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Circuit in question

I need to find the Thevenin Equivalent left of the >> symbols. Due to the lack of voltage sources I tried finding the Norton Equivalent first and came up with \$Rn=R_O\$, \$v_{th}=i_S \cdot R_O\$. Is that correct? If not how does one handle this case?

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  • \$\begingroup\$ It is trivial to go from Norton to Thevenin and vice versa. \$\endgroup\$ – Oldfart Apr 11 '18 at 15:51
  • \$\begingroup\$ Also, I guarantee you that \$R_N = R_O\$ is incorrect. You haven't accounted for the VCCS, which is configured to act just like another resistor in parallel with \$R_O\$. \$\endgroup\$ – The Photon Apr 11 '18 at 15:55
  • \$\begingroup\$ -oldfart; I know but doing that in this case yields a result which is independent of g, so i suppose i am missing something \$\endgroup\$ – Manouil Apr 11 '18 at 15:56
  • \$\begingroup\$ what you've tried is correct but it's just an intermediate result: a Thevenin source connected to a dependent current source. Now use it to determine short circuit current and open circuit voltage \$\endgroup\$ – Curd Apr 11 '18 at 15:57
  • \$\begingroup\$ Can you find the open-circuit output voltage and short-circuit output current of the circuit? \$\endgroup\$ – The Photon Apr 11 '18 at 15:58
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enter image description here

  • To find Rth, open circuit all independent current sources, short all independent voltage sources.

  • Connect a fictitious 1V source at the open ckt terminal.

  • Find the current driven by 1V source.

  • If I is the current driven by 1V source, then \$\frac{V}{I} = \frac{1}{I}\$ is the load seen by the voltage source. i.e., Rth. $$I = I_{Ro} + gv_x$$ vx = voltage across Ro = 1V $$\implies I = \frac{1}{Ro}+g = \frac{(1+gRo)}{Ro}$$ $$\therefore R_{th} = 1/I = \frac{Ro}{(1+gRo)} $$

  • To find Vth, go back to the original ckt. enter image description here

Vth is the voltage drop across Ro which is same as vx.

  • Write ohms law equation for voltage across Ro $$(i_s-gv_x)Ro= v_x$$ $$\implies v_x = i_s\frac{Ro}{(1+gRo)} = V_{th}$$
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