# Need help finding my mistake using Thenevin's Theorem for this circuit (not homework!)

So I have the following circuit (an example from my textbook). The answer for V_o is 3.88V. I got the wrong answer and I'm not sure where I went wrong.

I know I could solve for V_o using other techniques (perhaps mesh or nodal analysis), however, I specifically need to work this problem to practice Thenevin's theorem ("any linear electrical network with voltage and current sources and only resistances can be replaced at terminals A-B by an equivalent voltage source Vth in series connection with an equivalent resistance Rth.")

Will someone please follow my procedure and tell me what I did wrong? Thank you in advance. My steps:

1) Remove the 1k ohm resistor, resulting in the following: 2) Now I have to find E_Th which is the voltage across the two terminals
that I just opened.
3) I will choose to use mesh analysis for the three remaining loops
since I have a couple independent current sources. 4) The mesh equation (KVL) for I_2 is: 3k*(I2 - 8/1k) + 6K(I2-2/1k) + 6K*I2 + 12 = 0

   Simplified:

3k*I2 - 24 + 6k*I2 - 12 + 6K*I2 + 12 = 0

15k*I2 = 24

I2 = 24/15K = 8/3K A = 2.667 mA

5) Now that I know I_2 is 2.667 mA, I can calculate the voltage across
6K ohm resistor, which is 2.667mA * 6k = 16V

6) Calculating the voltage across the 2K ohm resistor:
2k*2mA = 4V

7) Now this is where I think I'm making a mistake.  I'm just
summing the voltages: 4V + 16V + 12V = 32V.  I don't think that's right.
If it is right, then E_Th = 32V.

8)  Removing all the independent sources by replacing the current sources
with opens and the voltage source with a short, I get the following: 9) From this, I need to find the equivalent resistance:
[(6k in series with 3k) || 6k ] + 2k

6k + 3k = 9k
9k || 6k = 6*9/6+9 = 54/15
54/15 k + 2k = 5.6k ohms = R_Th

10) Now, building the Thenevin circuit with E_Th, R_Th and adding the removed
1k ohm resistor from step 1: 11)  Now, voltage division:

V_1k = 1k/(1k + 5.6k) * 32
= 4.84 V = V_o


So, that's obviously wrong... where did I mess up, and how do I fix this?

• Are you sure the $1k\Omega$ resistor is actually supposed to be the load? Jun 27, 2014 at 15:05

Calculation mistake at Step number 4. You wrote:

I2 = 24/15K = 8/3K A = 2.667 mA


But 24/15k = 8/5k not 8/3k. So I2 is

I2 = 24/15k = 8/5k A = 1.6mA


You will get Vth = 25.6 V.

• I2 = 24/15K = --> 8/3K <-- Jun 27, 2014 at 15:25
• @sherrellbc please see the edited answer. 24/15 $\ne 8/3$ Jun 27, 2014 at 15:27
• Oh yeah, sorry I did not intend to make that appear as if it were directed at you. I was just making it explicit for anyone reading I guess - albeit unnecessary. Jun 27, 2014 at 15:35
• @sherrellbc sorry about that misunderstanding :) Jun 27, 2014 at 15:53