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schematic

simulate this circuit – Schematic created using CircuitLab

How to solve this circuit for Thevenin and Norton circuit given RL in the circuit. My guess is to remove RL (open circuit). But I'm unsure where the Voc will be measured.

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  • \$\begingroup\$ do you mean you want the norton and thevenin equivalent between the terminals of Rl? \$\endgroup\$ – Juan May 31 at 7:39
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To determine the Thévenin voltage \$V_{th}\$, express the voltage \$V_x\$. It can be obtained by writing: \$V_x=-R_x(2V_x-I_1)\$. In this expression, 2 is a conductance and its unit is the siemens. Solving for \$V_x\$ gives \$V_x=\frac{I_1R_x}{1+2R_X}=333\;mV\$. We can now immediately determine the Thévenin voltage across \$R_L\$ which is considered infinite here: \$V_{RL}=2V_xR_1+V_1=6.33\;V\$. A quick sim shows this is the correct answer: enter image description here

To obtain the output resistance \$R_{th}\$ also called the incremental resistance or Thévenin resistance, we turn all excitation sources off and we "look" through the output terminals (across \$R_L\$ connecting terminals) to determine the resistance we see. Exactly as when determining time constants with the fast analytical circuits techniques (FACTs):

enter image description here

By inspection, we see that the resistance is \$R_1\$ and equal to \$2\;\Omega\$. To verify our answer, we can ask SPICE to calculate the transfer function using the first schematic diagram linking V(4) to source \$V_1\$ and invoking a .TF primitive function:

***** SMALL SIGNAL DC TRANSFER FUNCTION

output_impedance_at_V(4)2.000000e+000
v1#Input_impedance      1.000000e+020
Transfer_function       1.000000e+000

We could also determine the short circuit current \$I_{sc}\$ and a quick sim tells us this current is 3.166 A:

enter image description here

The resistance we want is obtained by dividing \$V_{th}\$ by the short circuit current and we find our \$2\;\Omega\$ again.

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first don't fear from that Dependent source and try to solve it using KVL and KCL as follows :

enter image description here

if you try solving Rth by test source method it will cause trouble (why?)

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    \$\begingroup\$ The image in your answer seems to be too small. Yes, one can enlarge it, but I think it should be easily readable without any zoom. \$\endgroup\$ – Daniel Tork May 31 at 8:32

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