Thevenin's Theorem with Source Transforms

In summary, the homework statement says that there is a Thevenin equivalent between A and B using loop analysis and Ohm's law. The Thevenin equivalent was ##V_{th} = -4.0 V## in series with ##R_{th} = 0.8 k##. Next, the post states that source transforms are a sort of "corollary" to Thevenin and Norton's theorems, so I was wondering if it was possible to use those in solving this problem. After realizing my mistake, I was able to solve the problem using different approaches. Finally, the post says that superposition would be useful from this point, but I want to know if it's possible with source transforms. I
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Homework Statement



I was able to find the Thevenin equivalent between ##A## and ##B## using loop analysis and then Ohm's law:

Screen Shot 2014-12-12 at 3.38.19 PM.png


The Thevenin equivalent was ##V_{th} = -4.0 V## in series with ##R_{th} = 0.8 k##.

Now I had a question about solving the problem using a different approach. I realized now source transforms are a sort of "corollary" to Thevenin and Norton's theorems. So I was wondering if it was possible to use those in solving this problem.

Homework Equations

The Attempt at a Solution



From what I can understand, first I should use ##V = IR \Rightarrow I = \frac{5}{2} mA##. Then I could have a current source in parallel with the ##2k## on the left.

Now ##R_{eq} = (\frac{1}{2k} + \frac{1}{4k})^{-1} = \frac{4k}{3}##.

So I have that ##\frac{5}{2} mA## in parallel with ##\frac{4k}{3}##.

From this point I have not been able to continue using source transforms to simplify to the correct answer. Is it just not possible? Or is it a combination I'm not seeing?

I see that superposition would be useful from this point as well, but I want to know if it's possible with source transforms.
 
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Zondrina said:
So I have that ##\frac{5}{2} mA## in parallel with ##\frac{4k}{3}##.

From this point I have not been able to continue using source transforms to simplify to the correct answer. Is it just not possible? Or is it a combination I'm not seeing?

I see that superposition would be useful from this point as well, but I want to know if it's possible with source transforms.

Convert ##\frac{5}{2} mA## in parallel with ##\frac{4k}{3}## back into a voltage source of 10/3 volts in series with a resistance of 4000/3 ohms. Now you have 2 voltage sources in series with two resistors. Slide the 4000/3 ohm resistor to the right so it's just above the 2k output resistor. Those two resistors form a voltage divider driven by the difference in the two voltage sources.

I get Vth = -4V
 
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  • #3
The Electrician said:
Convert ##\frac{5}{2} mA## in parallel with ##\frac{4k}{3}## back into a voltage source of 10/3 volts in series with a resistance of 4000/3 ohms. Now you have 2 voltage sources in series with two resistors. Slide the 4000/3 ohm resistor to the right so it's just above the 2k output resistor. Those two resistors form a voltage divider driven by the difference in the two voltage sources.

I get Vth = -4V

I tried this earlier, but I messed the numbers up and it didn't spit the right answer out.

Reading your post and going back to it, I face-palmed and realized my mistake. I now get the same answer in 10 different ways so I should be good.

Thank you.
 

1. What is Thevenin's Theorem with Source Transforms?

Thevenin's Theorem with Source Transforms is a method used in circuit analysis to simplify complex electrical circuits into a single equivalent circuit. It states that any linear electrical network with voltage and current sources can be represented by an equivalent circuit with a single voltage source and a single resistance.

2. How is Thevenin's Theorem with Source Transforms applied in circuit analysis?

To apply Thevenin's Theorem with Source Transforms, the circuit is first simplified by removing all sources. Then, the equivalent Thevenin voltage and resistance are calculated by finding the open-circuit voltage and short-circuit current across two terminals of the circuit. The original circuit can then be replaced by the Thevenin equivalent circuit without changing the behavior of the circuit.

3. What are the assumptions of Thevenin's Theorem with Source Transforms?

The main assumptions of Thevenin's Theorem with Source Transforms are that the circuit is linear, contains only independent sources, and does not contain any dependent sources. It also assumes that the circuit is in steady state and that the load connected to the circuit does not affect the behavior of the circuit.

4. What are the advantages of using Thevenin's Theorem with Source Transforms?

Thevenin's Theorem with Source Transforms is a powerful tool in circuit analysis as it simplifies complex circuits into an equivalent circuit with only a single voltage source and a single resistance. This makes it easier to analyze and understand the behavior of the circuit. It also allows for easier calculations of circuit parameters such as voltage, current, and power.

5. Are there any limitations to Thevenin's Theorem with Source Transforms?

Thevenin's Theorem with Source Transforms is not applicable to non-linear circuits, circuits with dependent sources, or circuits that are not in steady state. It also assumes that the load connected to the circuit does not affect the behavior of the circuit, which may not always be the case. Additionally, Thevenin's Theorem with Source Transforms is only accurate for linear circuits and may not accurately represent the behavior of non-linear circuits.

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