How can I determine Rth in Thevenin with a transconductance source?

[qwertyuiop]

This is the question

First,I use thevenin theorem,when I close the voltage source, i apply nodal analysis to find rth,is my answer correct?

 

[Futurestar33]

Just a question but aren't those all resistors if the units are kila-ohms?

 

[qwertyuiop]

Just a question but aren't those all resistors if the units are kila-ohms?

I don't understand what you mean

 

[NascentOxygen]

Dependent source 0.003v_o should, with your nomenclature, be written 0.003V₁

 

[NascentOxygen]

Note: we are not able to see what's at your facebook link, if it's applicable

 

[gneill]

Note: we are not able to see what's at your facebook link, if it's applicable

Moderator's note: I've restructured the original post to make the relevant images visible, and removed the facebook link as it was not accessible. @qwertyuiop is advised to upload relevant content rather than link to a facebook page.

 

[The Electrician]

Halfway down the page of your calculations you have:

(V1-V2)/22k + 0.003 V2 +1 + V2/30k = 0

This should be:

(V1-V2)/22k + 0.003 V1 +1 - V2/30k = 0

Which should then become:

30 V1 - 30 V2 + 1980 V1 + 660000 - 22 V2 = 0

Which becomes:

2010 V1 - 52 V2 = -660000

 

[qwertyuiop]

Dependent source 0.003v_o should, with your nomenclature, be written 0.003V₁

Thank you for you helping,now i know my mistake

 

[qwertyuiop]

Moderator's note: I've restructured the original post to make the relevant images visible, and removed the facebook link as it was not accessible. @qwertyuiop is advised to upload relevant content rather than link to a facebook page.

Thank you

 

[qwertyuiop]

Halfway down the page of your calculations you have:

(V1-V2)/22k + 0.003 V2 +1 + V2/30k = 0

This should be:

(V1-V2)/22k + 0.003 V1 +1 - V2/30k = 0

Which should then become:

30 V1 - 30 V2 + 1980 V1 + 660000 - 22 V2 = 0

Which becomes:

2010 V1 - 52 V2 = -660000

Thank you for your help.
I don't understand why - V2/30k ?

 

[The Electrician]

Thank you for your help.
I don't understand why - V2/30k ?

Because you have chosen to define currents into a node as positive. The other choice people make is to define current out of a node as positive. But whichever you choose, you must be consistent.

The expression V2/30k gives the current out of the node, so its sign must be negative.

 

[qwertyuiop]

Because you have chosen to define currents into a node as positive. The other choice people make is to define current out of a node as positive. But whichever you choose, you must be consistent.

The expression V2/30k gives the current out of the node, so its sign must be negative.

ok, thank you for your help. In this circuit, I find the rth = -1363ohm. Is possible rth is negative? why?

 

[The Electrician]

ok, thank you for your help. In this circuit, I find the rth = -1363ohm. Is possible rth is negative? why?

The dependent source in this circuit is a transconductance source. The value 0.003 is the transconductance. Substitute a variable x for the value 0.003 and solve the circuit for Rth. You will get this expression: Rth = 3.75/(.00025-x). You can see that if x is greater than .00025, Rth will be negative; since .003 is greater than .00025, we get a negative value for Rth.

The circuit of this thread is an example of a circuit with an active element connected in such a way as to generate a negative resistance. There are other well known circuits that can do this, for example: https://en.wikipedia.org/wiki/Negative_impedance_converter

 

[qwertyuiop]

The dependent source in this circuit is a transconductance source. The value 0.003 is the transconductance. Substitute a variable x for the value 0.003 and solve the circuit for Rth. You will get this expression: Rth = 3.75/(.00025-x). You can see that if x is greater than .00025, Rth will be negative; since .003 is greater than .00025, we get a negative value for Rth.

The circuit of this thread is an example of a circuit with an active element connected in such a way as to generate a negative resistance. There are other well known circuits that can do this, for example: https://en.wikipedia.org/wiki/Negative_impedance_converter

Thank you