Thévenin and Norton circuits Exercise

AI Thread Summary
The discussion centers on solving a Thévenin and Norton circuit exercise, with the user presenting their calculations for Thévenin voltage (V_TH) and resistance (R_TH). The user arrives at V_TH = 2 V and R_TH = 2.4 Ω, but expresses confusion over the calculations and seeks clarification. Another participant suggests focusing on the supermesh involving only the left and middle loops, and advises against equating voltage with resistance. They also recommend simplifying the circuit by replacing two parallel resistors with a single equivalent resistor for easier analysis.
Guillem_dlc
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Homework Statement
In the circuit of the figure, determine:

1. Equivalent Thévenin and Norton circuits, viewed from terminals A and B.

2. Value of the load resistance ##R_C## connected between A and B for the circuit to work at maximum power transfer conditions.

3. The power in the load resistor deduced in the previous section.

4. The power in the load resistor in the event that its value is twice that calculated in section 2.
Relevant Equations
Thévenin and Norton
Figure:
Captura de pantalla_2022-11-15_20-35-41.png


My attempt at a solution:
Captura de pantalla_2022-11-15_20-36-27.png

a) ##\boxed{V_{TH}}##
  • Current source equations ##\rightarrow \boxed{2=I_2-I_1}##
  • Control variable equations ##\rightarrow V_1=4I_2\rightarrow \boxed{0=4I_2-V_1}##
  • Super Mesh ##\rightarrow -0,5V_1+2I_1+I_2(4+8)-8I_3=0\rghtarrow \boxed{0=2I_1+12I_2-8I_3-0,5V_1}##
  • Mesh 3 ##\rightarrow 16I_3-8I_2=0\rightarrow \boxed{-8I_2+16I_3=0}##
$$\boxed{I_1=-1,5\, \textrm{A}\qquad I_2=0,5\, \textrm{A}\qquad I_3=0,25\, \textrm{A}\qquad V_1=2\, \textrm{V}}$$
Captura de pantalla_2022-11-15_20-40-38.png

We know that ##\rightarrow V_B=V_{TH}##
$$V_A=0+1,5\cdot 2=3\, \textrm{V}$$
$$V_B=V_A+1-0,5\cdot 4=2\, \textrm{V}$$
$$\boxed{V_{TH}=2\, \textrm{V}}$$
##\boxed{R_{TH}}##
Captura de pantalla_2022-11-15_20-42-44.png

$$\dfrac{1}{R_{TH}}=\dfrac16+\dfrac18+\dfrac18 \rightarrow \boxed{R_{TH}=2,4\, \Omega}$$
Captura de pantalla_2022-11-15_20-43-47.png

$$V_1\rightarrow -1+I_1+I_2+I_3=0\rightarrow$$
$$\rightarrow 1=\dfrac{V_1}{8}+\dfrac{V_1}{8}+\dfrac{V_1}{6}\rightarrow$$
$$\rightarrow 12=3V_1+2V_1=5V_1\rightarrow$$
$$\boxed{V_1=\dfrac{12}{5}=2,4\, \Omega}$$
$$V_1=R_{TH}\rightarrow \boxed{R_{TH}=2,4\, \Omega}$$
##\boxed{I_N}##
Captura de pantalla_2022-11-15_20-45-53.png

  • Current equations: ##\boxed{2=I_N-I_2}##
  • ##S\rightarrow \boxed{4I_N+2I_2-0,5V_1=0}##
  • ##V_1=4I_1\rightarrow \boxed{0=4I_N-V_1}##
$$\boxed{I_N=1\, \textrm{A}}$$
$$R_{TH}=\dfrac{V_{TH}}{I_N}=2\, \Omega \rightarrow \boxed{R_{TH}=2\, \Omega}$$
I was doing this exercise and the time to calculate the intensity and the Norton resistance doesn't add up... I've done it in different ways and I don't know where I fail. I don't know which one is right.
 
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Well, since you've no replies yet and we don't want you to feel neglected...

I’m no expert and I haven’t checked much of your work. But this might help…

The supermesh (with my limited knowledge) will consist of the left and middle loops only - because they share the current source. The right-hand loop is not part of the supermesh.

So. my supermesh equation would be (starting on the negative side of the voltage source and going clockwise):

##0.5V_1 - 4I_2 -8(I_2 – I_3) - 2I_1= 0##

Also, avoid writing down things like ##V_1 = 2.4Ω##. A voltage can't equal a resistance. It suggests you are confused.

Note you could simplify the problem. Replace the two parallel 8Ω resistors by a single 4Ω; this gives you an easier circuit to analyse.

If you are struggling with supermesh analysis, there are some good YouTube videos.
 
I have found the error. It's working fine now.
 
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