Finding Norton's equivalent

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In summary, the Norton's resistance is 1.58k Omega and the equivalent circuit for a 2-terminal network is feasible and trivial.
  • #1
johnsy1312
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Find the Norton's equivalent for the following circuit attached







I have found the Norton's resistance:
[itex]RN=\frac{1}{1/5.6 + 1/2.2}=1.58k\Omega[/itex]
I am confused on how to approach finding the current because of the +16v variable, what do i do with that variable? do i just treat that as any other voltage source? Is there another way i could draw that to make things more familiar?
 

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  • #2
The circuit doesn't show any ground (or common) connection, and so is just "floating" off the end of a supposed 16 V potential for which we have no reference point. No reference point makes the 16 V designation meaningless.

It also means there's no path from any part the circuit under study back to the "bottom" of the alleged 16 V source. What does that tell you about the current in that leg (the 2.2 K + 16 V leg) and its influence on the circuit?
 
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  • #3
Yah - the Norton equivalent is for a 2-terminal network.
This network only has one.

The problem statement does not even say which two points to take for the equivalent.
 
  • #4
Simon Bridge said:
Yah - the Norton equivalent is for a 2-terminal network.
This network only has one.

The problem statement does not even say which two points to take for the equivalent.

Presumably the R between the two "terminal connection" circles is to be taken as the load, so as a stand-alone subcircuit a Norton equivalent is viable (and trivial!).
 
  • #5
Hmm-hmmm... could also assume the bottom rail is ground.
This makes the voltage note as one end of a voltage supply.
Then it is not clear of the supply is to be part of the equivalent.

With nothing else to go on I'd ignore the voltage note to find the Norton.
Which does leave the R as the load yep.
 
  • #6
Simon Bridge said:
Hmm-hmmm... could also assume the bottom rail is ground.
This makes the voltage note as one end of a voltage supply.
Then it is not clear of the supply is to be part of the equivalent.
I would disagree with being able to assume the shown bottom rail is ground. If you have labels for external sources then you must have labels for the common return. One without the other is not correct as the choice of which node to make common is actually arbitrary despite the temptation to make the assumption based on recognizing a familiar pattern in the drawing layout.

The only correct assumption one can make from the diagram as given is that the subcircuit is floating without a return path for the "16 V leg".
 
  • #7
Yes, I agree that would be "correct" from the available data.
What is "correct" and what I would do are sometimes different things.
We need context for the diagram don't we?
Lets let OP fill in the gaps ;)
 

1. What is Norton's equivalent?

Norton's equivalent is a theoretical circuit model that is used to simplify complex circuits into an equivalent circuit that is easier to analyze. It consists of a current source and parallel resistor that has the same current-voltage relationship as the original circuit.

2. How is Norton's equivalent different from Thevenin's equivalent?

Norton's equivalent is derived from Thevenin's equivalent by using the Norton's theorem, while Thevenin's equivalent is derived using Thevenin's theorem. Both are used to simplify complex circuits, but Norton's equivalent is represented by a current source and parallel resistor, while Thevenin's equivalent is represented by a voltage source and series resistor.

3. How do you find Norton's equivalent?

To find Norton's equivalent, you need to follow these steps:

  1. Remove the load resistor from the circuit.
  2. Solve for the short-circuit current (Isc) by connecting a wire between the load resistor terminals.
  3. Calculate the equivalent resistance (Req) by looking into the circuit from the load resistor terminals.
  4. The Norton's equivalent current source (In) is equal to Isc, and the parallel resistor (Rn) is equal to Req.

4. Why is Norton's equivalent useful?

Norton's equivalent is useful because it simplifies complex circuits into a single current source and resistor, making it easier to analyze and design circuits. It also allows for the calculation of different circuit parameters, such as voltage and power, by using simple equations and Ohm's law.

5. What are the applications of Norton's equivalent?

Norton's equivalent is commonly used in circuit analysis and design, particularly in power systems and electronic circuits. It is also used in troubleshooting and testing circuits, as well as in computer-aided design (CAD) tools to simulate and optimize circuits.

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