# Thevenin and Norton Equivalent

1. Feb 15, 2014

### Tekneek

1. The problem statement, all variables and given/known data

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.

3. 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 :)

#### Attached Files:

• ###### Thevenin.jpg
File size:
8.2 KB
Views:
85
2. Feb 15, 2014

### Staff: Mentor

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. Feb 15, 2014

### Tekneek

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

4. Feb 15, 2014

### Staff: Mentor

Current divider, not voltage divider.

5. Feb 15, 2014

### Tekneek

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

Thanks!

6. Feb 15, 2014

### Staff: Mentor

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:

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

File size:
3.2 KB
Views:
143