Understanding Norton Current and Impedance in Point A to B Circuit

In summary, the homework statement is trying to find the impedance between points A and B. The student is unfamiliar with how to find Thevenin voltage and impedance and is asking for help. The student is given a technique for finding the Thevenin voltage and impedance and is asked to use it to solve for the impedance.
  • #1
crom1
57
1

Homework Statement


Find Norton current and impedance between points A and B.

Homework Equations

The Attempt at a Solution


I really don't have idea how to find impedance when points A and B are positioned like that in the diagram. And if let's say I found the impedance , when trying to look for Thevenin voltage(I don't 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.
 

Attachments

  • zad2.png
    zad2.png
    2.9 KB · Views: 362
Physics news on Phys.org
  • #2
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:
  • #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?
 
  • #4
crom1 said:
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?
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.
 
  • #5
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?
 
  • #6
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.
 
  • #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?
 
  • #8
Zth is a matter of series parallel impedances. Replace the source by a short and proceed.
 
Last edited:
  • Like
Likes crom1
  • #9
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?
 
  • Like
Likes crom1
  • #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?
 
  • #11
crom1 said:
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?
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.
 
  • Like
Likes crom1
  • #12
I meant why Zt is independant of that R, but I get it now,thanks.
 

1. What is Norton current and impedance?

Norton current and impedance are important concepts in electrical circuit analysis. Norton current is the equivalent current that flows through a circuit when a load is connected across its output terminals. It is named after American engineer Edward Norton. Impedance, on the other hand, is a measure of the opposition to current flow in a circuit. It is represented by the letter Z and is measured in ohms (Ω).

2. How is Norton current calculated?

Norton current is calculated by dividing the open-circuit voltage by the total resistance of the circuit. This can be represented by the formula IN = VOC / RT, where IN is the Norton current, VOC is the open-circuit voltage, and RT is the total resistance of the circuit.

3. What is the difference between Norton current and Thevenin current?

Norton current and Thevenin current are two equivalent circuit analysis techniques. The main difference between them is that Norton current is the current that flows through a short circuit across the output terminals, while Thevenin current is the current that flows through a load connected across the output terminals. Additionally, Norton current is represented by a current source, while Thevenin current is represented by a voltage source.

4. How can I convert a Norton equivalent circuit to a Thevenin equivalent circuit?

To convert a Norton equivalent circuit to a Thevenin equivalent circuit, you can use the following steps:

1. Calculate the Thevenin voltage by multiplying the Norton current with the total resistance of the circuit.

2. Replace the Norton current source with a Thevenin voltage source with the calculated value from step 1.

3. Remove the load resistor and calculate the Thevenin resistance by using the formula RT = VTH / IN, where VTH is the Thevenin voltage and IN is the Norton current.

4. Connect the load resistor back to the circuit across the output terminals.

5. How is impedance related to Norton current?

Impedance is related to Norton current through Ohm's law, which states that the current flowing through a circuit is directly proportional to the voltage and inversely proportional to the impedance. This means that the higher the impedance, the lower the Norton current will be. In other words, as impedance increases, the current flowing through the circuit decreases.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
441
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
235
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
889
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
974
Back
Top