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schematic

simulate this circuit – Schematic created using CircuitLab

I1 in the Kirchoff Voltage Law loop in the first circuit is equal to the I in the second circuit.I1 is also negative I0.

First Loop: $$2(I_1-I_2)+1=0$$

Second Loop: $$4I_2+2(I_2-I_1)-3I_1=0$$ (3I from the current controlled voltage source is equal to 3I_1 ).

You solve the equations and get I_1==-3A. And I_0=3A. R thevenin would be V_0/I_0=1/3 Ohm.

MY questions:

  1. Why do we have to replace the inductor with a voltage source of 1V (Or Current source of 1A)?
  2. Why can't we simply find R thevenin by short ciruciting the voltage source and find the equivalent resistance acroos the terminals of the inductor. In that case, the equivalent would be equal to 4 Ohm|| 2 Ohm =1.333 Ohm.
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  • \$\begingroup\$ It seems to me that you have to determine the resistance "seen" from the inductor terminals. This is will give you the time constant involving the inductance and the resistance you have determined. The normal way to determine a resistance or an impedance is to place a test current source \$I_T\$ biasing the terminals under test, the inductor's terminals here. By using KVL and KCL, you determine the voltage \$V_T\$ across the current source. The resistance you want is \$\frac{V_T}{I_T}\$. In a linear circuit, if the test source is 1 A, then the voltage across its terminals is the resistance. \$\endgroup\$ – Verbal Kint Oct 10 '17 at 7:01
  • \$\begingroup\$ This seems to be a variation on a technique to determine the impedance of the rest of the circuit. However, the technique I have usually seen is to use a voltage source of Vx, then divide that out. Makes it a bit more obvious than to use a fixed value (IMHO, anyway). \$\endgroup\$ – alex.forencich Oct 10 '17 at 17:15
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What you are asked to do, provided I understand your request correctly, is to determine the resistance "seen" by the inductor when you temporarily disconnect it from its connecting terminals. A resistance or an impedance is a transfer function linking a response - the voltage \$V_T\$ across the current source - to a stimulus, the current source \$I_T\$. You will find more details here about these transfer functions. When you are asked to determine a resistance or an impedance from two terminals, install a test current generator \$I_T\$ and determine the voltage \$V_T\$ that appears across its terminals. The resistance is therefore \$R=\frac{V_T}{I_T}\$ or the impedance is \$Z(s)=\frac{V_T(s)}{I_T(s)}\$. The below sketch shows how to do it with your circuit:

enter image description here

Then, by invoking KCL and KVL, determine the relationship linking \$V_T\$ to \$I_T\$. If everything goes well, you should find \$R=\frac{R_1(R_2-3)}{R_1+R_2}\$. Please note that the 3 in this equation has the dimension of ohms. Applying your component values leads to a resistance of 0.333 \$\Omega\$ as confirmed by the dc operating point below. The test generator is 1 A so the voltage across its terminals divided by 1 A is the resistance you want in this linear circuit (Thanks Mr Morton!):

enter image description here

You would install a voltage source instead of a current source in case you would have to determine an admittance defined as \$Y=\frac{I_T}{V_T}\$. In this case, the response is the current \$I_T\$ while the stimulus is the voltage source \$V_T\$.

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    \$\begingroup\$ I agree, this is a bit awkward, thanks for pointing this out. I meant the displayed node voltage divided by 1 A is the resistance you want. Hey, nice-looking animal this aardvark, I did not know it! : ) \$\endgroup\$ – Verbal Kint Oct 10 '17 at 14:11

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