Are My Thevenin and Norton Equivalent Calculations Correct?

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Discussion Overview

The discussion revolves around the calculations for Thevenin and Norton equivalents, specifically focusing on the open circuit voltage (Voc) and short circuit current (Isc) in a given circuit. Participants are examining their approaches to these calculations and seeking validation of their methods and results.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant, Tekneek, presents their calculations for Voc and Isc, asserting confidence in their open circuit voltage results but expressing uncertainty about the short circuit current calculations.
  • Tekneek assumes that V2 is Voc due to the parallel configuration of the components.
  • Another participant suggests that Tekneek's calculations appear correct and mentions the possibility of using the current divider formula for the short circuit current calculation.
  • Tekneek seeks clarification on how to apply the voltage divider formula, but a subsequent reply corrects this to the current divider formula.
  • Tekneek later realizes that the controlled current source remains in the circuit during the short circuit, which may affect the current calculations.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the initial calculations, but there is uncertainty regarding the impact of the controlled current source on the short circuit current. The discussion does not reach a consensus on the final calculations or methods.

Contextual Notes

Participants have not fully resolved the implications of the controlled current source on the short circuit current calculations, and there are assumptions about the circuit configuration that may affect the results.

Tekneek
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Homework Statement



For this problem (pic attached) I did the open circuit calculations and found V1 (6v) and V2 (4v) and I know they are correct. However I am not sure if my Voc and Isc (open circuit voltage and short circuit current) calculations are correct.

The Attempt at a Solution



I am assuming V2 is Voc since they are in parallel? Also after creating a short circuit to find Isc, I used KCL at Node 1:

(V1/10)+(V1/5)=1
V1=3.33

So, Isc = V1/5ohms = 0.67 Amps (Because the current wouldn't flow through the branches with current dependent current source and 20ohms resistor)

Thanks for any help. Long time reader first time poster :)
 

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Tekneek said:

Homework Statement



For this problem (pic attached) I did the open circuit calculations and found V1 (6v) and V2 (4v) and I know they are correct. However I am not sure if my Voc and Isc (open circuit voltage and short circuit current) calculations are correct.

The Attempt at a Solution



I am assuming V2 is Voc since they are in parallel? Also after creating a short circuit to find Isc, I used KCL at Node 1:

(V1/10)+(V1/5)=1
V1=3.33

So, Isc = V1/5ohms = 0.67 Amps (Because the current wouldn't flow through the branches with current dependent current source and 20ohms resistor)

Thanks for any help. Long time reader first time poster :)

Hi Tekneek, Welcome to Physics Forums.

Your calculations look fine. For the short circuit current calculation you could also have used the current divider formula, since the 1A divides between the 10 Ω and 5 Ω resistors.
 
gneill said:
Hi Tekneek, Welcome to Physics Forums.

Your calculations look fine. For the short circuit current calculation you could also have used the current divider formula, since the 1A divides between the 10 Ω and 5 Ω resistors.

Thanks for the reply. How would you get the current using the Voltage divider formula?
 
Tekneek said:
Thanks for the reply. How would you get the current using the Voltage divider formula?

Current divider, not voltage divider.
 
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gneill said:
Current divider, not voltage divider.

Yeah got it. I = [(Add 10 and 5 in parallel)/5ohms]*1amp = 0.67 Amp

Thanks!
 
Tekneek said:
Yeah got it. I = [(Add 10 and 5 in parallel)/5ohms]*1amp = 0.67 Amp

Thanks!

Whoops! I just realized that the controlled current source is still in the circuit even though the output is shorted! The path for the short circuit current is separate... so it'll look like:

attachment.php?attachmentid=66653&stc=1&d=1392495914.gif


The controlled current source is going to "steal" half the current before it gets to where the short circuit is. Very sorry about that.
 

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