CircuitLab voltage source question
The system you have drawn actually has an in-determinant answer for what \$V_{off}\$ is with respect to ground because it is galvanically isolated from ground.

Having a perfect open circuit in the system is problematic in how many circuit simulators are implemented, so the way this is often handled is to use a very high (but finite) resistance to model open circuits. With this very high but finite resistance, the solution becomes \$V_{off} = 5V\$.

To test if CircuitLab does use a finite high resistance, I built a "voltage divider" circuit with an open switch and a very large resistance. The DC simulator tells me the output voltage is 333.33mV, which means that \$R_{open} = 5\cdot10^{14} \Omega\$







simulate this circuit &ndash; Schematic created using CircuitLab