Thevenin Equivalent: HW Problem Solution Check

In summary: But even if i don't remove the load resistor, and use the same technique to find V_AB, then id still have the same equations because there would still be no current through the 10k resistor.
  • #1
mugzieee
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0
i have an attempted solution to a HW problem, and I want to know if it is right or wrong. Could you guys please help me out. I have attached the image as a pdf
 

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  • #2
Before you can actually start solving the problem, you will need to identify firstly the two points A and B for which the Thevenin equivalent circuit is defined. So where are these points upon which you want to find the Thevenin equivalent circuit? Across the 10k resistor?
 
  • #3
doodle said:
Before you can actually start solving the problem, you will need to identify firstly the two points A and B for which the Thevenin equivalent circuit is defined. So where are these points upon which you want to find the Thevenin equivalent circuit? Across the 10k resistor?
if you look at the second circuit i have labeled the two nodes V_A and V_AB, those are the two nodes in which i am going to do nodal analysis in.
 
  • #4
I didn't calculate the Vth, but your work nodal analysis looks correct to me. On a side note, you didn't really need to convert the mA to A or the kohms to ohms. The prefixes will cancel giving you V during the calculations. Now find Rth to complete the problem.
 
  • #5
teknodude said:
I didn't calculate the Vth, but your work nodal analysis looks correct to me. On a side note, you didn't really need to convert the mA to A or the kohms to ohms. The prefixes will cancel giving you V during the calculations. Now find Rth to complete the problem.
GREAT! thanks teknodude!
 
  • #6
mugzieee said:
if you look at the second circuit i have labeled the two nodes V_A and V_AB, those are the two nodes in which i am going to do nodal analysis in.

I have a few concerns.

1. Why remove the 10k resistor? Shouldn't the load resistor (if there should be one) be lying between the two points where the Thevenin equivalent circuit is defined?

2. I don't think [tex]V_{AB} = V_{Th}[/tex] as you have written in the solution. If the Thevenin equivalent circuit is to be determined across nodes AB and A, then [tex]V_{Th} = V_{AB} - V_A[/tex].
 
  • #7
doodle said:
I have a few concerns.

1. Why remove the 10k resistor? Shouldn't the load resistor (if there should be one) be lying between the two points where the Thevenin equivalent circuit is defined?

2. I don't think [tex]V_{AB} = V_{Th}[/tex] as you have written in the solution. If the Thevenin equivalent circuit is to be determined across nodes AB and A, then [tex]V_{Th} = V_{AB} - V_A[/tex].

you may be right about the load resistor lying betwen the two points. But even if i don't remove the load resistor, and use the same technique to find V_AB, then id still have the same equations because there would still be no current through the 10k resistor.
 
  • #8
That's true. But you should let the 10k resistor be and [tex]V_{Th}[/tex] should be as I stated above.
 

Related to Thevenin Equivalent: HW Problem Solution Check

1. What is Thevenin Equivalent?

Thevenin Equivalent is a method used to simplify complex electrical circuits into a single equivalent circuit for analysis. It is named after French physicist Leon Charles Thevenin.

2. How is Thevenin Equivalent calculated?

Thevenin Equivalent is calculated by finding the equivalent resistance of the circuit, which is the ratio of the voltage difference to the current flowing through the circuit. The equivalent voltage is then determined by measuring the voltage at the load terminals with the load disconnected.

3. Why is Thevenin Equivalent important?

Thevenin Equivalent is important because it allows for easy analysis and understanding of complex circuits. It also helps in designing and troubleshooting circuits, as well as predicting the behavior of a circuit under different load conditions.

4. What are the limitations of Thevenin Equivalent?

Thevenin Equivalent is only applicable to linear circuits, meaning that the relationship between voltage and current remains constant. It also assumes that there are no independent sources in the circuit and that the load is purely resistive.

5. How is Thevenin Equivalent used in real-world applications?

Thevenin Equivalent is commonly used in circuit design, power system analysis, and electronic device testing. It is also used in telecommunications, where it allows for the transmission of complex signals over long distances by simplifying the circuit.

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