Thevenin equivalent phasor domain dependent voltage source

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Discussion Overview

The discussion revolves around determining the Thevenin equivalent of a circuit at specified terminals (a,b), focusing on the calculation of the Thevenin voltage (V(th)) and Thevenin resistance (R(th)). The context includes homework-related problem-solving in the phasor domain with dependent sources.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant states that V(th) is the voltage across a 2-ohm resistor, suggesting it can be expressed as 2*I, but expresses uncertainty about finding I due to the absence of independent sources.
  • Another participant points out that if there are no independent sources to stimulate the dependent source, an external voltage source (V(ex)) must be added to find V(th).
  • A participant clarifies that they meant a 0.2-ohm resistor instead of a 2-ohm resistor and questions whether adding an external voltage source is necessary to find V(th).
  • Another response indicates that adding a voltage source at terminals (a,b) would fix the output voltage, complicating the determination of V(th) as a variable, but suggests that R(th) can still be calculated.
  • One participant proposes a KCL equation involving V(ex) and attempts to derive I(ex), leading to a calculation for R(th) as 0.066 ohm, but seeks confirmation on the correctness of this result.
  • A later reply cautions that placing a voltage source at terminals (a,b) means the node voltage will not simply equal V(ex) due to the presence of other components in the circuit.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and implications of adding an external voltage source to find V(th). There is no consensus on the correctness of the calculations or the interpretation of the circuit components.

Contextual Notes

There are unresolved assumptions regarding the circuit configuration and the role of dependent sources. The discussion reflects uncertainty about the implications of adding external sources on the calculations of V(th) and R(th).

asdf12312
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Homework Statement


determine thevenin equivalent of circuit at terminals (a,b):
cqeer.png


Homework Equations


Z(L)=jwL
Z(C)=-j/wC
V(th)=V(oc)
R(th)=V(ex)/I(ex)

The Attempt at a Solution


actually 1st i need help with finding V(th). i do know that V(th) would be the voltage across the 2ohm resistor, so 2*I. i suppose i just need to find I, but i have no idea how to do that, since there's no independent voltage/current sources. i have tried mesh analysis for instance: 0.2I+ 0.2I + 0.2I = 0. maybe i am trying the wrong method. for R(th) i know i would have to try the V(ex) method, but 1st i need to find V(th).
 
Last edited:
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asdf12312 said:

Homework Statement


determine thevenin equivalent of circuit at terminals (a,b):
cqeer.png


Homework Equations


Z(L)=jwL
Z(C)=-j/wC
V(th)=V(oc)
R(th)=V(ex)/I(ex)

The Attempt at a Solution


actually 1st i need help with finding V(th). i do know that V(th) would be the voltage across the 2ohm resistor, so 2*I. i suppose i just need to find I, but i have no idea how to do that, since there's no independent voltage/current sources. i have tried mesh analysis for instance: 0.2I+ 0.2I + 0.2I = 0. maybe i am trying the wrong method. for R(th) i know i would have to try the V(ex) method, but 1st i need to find V(th).


I don't see a 2 Ohm resistor in the diagram.

If there are no independent sources to stimulate the dependent source, you've got no choice but to add one...
 
sorry meant 0.2ohm resistor. and does that mean i need to add an external voltage source V(ex) in order to find v(th)?
 
asdf12312 said:
sorry meant 0.2ohm resistor. and does that mean i need to add an external voltage source V(ex) in order to find v(th)?

Well, if you pin the output with a fixed voltage then you lose the Vth as a variable. But you can find the Rth that way by dividing the applied potential by the current driven into the port.

On the other hand, it looks as though the Vth should be zero since the open-circuit output voltage is zero.
 
OK, so I add an external voltage source at terminals (a,b) on the right side so there is no open circuit anymore. at the top node the KCL is:
V(ex)/0.2 + (V(ex)-0.2I)/0.2 - I(ex) = 0

also, since I= -V(ex)/0.2:
V(ex)/0.2 + (V(ex)+V(ex))/0.2 = I(ex)
I(ex) = (3*V(ex))/0.2 = 15*V(ex)

R(th)= V(ex)/I(ex) = 1/15= 0.066ohm. is that right?
 
Last edited:
If you put a voltage source Vex at a-b the node voltage won't also be Vex; there are components between the node and Vex.
 

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