Why is the Voltage Negative in this Superposition Theorem Circuit?

[Iyafrady]

Why is this voltage negative??

http://qaboard.cramster.com/answer-board/image/96937de22e44b1524804f2cf580462fe.jpg
Using superposition theorem
With voltage source only,
Vth1 = (3)(5000) / (2000+5000) = 2.14V
With current source only
Current through 5kΩ resistor
I = (0.007)(2000) / (2000+5000) = 2mA
The voltage across it will be
Vth2 = - (0.002)(5000) = -10V why is this voltage negative??
The resultant voltage is
Vth = Vth1 + Vth2
Vth = 2.14 - 10 = - 7.86V

 

[MATLABdude]

http://qaboard.cramster.com/answer-board/image/96937de22e44b1524804f2cf580462fe.jpg
Using superposition theorem
With voltage source only,
Vth1 = (3)(5000) / (2000+5000) = 2.14V
With current source only
Current through 5kΩ resistor
I = (0.007)(2000) / (2000+5000) = 2mA
The voltage across it will be
Vth2 = - (0.002)(5000) = -10V why is this voltage negative??
The resultant voltage is
Vth = Vth1 + Vth2
Vth = 2.14 - 10 = - 7.86V

The direction of current flow through the resistor (due to the current source) is the opposite of that due to the voltage source. As the voltage measurement is (implicitly) with the "-" terminal on the shared node on the bottom, this makes the voltage due to the current source negative.

EDIT: Not implicit measurement, I missed the ground symbol on the bottom node. The more intuitive way of realizing that there is a negative voltage is that current flows from higher potential to lower. And then realize that if the higher potential is zero (ground) then the lower potential must be negative.

 

[piercas]

In this circuit, the voltage is negative because the current source is producing a current that is flowing in the opposite direction of the voltage source. This means that the voltage drop across the 5kΩ resistor is in the opposite direction of the voltage source, resulting in a negative voltage. This can also be seen in the calculation of Vth2, where the negative sign is used to indicate the direction of the voltage drop. Additionally, the negative voltage can also be interpreted as a voltage drop rather than a voltage source, since it is being caused by the current flowing through the resistor.