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Can you help me find the Thévenin equivalent between terminals A and B with all the steps?

This is the circuit

$$R_{eq}=\frac{4}{3}~\Omega\\ V_{eq}=\frac{5}{3}~\mathrm{V}$$

In the first image is the circuit, and below there are the solutions. I need the calculations for the solution of Req.

Here is my attempt:

Req = ((2//3)+3)//1 = 21/26

Veq = $$\left[\begin{matrix} 2+1+3+3&-3 \\\\ -3&3 \\\\ \end{matrix}\right] \left[\begin{matrix} I_1\vphantom{\frac{-1}{R_3}} \\\\ I_2\vphantom{\frac{-1}{R_3}} \\\\ \end{matrix}\right] = \left[\begin{matrix} 2\:\text{V}\vphantom{\frac{-1}{R_3}} \\\\ 3\:\text{V}\vphantom{\frac{-1}{R_3}} \\\\ \end{matrix}\right]$$

This solves to:

$$ \left[\begin{matrix} I_1=\frac56\:\text{A} \\\\ I_2=\frac{11}{6}\:\text{A} \\\\ \end{matrix}\right]$$

So

$$V_{eq}=\frac{5}{6}\times2=\frac{5}{3}V$$

Req = enter image description here

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    \$\begingroup\$ Homework questions need to show an attempt at an answer in order to demonstrate where you are getting stuck and prevent the possibility you are using this site during an exam. \$\endgroup\$
    – Andy aka
    Commented Jan 27, 2023 at 13:38
  • \$\begingroup\$ Ok , sorry about that. I've just added my attempt. \$\endgroup\$ Commented Jan 27, 2023 at 14:14
  • \$\begingroup\$ That's just a few out of focus and largely illegible equations that doesn't indicate what you are trying to specifically find. There are two things you are looking for <-- do you understand why I say that? Do you know how to use latex as in formula stuff like this: \$x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}\$? = $$$$ \$x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}\$ <-- it would help a lot if you used it. \$\endgroup\$
    – Andy aka
    Commented Jan 27, 2023 at 14:16
  • \$\begingroup\$ Never used sorry. I'll try to edit the post using Latex. \$\endgroup\$ Commented Jan 27, 2023 at 14:24
  • \$\begingroup\$ That might be a big haul. Do you understand the two things you are trying to find? Do you understand that redrawing the circuit can make it look simpler: i.sstatic.net/7K5Iu.png I can work Req out in my head. \$\endgroup\$
    – Andy aka
    Commented Jan 27, 2023 at 14:27

1 Answer 1

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These are just hints to get the OP down the right path

The problem is to find the Req

Start like this: -

enter image description here

The resistance (\$R_{AB}\$) is easy now.

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  • \$\begingroup\$ What are the resistence in parallel and in series? Because I don't know how to solve it... \$\endgroup\$ Commented Jan 27, 2023 at 17:05
  • \$\begingroup\$ I'm trying to calculate it this way: 2 in series with 3 in parallel with 3 all in parallel with 1 but the result is different from the solution. \$\endgroup\$ Commented Jan 27, 2023 at 17:22
  • \$\begingroup\$ What does the short in parallel with the left 3 ohm resistor make that part of the circuit? Redraw it to understand it. Then redraw with the two series resistors combined into one resistor. Then, what are you left with in parallel with the 2 ohm resistor? \$\endgroup\$
    – Andy aka
    Commented Jan 27, 2023 at 17:26
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    \$\begingroup\$ OK, I did it. The resistor on the short circuit is 0. Then the 3 ohms resistor is in series with the 1 Ohm resistor, and then what is left is in parallel with the 2 Ohms resistor at the top. So Req= (3+1)//2=4//2= 4/3 Ohms Thanks for the help! \$\endgroup\$ Commented Jan 28, 2023 at 10:47
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    \$\begingroup\$ Ok, I accepted the answer. \$\endgroup\$ Commented Jan 28, 2023 at 12:16

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