Thevenin and Norton Equivalent

In summary: 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:(V1/10)+(V1/5)=1V1=3.33So, Isc = V1/5ohms = 0.67 Amps (Because the current wouldn't flow through the branches with current dependent current source and 20ohms resistor)
  • #1
Tekneek
70
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
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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  • #5
gneill said:
Current divider, not voltage divider.

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

Thanks!
 
  • #6
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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1. What is Thevenin and Norton Equivalent?

Thevenin and Norton Equivalent are two equivalent circuit models used to simplify complex circuits by replacing them with a single voltage source and a single current source, respectively. They are used to analyze the behavior of circuits and determine their parameters.

2. How are Thevenin and Norton Equivalent calculated?

The Thevenin Equivalent is calculated by finding the open-circuit voltage and the internal resistance of the circuit. The Norton Equivalent is calculated by finding the short-circuit current and the internal resistance of the circuit.

3. What is the difference between Thevenin and Norton Equivalent?

The main difference between Thevenin and Norton Equivalent is the type of source used. Thevenin Equivalent uses a voltage source, while Norton Equivalent uses a current source. They are mathematically equivalent and can be converted into one another.

4. When is it useful to use Thevenin and Norton Equivalent?

Thevenin and Norton Equivalent are useful when analyzing complex circuits, especially when the circuit contains multiple sources and resistors. They simplify the circuit and make it easier to calculate parameters such as voltage, current, and power.

5. What are the limitations of Thevenin and Norton Equivalent?

The limitations of Thevenin and Norton Equivalent are that they are only valid for linear circuits, and they do not take into account non-linear elements such as diodes and transistors. Additionally, they are only accurate for a specific range of frequencies and may not accurately represent the behavior of the circuit at higher frequencies.

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