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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.

enter image description here

My steps:

1) Remove the 1k ohm resistor, resulting in the following:

enter image description here

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.

enter image description here 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:

enter image description here

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:

enter image description here

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?

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  • \$\begingroup\$ Are you sure the \$1k\Omega\$ resistor is actually supposed to be the load? \$\endgroup\$ – Matt Young Jun 27 '14 at 15:05
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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.

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  • \$\begingroup\$ I2 = 24/15K = --> 8/3K <-- \$\endgroup\$ – sherrellbc Jun 27 '14 at 15:25
  • \$\begingroup\$ @sherrellbc please see the edited answer. 24/15 \$\ne 8/3\$ \$\endgroup\$ – nidhin Jun 27 '14 at 15:27
  • \$\begingroup\$ 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. \$\endgroup\$ – sherrellbc Jun 27 '14 at 15:35
  • \$\begingroup\$ @sherrellbc sorry about that misunderstanding :) \$\endgroup\$ – nidhin Jun 27 '14 at 15:53

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