# Thévenin and Norton circuits Exercise

• Guillem_dlc
In summary, the author is trying to solve an equation that has three unknowns. The first two are the current source and the control variable. The third is the mesh resistance. The first two are solved and the third is given as 2Ω.
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.

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.

## 1. What is the purpose of Thévenin and Norton circuits?

Thévenin and Norton circuits are used to simplify complex circuits into simpler equivalent circuits, making it easier to analyze and understand their behavior.

## 2. How do you calculate the Thévenin equivalent resistance?

The Thévenin equivalent resistance is calculated by removing all sources from the original circuit and calculating the resistance between the two terminals of the circuit.

## 3. How do you calculate the Thévenin voltage and Norton current?

The Thévenin voltage is calculated by measuring the voltage across the two terminals of the circuit with all sources removed. The Norton current is calculated by dividing the Thévenin voltage by the Thévenin equivalent resistance.

## 4. What is the main difference between Thévenin and Norton circuits?

The main difference is that Thévenin circuits use a voltage source in series with a resistance, while Norton circuits use a current source in parallel with a resistance.

## 5. Can Thévenin and Norton circuits be used for any circuit?

Yes, Thévenin and Norton circuits can be used for any linear circuit with resistors, independent sources, and dependent sources.

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