Basic Circuit Homework: Is Vx Connected?

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SUMMARY

The discussion centers on the interpretation of Vx in a circuit connected to a 14V source and a 3.9kΩ resistor. Participants conclude that Vx represents the voltage difference between two points in an open circuit, indicating that no current flows through the 3.9kΩ resistor, resulting in no voltage across it. The voltage at Vx is contingent upon the load connected to the circuit, and the analysis involves calculating the resultant voltage and Thevenin resistance of other components.

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Homework Statement


[PLAIN]http://img841.imageshack.us/img841/1456/pforum.png


Is the Vx in this connected to the 14[V] source and the 3.9 [kΩ]? I basically just want to look at it like a voltage source but I think it is incorrect. I mean can it still be considered a closed loop? :redface:
 
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Neophyte said:

Homework Statement


[PLAIN]http://img841.imageshack.us/img841/1456/pforum.png


Is the Vx in this connected to the 14[V] source and the 3.9 [kΩ]? I basically just want to look at it like a voltage source but I think it is incorrect. I mean can it still be considered a closed loop? :redface:

I would interpret it as just a label for the voltage difference between those two points, unless the problem statement says something otherwise. So for example, since it is an open circuit (not a voltage source), there is no current flowing in the 3.9k resistor at the bottom, and therefore no voltage difference across it.
 
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Vx looks like an output, so you would include the 14 V supply and 3.9K resistor after you had worked out the resultant voltage and Thevenin resistance of the other components.

The actual voltage at Vx would depend on what load you connect to these points.
 

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