# Thevenin equivalent of circuit with no voltage sources 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?

• It is trivial to go from Norton to Thevenin and vice versa. – Oldfart Apr 11 '18 at 15:51
• 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 $R_O$. – The Photon Apr 11 '18 at 15:55
• -oldfart; I know but doing that in this case yields a result which is independent of g, so i suppose i am missing something – Manouil Apr 11 '18 at 15:56
• 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 – Curd Apr 11 '18 at 15:57
• Can you find the open-circuit output voltage and short-circuit output current of the circuit? – The Photon Apr 11 '18 at 15:58

## 1 Answer • 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. Vth is the voltage drop across Ro which is same as vx.

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