Thevenin Equivalent Circuits with Dependent Source

[simba9071]

May anyone please help with the method of getting Thevenin Equivalent resistance (R_TH) for the attached circuit. I have already found V_TH to be 8V.

View attachment ECA.doc

Sim

 

[Metaleer]

I haven't looked at the circuit, but since you say it has dependent sources, the usual way to go about this is to connect a voltage source of generic EMF $$e_g$$ between the two points, turn off all the independent sources, and then solve this circuit. Now, you use $$R_\text{Th} = \frac{e_g}{i_g}$$ where $$i_g$$ is the current going through the voltage source, where active references are used for voltage and current. If you've done everything right, $$e_g$$ will be a linear function of $$i_g$$ and you can simplify when you do the division.

This is for the Thévenin resistance - for the Thévenin EMF, all you need is the voltage drop between the two points when you open the circuit, as always.

Hope this helps. :)

 

[Neandethal00]

I haven't looked at the circuit, but since you say it has dependent sources, the usual way to go about this is to connect a voltage source of generic EMF $$e_g$$ between the two points, turn off all the independent sources, and then solve this circuit. Now, you use $$R_\text{Th} = \frac{e_g}{i_g}$$ where $$i_g$$ is the current going through the voltage source, where active references are used for voltage and current. If you've done everything right, $$e_g$$ will be a linear function of $$i_g$$ and you can simplify when you do the division.

This is for the Thévenin resistance - for the Thévenin EMF, all you need is the voltage drop between the two points when you open the circuit, as always.

Hope this helps. :)

I'm not sure if we have to treat a dependent source differently from a independent source when measuring resistance. Because you can also find a non-zero resistance in independent sources using your formula
$$R = \frac{V}{I}$$.
Dependent source definitely contributes to $$V_\text{Th}$$.
For the circuit I see,
R_th = 2//8 = 1.6 ohms.

 

[Metaleer]

I'm not sure if we have to treat a dependent source differently from a independent source when measuring resistance. Because you can also find a non-zero resistance in independent sources using your formula
$$R = \frac{V}{I}$$.
Dependent source definitely contributes to $$V_\text{Th}$$.
For the circuit I see,
R_th = 2//8 = 1.6 ohms.

I'm not sure what you mean. All independent sources need to be switched off to measure the equivalent Thévenin resistance between two terminals, and dependent sources need to be left there.