Engineering Basic Circuit Homework: Is Vx Connected?

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Vx is interpreted as a label for the voltage difference between two points rather than a voltage source. The circuit is considered open, meaning no current flows through the 3.9 kΩ resistor, resulting in no voltage difference across it. To analyze Vx, one should calculate the resultant voltage and Thevenin resistance of other components first. The actual voltage at Vx will vary depending on the load connected to those points. Understanding these relationships is crucial for solving the circuit homework effectively.
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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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