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I want to find the Thévenin resistance from the terminals a-b. I've been trying out various methods that look correct to me but none of them seem to be the right answer: 4/9 kΩ.

First, according to KCL at node A I've considered that the current going through the right loop is 5I1 since the current going through the left loop is just I1 and the dependent source provides 6I1.

With this in mind I tried:

  1. Transforming the dependent current source into a dependent voltage source, which gave me two resistors in series R3 + R1, which I added and then I transformed it back to a current dependent source. Now the Thévenin-Norton resistance should be the parallel resistors R3+R1 // R2 = 2/3 which is wrong (not 4/9);
  2. Then I tried calculating the Thévenin voltage at a-b and the Norton current by short-circuiting a-b and then apply Rth = Vth/In. In = 5I1 and Vth=5I1x1 and of course is not right either.

How can I calculate Rth in this setup? Also, I don't understand why the methods I've applied are wrong.

enter image description here

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  • \$\begingroup\$ I think that one problem is there is not really a Thevenin voltage. Without an external source driving this circuit, nothing happens inside. It's like a transistor amplifier without a power supply. \$\endgroup\$
    – Bart
    Aug 24 at 8:55
  • \$\begingroup\$ Just apply a voltage U to A and B, calculate I1, and divide U by (-5R1) to get the current. I'm getting 800 ohms for the resistance without considering R2. Paralleling R2 gives 4/9k. \$\endgroup\$
    – Bart
    Aug 24 at 9:03
  • \$\begingroup\$ It is not Thevenin you apply here. Simply equivalence of two circuits (OP and a new one which is resistance only). \$\endgroup\$
    – Antonio51
    Aug 24 at 9:35
  • \$\begingroup\$ Ludicrous, the circuit is degenerate. If you have I1 passing through R3 then the voltage at node A must be I1*R3. But by KCL at node A, there must be 5*I1 flowing through R1+R2. But R1+R2=2*R3. So the voltage at node A must be 2*R3*5*I1 = 10*R3*I1. But we just said node A must be I1*R3 from the left side. How can node A be both I1*R3 and 10*I1*R3, unless I1 itself is 0. Therefore, I1=0 A and VA=0 V. Right? \$\endgroup\$
    – jonk
    Aug 29 at 19:23
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    \$\begingroup\$ Ludicrous - are we done here? Is the question answered to your satisfaction? If not then raise a comment under the most appropriate answer for getting feedback. Leaving the question open is not really a good thing; people get less interested in helping you when there is no clear reason for your inaction. \$\endgroup\$
    – Andy aka
    Sep 1 at 20:14

2 Answers 2

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I want to find the Thévenin resistance from the terminals a-b.

enter image description here

This is because if you apply 1 volt then \$I1\$ is 1 mA and, due to the current source, 6 mA is returned to the 1 volt giving a net load current of -5 mA hence, its equivalent resistance is -200 Ω.

Then if you simply did the analysis of R1 (1 kΩ) in series with -200 Ω all in parallel with R2 (1 kΩ) you get 444.444 Ω.

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Applying a voltage Uab to a and b, we can find the following relation:

R3⋅I1-5R1⋅I1 = Uab -> Rth = Uab/(-5I1) = 800 Ω.

This is not considering R2. Paralleling R2 with 0.8kΩ yields the answer 4/9kΩ.

Without assuming an external voltage all currents and voltages inside this circuit will be zero. That's why there's not really a Thevenin voltage or Norton current.

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