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 [tex]e_g[/tex] between the two points, turn off all the independent sources, and then solve this circuit. Now, you use [tex]R_\text{Th} = \frac{e_g}{i_g}[/tex] where [tex]i_g[/tex] is the current going through the voltage source, where active references are used for voltage and current. If you've done everything right, [tex]e_g[/tex] will be a linear function of [tex]i_g[/tex] 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 [tex]e_g[/tex] between the two points, turn off all the independent sources, and then solve this circuit. Now, you use [tex]R_\text{Th} = \frac{e_g}{i_g}[/tex] where [tex]i_g[/tex] is the current going through the voltage source, where active references are used for voltage and current. If you've done everything right, [tex]e_g[/tex] will be a linear function of [tex]i_g[/tex] 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
[tex]R = \frac{V}{I}[/tex].
Dependent source definitely contributes to [tex]V_\text{Th} [/tex].
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
[tex]R = \frac{V}{I}[/tex].
Dependent source definitely contributes to [tex]V_\text{Th} [/tex].
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.

 

[pmacias]

ulating circuits using Thevenin equivalent circuits can be a useful tool for analyzing complex circuits. In order to find the Thevenin equivalent resistance (RTH) for a circuit with dependent sources, the following steps can be followed:

1. Remove the load resistor from the original circuit and label the open circuit voltage as VTH. In this case, VTH is already given as 8V.

2. Short all independent voltage sources and open all independent current sources in the circuit.

3. Find the equivalent resistance looking into the open-circuited load terminals. This can be done by using the voltage divider rule or by applying a test voltage and calculating the resulting current.

4. Once the equivalent resistance (RTH) is found, it can be used to create the Thevenin equivalent circuit, which will consist of a voltage source with VTH as its value and a series resistor with RTH as its value.

In summary, the method for finding the Thevenin equivalent resistance for a circuit with dependent sources involves simulating the circuit with open-circuited load terminals and finding the equivalent resistance. This value can then be used to create the Thevenin equivalent circuit, which can be used for further analysis and simplification of the original circuit.