Need Help With Thevenin Equivalent Question

In summary, galaxy_twirl attempted to solve a question on Thevenin's equivalent, but got different answers when trying to do it another way. She used 2.5V as the result for V2, which checked okay. However, she failed to include the Thévenin resistance in her S/C diagram, which resulted in a wrong answer for VTH. Her other working was erroneous.
  • #1
galaxy_twirl
137
1
Hi everyone. I need some help with regards to a question on Thevenin's equivalent. I attempted to solve it in 2 ways but I got different answers when I tried to do it another way. However, I am not sure of my answer as I do not have the answer key at the moment. The picture below shows one of my attempts at solving it, arriving at the answer, V2 = 2.5V, which gives VOC = -2.5V. My workings are bordered by the red line. Pardon me for the messy working as I was in a rush.

11j7mvk.jpg


The question is below: (Please ignore the red markings and the pencil markings.)

5nj1cg.jpg


Thank you in advance for all your help. :)
 
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  • #2
Hi galaxy_twirl. You need to use the posting template when you create a thread in the homework sections of Physics Forums. Posts without the template will usually be deleted. Please keep that in mind for future posts.
Since you've shown a good effort in your attempt at solution I'll relax the format requirement this time.

The result that you've achieved for the Voc in your work looks good.
 
  • #3
The 2.5V figure for V2 checks okay.

In your S/C diagram I can't see that 5Ω resistance coming from anywhere.
 
  • #4
gneill said:
Hi galaxy_twirl. You need to use the posting template when you create a thread in the homework sections of Physics Forums. Posts without the template will usually be deleted. Please keep that in mind for future posts.
Since you've shown a good effort in your attempt at solution I'll relax the format requirement this time.

The result that you've achieved for the Voc in your work looks good.

Hi gneill. Thank you. I didn't know that we need to follow the template, as I thought as long as I show the question and my attempts, the post should be good. I will keep that in mind in the future. Just wondering, is the template the bolded words shown in the box whenever I create a new thread?

Oh dear. That means my other working is wrong. Is it okay if I post my other (wrong) working here? I did both workings by myself.

Thank you! :)
 
  • #5
NascentOxygen said:
The 2.5V figure for V2 checks okay.

In your S/C diagram I can't see that 5Ω resistance coming from anywhere.

Hi NascentOxygen. Thank you. What do you mean by the 5Ω resistance? I didn't write any 5Ω resistance in my working, at least for the workings enclosed by the red line. :)
 
  • #6
The Thévenin voltage is only half the answer. You need to come up with the Thévenin resistance, too.
 
  • #7
NascentOxygen said:
The Thévenin voltage is only half the answer. You need to come up with the Thévenin resistance, too.

Ah yes, I got the RTH during the test, but I got a mistake when doing VTH, hence I re-did the question after my teacher gave me some hints.

I think I should post my other solution which was erroneous, so I can learn from my mistake as well. :)
 
  • #8
If you can't find the flaw in your alternative approach, post it here.
 
  • #9
NascentOxygen said:
If you can't find the flaw in your alternative approach, post it here.

Alright. Thanks! I shall post it below. :D
 
  • #10
Here is my alternative approach, which was what my teacher wrote on the board, but I got a different answer for V2, which is supposed to be 2.5V. May I know what is wrong with my approach? I suspect it has to do with me putting GND between the 2 10 Ohm resistors.

w6xzs4.jpg


Thanks! :)

(Do tell me if the above image is hard to read. I will try to use a better camera to snap my working.)
 
  • #11
V2 is the voltage at an internal node, it is not the voltage measured across the terminals. You have yet to determine the terminal voltage here.
 
  • #12
NascentOxygen said:
V2 is the voltage at an internal node, it is not the voltage measured across the terminals. You have yet to determine the terminal voltage here.

Oh ya! I see my mistake now. >< Sorry. Thanks! :D
 
  • #13
galaxy_twirl said:
Just wondering, is the template the bolded words shown in the box whenever I create a new thread?
Yes.
 
  • #14
gneill said:
Yes.

Thank you. :)
 

FAQ: Need Help With Thevenin Equivalent Question

1. What is Thevenin's theorem and why is it useful?

Thevenin's theorem is a circuit analysis technique that allows us to simplify complex circuits into a single equivalent circuit. This is useful because it makes it easier to analyze and understand the behavior of the circuit, as well as make calculations for things like voltage, current, and resistance.

2. How do I find the Thevenin equivalent for a circuit?

To find the Thevenin equivalent, you will need to follow a few steps. First, remove the load (resistor, etc.) from the circuit and find the open circuit voltage (Voc) across the load terminals. Then, find the equivalent resistance (Req) by shorting all the voltage sources and opening all the current sources in the original circuit. Finally, draw the Thevenin equivalent circuit with the equivalent voltage source (Voc) and equivalent resistance (Req).

3. Can Thevenin's theorem be used for both DC and AC circuits?

Yes, Thevenin's theorem can be used for both DC (direct current) and AC (alternating current) circuits. However, the calculations for finding the Thevenin equivalent may differ slightly between the two types of circuits.

4. What are the limitations of Thevenin's theorem?

Thevenin's theorem is not applicable for circuits that contain nonlinear elements such as diodes and transistors. It is also not applicable for circuits with dependent sources. Additionally, the theorem only works for linear circuits, meaning that the current-voltage relationship must be linear.

5. Can Thevenin's theorem be used for circuits with multiple sources?

Yes, Thevenin's theorem can be used for circuits with multiple sources. However, it is important to keep in mind that the equivalent circuit will only be valid for the load connected to the original load terminals. If the load is connected to a different set of terminals, the Thevenin equivalent will be different.

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