#### dyampir

Joined Apr 1, 2008
7
I am having issues with Norton and Thevenin. I think I have the basic concepts, but I am still really unsure about how these are done.

I have read the other posts and because my circuit is different, I am not sure how to apply the info in those posts to my circuits.

Please review my work and see if I am on the right track.

****The attached Thumbnail is the circuit****

1.) Thevenin Voltage if R3 is the load
- I removed R3 and calculated voltage across it

V(th) = 10v/(R1//R2) + R4
V(th) = 10/ (1/(1/2+1/4))+2
V(th) = 9.5V

2.) Norton Current if R2 is the load

V(th) = 10V/8ohms * 6ohms = 7.5 V

R(th) = R1//(R4+R3)
R(th) = 1/2 + 1/6
R(th) = 1.5 ohms

I(norton) = V(th) / R(th)
I(n) = 7.5V/1.5ohms
I(n) = 5A

3.) Thevenin resistance if R4 is the load

V(resistor 2) = R2/(R1+R2) * Voltage
V(r2) = 4/6 * 10
V(r2) = 6.67 = V(th)

R(th) = R3 + (R1//R2) + R4
R(th) = 7.3 ohms

**P.S.**
Am I suppose to do anything with the Current source? I just left it out because one step of simplifying the circuit is to short voltage sources and open current sources

#### dyampir

Joined Apr 1, 2008
7
Okay.....am I to assume that my calculations are correct?

#### veritas

Joined Feb 7, 2008
167

#### JoeJester

Joined Apr 26, 2005
4,126

Attached are the diagrams I used for Question 1 so you can rethink your answer.

#### Attachments

• 25.1 KB Views: 32

#### dyampir

Joined Apr 1, 2008
7
Okay...with superposition I got this:

I did two matrixes (one with the 10V and one with the 2A)

loop 1 current(with 10V) = 2.86
loop 2 current (with 10V) = .143

loop 1 current(with 2A) = .714
loop 2 current (with 2A) = .286

Total current loop 1 = 3
Total current loop 2 = 1

so my total curent is 2A....but that's the value of my current source...right?

I know that E=IR.....so, if R3 is 4 ohms, then my V(th) would be 8V across R3....if R3 were the load.

I am honestly guessing here. I'm just not sure how to relate this information back to voltage...and the website that was listed for superposition help lists a table of E I R values. This really confused me because I couldn't figure out where they got their values.

In my notes I have superposition entered into the calculator as a matrix....but I thought it could only be used with DC sources.

I will be taking a Capstone test in 2 weeks and if I don't pass, I don't graduate.

I MUST understand these concepts. #### veritas

Joined Feb 7, 2008
167
You really don't need KVL/KCL analysis to solve this. You also lost me with your loop analysis because I don't know which problem you were trying to solve, or what loops are I1 and I2

JoeJester made some really clean and easy to understand diagrams of the steps for solving problem one for Vth. It is done using ohm's law to find the voltage across the open connection for each of the sources, and then adding the values from your two calculations.

I would suggest starting there, and showing us your calculations. We might then be able to point out any errors you make and how to correct them.

#### JoeJester

Joined Apr 26, 2005
4,126
JoeJester made some really clean and easy to understand diagrams of the steps for solving problem one for Vth.
Apparently not clear enough. I also have the diagrams ready for question 2 and 3. I have another pdf with the derived formulae for each question.

#### Attachments

• 40.5 KB Views: 18

#### dyampir

Joined Apr 1, 2008
7
1.) Thevenin voltage if R3 is the load Okay...different approach....Kirchhoff's Law

[-10 + 2(I) + 2(I+2)] = 0
-10 + 2I + 2I + 4 = 0
4I = 6
I = 6/4
I = 1.5A

V(th) = 1.5A * 2ohms
V(th) = 3V

2.) Norton Current if R2 is the load
I = 5A

At point of R2, current is I + 2A
I(N) = 7A

3.) Norton or Thevenin resistance if R4 is the load 1/R = (1/2 + 1/2) = 1

R = 1+2 = 3ohms

#### JoeJester

Joined Apr 26, 2005
4,126
Question 1 - INCORRECT.

Question 2 - CORRECT.

Question 3 - CORRECT.

#### dyampir

Joined Apr 1, 2008
7
Hmm...okay Let me recalculate 1

#### dyampir

Joined Apr 1, 2008
7
Question 1:

Thevenin voltage if R3 is the load:

[-10 + 2(i) + (i+2)2 = 0]
-10 + 2i + 2i + 4 = 0
4i = 10-4
i = 6/4
i = 1.5

I = 2A + 1.5A = 3.5A

3.5A * 2 ohms = 7 V(th)

#### JoeJester

Joined Apr 26, 2005
4,126
Correct.

Attached are two pdf's. One with the formulae's I used [superposition based] and the other, the simulations.

#### Attachments

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• 41.4 KB Views: 20

#### dyampir

Joined Apr 1, 2008
7
:O

I LOVE the way you broke that down ...

Thank you so much for all of the help #### JoeJester

Joined Apr 26, 2005
4,126
I LOVE the way you broke that down ...
I did it by the diagrams .... then reassembled it to one formula.