1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Norton current and impedance

  1. Jan 28, 2016 #1
    1. The problem statement, all variables and given/known data
    Find Norton current and impedance between points A and B.

    2. Relevant equations


    3. The attempt at a solution
    I really dont have idea how to find impedance when points A and B are positioned like that in the diagram. And if lets say I found the impedance , when trying to look for Thevenin voltage(I dont really like norton, I always try to find U_th, speaking of it, when exactly should I use Norton instead Thevenin?), I found that Ut=0, but I don't if that's correct, neither if way of finding it was correct.
     

    Attached Files:

  2. jcsd
  3. Jan 28, 2016 #2

    cnh1995

    User Avatar
    Homework Helper

    You can first convert the circuit into Thevenin equivalent and then by using source transformation, convert it into Norton equivalent. I always prefer this method. Thevenin impedance value is same as that of Norton impedance, but Zth comes in series with Vth and ZN comes in parallel with IN. Are you given the values of R, XL and Xc?
     
    Last edited: Jan 28, 2016
  4. Jan 28, 2016 #3
    But I still don't understand how to find Thevenin impedance (or is my Thevenin voltage correct). I do understand how to find Norton current, but for that I need Thevenin voltage and impedance and then I_N=U_t/Z_t. But when is advisable to use Norton over Thevenin?
     
  5. Jan 28, 2016 #4

    cnh1995

    User Avatar
    Homework Helper

    As per my knowledge, a circuit which can be solved by Norton's theorem can also be solved by Thevenin's theorem. It's a matter of choice. If you are comfortable with current sources, you can choose to go with Norton. Here,they are specifically asking for Norton current. To get that, you can proceed with Norton from the beginning, or you can first develop the Thevenin equivalent and convert it into Norton. It's all up to you.
    However, in some examples, you have to go with only Thevenin or Norton. But here, you have a choice.
     
  6. Jan 28, 2016 #5

    cnh1995

    User Avatar
    Homework Helper

    How did you get Vth=0? It would be 0 only when R2=XL*Xc. What are the values of R,Xc and XL?Could you show your working?
     
  7. Jan 29, 2016 #6

    gneill

    User Avatar

    Staff: Mentor

    If you're not working with actual component impedance values the math is probably going to get a bit messy. Be sure to have a large piece of paper handy :smile:

    You can find expressions for the potentials at A and B individually, then take the difference ##V_A - V_B## to find the open-circuit potential across AB. That's your Thevenin voltage ##V_{th}##. Similarly you can place a short across AB yielding three loops. Solve for the current through the shorting wire (mesh analysis comes to mind) to yield the Norton current ##I_N##. With both of those in hand you should know how to find the Thevenin (= Norton) impedance.
     
  8. Jan 29, 2016 #7
    I forgot to say U=100 V , R=XL=Xc=10 ohms. Can I conclude since this is wheatstone bridge , and R^2=Xl*Xc, that the I_N=0?
    But still, I do get for thevenin voltage for A=50+j50, and B=50+j50,i.e. Ut=0?
    How do I found impedance now when both of these are 0?
     
  9. Jan 29, 2016 #8

    cnh1995

    User Avatar
    Homework Helper

    Zth is a matter of series parallel impedances. Replace the source by a short and proceed.
     
    Last edited: Jan 29, 2016
  10. Jan 29, 2016 #9

    gneill

    User Avatar

    Staff: Mentor

    Here's a technique that can come in handy when dealing with bridge circuits such as this.

    First solve for the voltage at the top of the bridge. Let's call it V1. Use any method you're comfortable with. Looks like a natural for nodal analysis to me.:

    upload_2016-1-29_10-50-59.png

    Now, since both branches of the bridge are in parallel they must both have this voltage V1 across them. You can split V1 into two identical sources, each powering one of the branches:

    upload_2016-1-29_10-53-0.png
    Now can you find the Thevenin impedance?
     
  11. Jan 29, 2016 #10
    Yes,I do get Zt=10 ohms. One more question though, how do I justify the fact that I ignore the resistor R next to voltage source in the first picture?
     
  12. Jan 29, 2016 #11

    gneill

    User Avatar

    Staff: Mentor

    If you think about it, it wasn't ignored. It was taken into account when you determined V1. If that R had been different, V1 would have been different.

    If the operating frequency changes or the reactances are changed, V1 will change, too.
     
  13. Jan 29, 2016 #12
    I meant why Zt is independant of that R, but I get it now,thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted