Norton/Thevenin equivalents

  • Thread starter kawafis44
  • Start date
In summary, the homework statement says that to find the Thevenin equivalent of a circuit, you need to deactivate the ideal voltage source and simplify the new circuit. The Thevenin equivalent of the circuit can then be calculated by adding the voltages from the deactivated ideal current sources.
  • #1
kawafis44
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0

Homework Statement


1) Find Norton and Thevenin equivalents.
2) Use a series of Norton-Thevenin and series/parallel transformations to reduce a circuit shown into the single-loop circuit. Then, find current I. (Current I is between two nodes at the botom of (1) circuit!)

Homework Equations


no equations

The Attempt at a Solution


My way of solution is presented on these pictures.

I have got two exercises which I try to solve.
429388a4bf76161f.jpg

97100e909c904025.jpg

Thanks for help in advance!
 
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  • #2
For the first exercise, steps 1 and 2 look alright. The equivalent resistance is the equivalent resistance between the terminals (the resistance a load circuit would be exposed to if it were connected between the terminals). As a result, since out of the two resistors you have left, only the 2 Ω resistor is actually between the terminals (the 1 Ω is just dangling), the equivalent resistance is just 2 Ω.

So, the method of finding Rthev was correct. However, I am not sure what method you were using in steps 3, 4, 5, and 6. As far as I know, the Thévénin voltage is just the open circuit voltage, which you can calculate directly in this example. It's just the 2 A of current from the current source going through the rightmost resistor, so that 2 A * 2 Ω = 4 V.

Once you have the Thévénin equivalent, you can get the Norton. It's just Ithev = Vthev / Rthev = 4 V / 2 Ω = 2 A.

This is also obvious because the short circuit current (the current you'd measure across the leads if you shorted them) is 2 A (from the current source)
 
Last edited:
  • #3
I thought that Thevenin equivalence can be obtained if I at first deactivate ideal voltage source (3), simplify new circuit and calculate voltage, at second deactivate ideal current source (5), simplify new circuit and calculate voltage. Then add these two voltages and it will be Thevenin voltage. (Analogously I can calculate two currents and add them). But I do not know how to simplify (6) and (7).
What about the second example (Current I is between two nodes at the botom of (1) circuit!)?
 

What is a Norton/Thevenin equivalent?

A Norton/Thevenin equivalent is a simplified circuit model that represents a complex circuit, typically used to analyze and design electrical networks. It consists of a single voltage or current source and a single equivalent resistance, which can be used to replace the original circuit without changing its behavior.

How do you find the Norton/Thevenin equivalent?

To find the Norton/Thevenin equivalent, you need to first remove all the loads from the circuit and determine the open-circuit voltage and short-circuit current at the load terminals. Then, you can calculate the equivalent resistance by reducing the original circuit to a single voltage or current source. For Norton, you need to convert the equivalent resistance to a current source by dividing the open-circuit voltage by the equivalent resistance.

What is the difference between Norton and Thevenin equivalents?

The main difference between Norton and Thevenin equivalents is that Norton uses a current source as the equivalent element, while Thevenin uses a voltage source. Additionally, for Norton, the equivalent resistance is calculated by converting the open-circuit voltage to a current source, while for Thevenin, it is calculated by dividing the open-circuit voltage by the short-circuit current.

When is it useful to use a Norton/Thevenin equivalent?

A Norton/Thevenin equivalent is useful when analyzing a complex circuit that contains many components. It allows for simplified calculations and makes it easier to understand the behavior of the circuit. It is also commonly used in circuit design, as it can help determine the optimal values for resistors and other components.

What are some limitations of using Norton/Thevenin equivalents?

One limitation of using Norton/Thevenin equivalents is that they are only accurate for linear circuits, meaning that the components must have a linear relationship between voltage and current. Additionally, they do not take into account any time-varying or transient effects in the circuit. Therefore, they cannot be used for circuits that involve switches, capacitors, or inductors.

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