Thévenin and Norton circuits Exercise

[Guillem_dlc]

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:

My attempt at a solution:

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 $\rightarrow 16I_3-8I_2=0\rightarrow \boxed{-8I_2+16I_3=0}$

$$\boxed{I_1=-1,5\, \textrm{A}\qquad I_2=0,5\, \textrm{A}\qquad I_3=0,25\, \textrm{A}\qquad V_1=2\, \textrm{V}}$$

We know that $\rightarrow V_B=V_{TH}$
$$V_A=0+1,5\cdot 2=3\, \textrm{V}$$
$$V_B=V_A+1-0,5\cdot 4=2\, \textrm{V}$$
$$\boxed{V_{TH}=2\, \textrm{V}}$$
$\boxed{R_{TH}}$

$$\dfrac{1}{R_{TH}}=\dfrac16+\dfrac18+\dfrac18 \rightarrow \boxed{R_{TH}=2,4\, \Omega}$$

$$V_1\rightarrow -1+I_1+I_2+I_3=0\rightarrow$$
$$\rightarrow 1=\dfrac{V_1}{8}+\dfrac{V_1}{8}+\dfrac{V_1}{6}\rightarrow$$
$$\rightarrow 12=3V_1+2V_1=5V_1\rightarrow$$
$$\boxed{V_1=\dfrac{12}{5}=2,4\, \Omega}$$
$$V_1=R_{TH}\rightarrow \boxed{R_{TH}=2,4\, \Omega}$$
$\boxed{I_N}$

• 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.

 

[Steve4Physics]

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.

 

[Guillem_dlc]

I have found the error. It's working fine now.