth in that software is telling me Rth = 56000.001. I just took that value and tried to figure the work to get it.

    It now tells me (ya i didn't know how to setup the problem before so softwares answers and mine were totally wrong), that vth = (100 * r2*vs)/R1 = vth = (224/3) * vs. So that probably is the right answer but i'm still trying to unwind it to figure out the technique being used.
  5. Sep 10, 2008 #4
    Ok finally making progress here... the program is right. lol

    Basically the unknown variable "i" = -vs/r1
    so plug that into the dependent current source and i*b gives (-vs*b)/r1.

    Now flip the direction and sign of the dependant source and that takes off the negative so it's current is (vs*b)/r1.

    And to get the vth just multiply that current by R2

    so ((vs*b)/r1)=Vth and everthing but vs is known so vth = (224/3)*vs

    Now to get Ishort-circuit... assume a short on the end, and R2 falls off... that leaves the current just discovered at the dependant source to be Inorton = (vs*b)/r1. I believe it to be (1/750)*vs is the norton current.

    and ya Vth/Isc = Rth

    So is this correct? I mean is it really impossible to find the value for Vs?

    I'm gonna try a 1 amp test charge on the Isc version of the circuit to see if i can figure Vs... not sure at this time if i'll get anything
    Last edited: Sep 10, 2008
  6. Sep 10, 2008 #5
    Ya that 1 amp idea didn't fly to well...

    basically i took the thevenin circuit and wired on 1 amp and when working the problem back to get vs... i get 750V... but putting that into my result for the norton circuit is 750/750 = 1 amp and into the thevnin voltage = 56000 Volts... and i think this is a sign that it's not possible to get vs.
  7. Sep 11, 2008 #6


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    Yes I got that as well.

    Ok so far.

    That looks ok.

    Where did you measure 1A? What extra resistor did you connect and how was it connected? More important was this simulated or actual?
  8. Sep 11, 2008 #7
    O, well first off this was just done on paper (and not simulated with software or hardware), and basically i was trying to use the technique where you put a 1 Amp test charge across the output terminals (to measure the voltage across that point and use Vth / 1amp = Rth)... but bit more reading suggested this technique is only to be used for finding Rth and to only used if there are no independent sources present in the circuit (only dependent ones). So basically I was wrong to use that technique to find Vs lol.
  9. Sep 11, 2008 #8


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    Hmm, how did you put a 1A test current through the terminals? By connecting a current source of 1A? But then you wouldn't know the potential drop across the current source since you don't have Vs.
  10. Apr 23, 2009 #9
    Hi, In your picture, although the problem is asking for the Norton equivalent circult you see the labels vth and Rth. But you don't need to find vth. You can find Isc directly by shorting the terminals:

    Isc is related directly to i, and i is related directly to vs.

    So Isc is a function of vs. vs will always be an unknown.

    Your Rth is correct.
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