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I am looking for the Thevenin Resistance here. However, I have no clue what to do with \$i_b\$ and the source dependent on it. This is the entire circuit. \$i_b\$ is the current through the 5\$k\Omega\$ resistor. In this case, There is an open circuit across terminals a and b and so there is no current flowing through the 10\$k\Omega\$ resistor. Meaning that all of the current flowing in the circuit is \$i_b\$. That presents this problem, because there's a dependent source that is dependent on the current \$i_b\$ and they are in series. Hence, the confusion and the question.

Equations: KVL along outer loop CCW: enter image description here

Sidenote: \$I_{sc}\$, short circuit current across terminals a & b, is easily obtainable by a KVL on the outer loop, the relationship of the current through the 10\$k\Omega\$ resistor and \$i_b\$, (Which is \$4i_b = i_b + 3i_b\$ :: KCL in the middle top node), and it is -1.8mA.

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    \$\begingroup\$ It is not dependant on itself, but on the current in another place of the circuit. Did you try to simply write the equations? \$\endgroup\$ – dim Sep 27 '18 at 5:21
  • \$\begingroup\$ Homework needs an attempt at a solution or it is off topic, please edit your question and try to put in some equations. Yes this is solvable \$\endgroup\$ – Voltage Spike Sep 27 '18 at 5:22
  • \$\begingroup\$ @dim That i_b on the 5kOhm resistor is the current that the source is dependent on. I am looking for the thevenin resistance, in this case, since there are both Dependent and independent sources in the circuit, I was taught to find the short circuit current Isc and the open circuit voltage Voc. Isc is easily obtainable through KVL in the outer loop, which equates to -1.8A (assuming the direction is downward). This part stumps me since I'm not sure what to make of it. No current will pass through the 10kOhm resistor, and so I am left with this. \$\endgroup\$ – zodkc Sep 27 '18 at 5:41
  • \$\begingroup\$ With only one current loop you have two inconsistent defined currents ib and 4ib which cannot be equal. so the output must have a load to work. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 27 '18 at 6:01
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    \$\begingroup\$ Read the lecture notes. \$\endgroup\$ – Chu Sep 27 '18 at 6:59
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There are several ways of determining the Thevenin equivalent resistance. One way is to open the output and measure measure the open circuit voltage and resistance. However, that doesn't work here because you end up with a nonsense term 4 ib = ib. So you have to use a different method. I think they penalty want you to use the "test source" technique here, where you attach a test voltage source to the output.

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  • \$\begingroup\$ I was thinking of doing that. Hence I checked the notes and it said that that particular method is used for when there are only dependent sources in the circuit. I think I'll do that. Not sure if it's an unbreakable rule. Thanks. \$\endgroup\$ – zodkc Sep 27 '18 at 12:50

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