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.
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?