Thevenin/Norton Equivalents

  • Thread starter mathman44
  • Start date
In summary, the current source supplies the current written on it no matter what is the voltage across its terminals. To find the Thevenin/Norton equivalents, one must replace the 20K resistor and the 12-V source with their equivalent Norton's. The emf of the voltage source is equal to the open circuit voltage Uo of the "box", the current of the Norton equivalent source is equal to the short-circuit current flowing through the terminals. The internal resistance is Uo/Is. To obtain the internal resistance, one must determine the resultant resistance between the output terminals and then use that information to find the current source's internal resistance.
  • #1
mathman44
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Homework Statement



Asn3.jpg


I am being asked to draw the Thevenin/Norton equivalents for this circuit. The current sources are throwing me off... how do I deal with these?
 
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  • #2
The current source supplies the current written on it no matter what is the voltage across its terminals.

ehild
 
  • #3
I understand that, I should have been more specific. The problem is turning this into a Thevenin equivalent.
 
  • #4
I need help with this one too
 
  • #5
You have to show some effort. What is the Thevenin-equivalent of an active two-pole?
 
  • #6
I can't even say I know what an active two-pole is... any hints to get this started? I would use superposition if there were only one current source, but even still this wouldn't help me in finding the Thevenin/Norton equivalents..?
 
  • #8
...no mention of an active two-pole there either. I know what Thevenin/Norton theorems are, it's the application that is causing confusion. Like I said before I would use superposition to find the currents in all the loops, if there was only one current source.
 
  • #9
Start by replacing the 20K resistor and the 12-V source with its Norton equivalent.
 
  • #10
You have a box, with resistors and sources inside and an output or two terminals, it is equivalent either with a voltage source or a current source.
The emf of the voltage source is equal to the open circuit voltage Uo of the "box", the current of the Norton equivalent source is equal to the short-circuit current flowing through the terminals. Th einternal resistance is Uo/Is.
You need to obtain the open-circuit voltage: The voltage across the terminals when nothing is connected there. Assume that the upper terminal is at potential Uo with respect to the ground. The 9 mA current from the right current source flows through the 4 kohm resistor and produces 36 V drop across it. At the green node, the potential is Uo-36 V. Here, 3 mA flows downward and 6 mA flows through the voltage source where the potential rises by 12 V up to U-36+12. 6 mA flows through the 20 kohm resistor, causing potential drop of 120 V and here we arrive to zero potential: Uo-36+12-120=0 --> Uo=144 V, so the emf of the Thevenin equivalent is 144 V.

We need the short-circuit current Is which flows through a resistance-free wire connected across the output terminals, both of which are at zero potential now. Assume downward current. Then we have 9-Is current flowing through the 4 kohm resistor, at a potential drop of (9-Is)*4. At the green node, 3 mA current flows through the current source, so 6-Is flows through the voltage source and through the 20 kohm resistor. The voltage source rises the potential by 12 V, the resistor lowers it, so the overall change of potential = -(9-Is)*4 +12-(6-Is)*20=0 -->Is=6 mA. The internal resistance of the Thevenin equivalent source is Ri= Uo/Is=144/6=24 kohm.

It is a bit easier procedure to get the internal resistance by determining the resultant resistance between the output terminals, replacing voltage sources by their internal resistance in series, and current sources by their parallel internal resistance. The sources are ideal here, so the resultant output resistance is 24 kohm.

As for the Norton equivalent, it is a current generator with current Is=6 mA and 24 kohm resistor connected in parallel with it.

ehild
 
  • #11
O.O

Thank you so much.
 

What is Thevenin/Norton equivalence?

Thevenin/Norton equivalence is a concept in electrical circuit analysis that states that any linear two-terminal circuit can be represented by an equivalent circuit consisting of a voltage source in series with a resistor, or a current source in parallel with a resistor. This allows for simplified analysis of complex circuits.

What are the advantages of using Thevenin/Norton equivalence?

The main advantage of using Thevenin/Norton equivalence is that it simplifies complex circuits into simpler equivalent circuits, making it easier to analyze and understand the behavior of the circuit. It also allows for easier troubleshooting and circuit design.

How do you calculate Thevenin and Norton equivalent circuits?

To calculate the Thevenin equivalent circuit, you need to find the Thevenin voltage, which is the open-circuit voltage across the terminals of the circuit, and the Thevenin resistance, which is the equivalent resistance seen by the circuit when all the independent sources are removed. To calculate the Norton equivalent circuit, you need to find the Norton current, which is the short-circuit current through the terminals of the circuit, and the Norton resistance, which is the equivalent resistance seen by the circuit when all the independent sources are removed.

Can Thevenin and Norton equivalent circuits be used for non-linear circuits?

No, Thevenin and Norton equivalence only applies to linear circuits. Non-linear circuits, such as those containing diodes or transistors, cannot be simplified using Thevenin and Norton equivalents.

What are some real-world applications of Thevenin and Norton equivalent circuits?

Thevenin and Norton equivalent circuits are commonly used in the design and analysis of electronic circuits, such as in power supplies, amplifiers, and filters. They are also used in circuit simulation software and in the development of electronic devices.

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