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I tried to do an exercise from my textbook where I have to apply the Thèvenin theorem, but can't solve it. I got stuck at one point.

Given the circuit in the picture find the I current by using the Thèvenin theorem.

schematic

simulate this circuit – Schematic created using CircuitLab

I found \$R_{Th}\$ and \$v_{Th}\$ first.

(a) To find \$R_{Th}\$, I powered off all independent voltage generators (that is \$E\$) and so I found \$R_{Th}=\frac{R*R_1}{R+R_1}\$

(b) To find \$v_{Th}\$ I applied the voltage divider: \$v_{Th}=E \frac{R_1}{R_1+R}\$

schematic

simulate this circuit

After this I got the Thèvenin equivalent circuit:

schematic

simulate this circuit

And now I don't know how to proceed. Maybe I'm wrong with the above calculations too. The text says to try to solve it with the Norton equivalent circuit too. I'll sorry in advance for the triviality, but I'm doing a basic course at college and I've never did it before. So, please, be comprehensive. Thanks who will reply.

EDIT: So, I managed to do it till the point to calculate \$R_{eq}\$ and \$v_{eq}\$. Finally I got this circuit:

schematic

simulate this circuit

Where \$R_{Th}=\frac{R R_1}{R+R_1}\$, \$v_{Th}=E\frac{R}{R+R_1}\$ and \$R=R_3+\frac{R_4R_5}{R_4+R_5}+R_6\$

Now I have: \$v_{AB}=R_{Th}i_1+v_{TH}=-R_{Th}i_2+v_{TH} => i_2=-\frac{v_{AB}-v_{Th}}{R_{Th}}\$, but the book report in the solution that \$I=i_2=\frac{v_{Th}}{R_{Th}+R_2}\$. Why is that?

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2 Answers 2

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I think you're misreading the problem. If I understand you right, you're breaking the circuit on the left side of R3 and R6, then finding the Thevenin equivalent of E, R1, and R2. But what you're supposed to do is break the circuit above and below R2, then find the Thevenin equivalent of E, R1, and R3-R6. You're "looking into" the same terminals that R2 is connected to.

Also, is R3 really 1000 ohms? That looks like a typo.

Edit: I can't go into much more detail without just giving you the answer. You already know the basic steps:

  1. Disconnect R2 from the circuit.
  2. Determine the voltage between nodes A and B. This is Vth
  3. Deactivate E, then determine the resistance between nodes A and B. This is Rth.
  4. Now you have a Thevenin equivalent circuit. Attach R2 to it, then compute the current.

Most of the actual work is combining series and parallel resistors, which you should already know how to do.

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  • \$\begingroup\$ Not a typo. Can you explain me step-by-step,please? \$\endgroup\$
    – sl34x
    Nov 13, 2014 at 22:41
  • \$\begingroup\$ I added some more detail to my answer. \$\endgroup\$
    – Adam Haun
    Nov 14, 2014 at 1:12
  • \$\begingroup\$ Could you please explain me that last thing, above after the "EDIT:" ? \$\endgroup\$
    – sl34x
    Nov 14, 2014 at 14:18
  • \$\begingroup\$ Your formulas for Rth and Vth are correct. You then got: $$i2 = \frac{Vth - VAB}{Rth}$$ which is true but not useful. You don't need to know VAB to find the current. You have a voltage source in series with two resistors, so: $$I = \frac{V}{\Sigma R}$$ That's what your book did. The equivalent circuit is a voltage divider, so if you want to find VAB you can use the standard formula. \$\endgroup\$
    – Adam Haun
    Nov 14, 2014 at 15:25
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Depending on the exact situation, you could be overthinking this.

If this were a real world problem, you would immediately notice that R3 is so large compared to all other resistances that the current thru R3 is irrelevant. You can therefore discard everything in series with R3, which leaves just the voltage source, R1, and R2. Now solving for the current becomes trivial. You don't even need to invoke Thevening to do it. You simply divide the voltage divided by the sum of the two resistors.

If this is a engineering course and was presented in a word problem intended to be more real-life, then this may be exactly the solution the professor is looking for. Note that assuming 1% accuracy of the resistors yields more slop than ignoring the current thru R3. The point may be more to teach engineering than the details of the electronics in this case. That is actually a important lesson that engineering students need to learn.

Even if the problem is more theoretical, you can use the above to get a good and quick approximation of the answer. If your ultimate theoretical answer isn't close, then you did something wrong.

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  • \$\begingroup\$ I didn't notice it. Thanks for the hint! \$\endgroup\$
    – sl34x
    Nov 14, 2014 at 19:01

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