Is this possible for simulation?

For testing purposes in a circuit simulation software is it possible to generate a result for a hanging circuit? please see my attachment for better explanation.

You can also see my link:
http://www.partsim.com/simulator/#12162

Thanks

The circuit will not simulate - it will error out. Even if you were able to simulate it or build it, what purpose would it serve? Unless I am missing something, there will be no current through the resistor and therefore nothing would happen. The only thing you can do (and simulate) is connect a voltmeter or ammeter between the resistor's open end and ground. It will then simply indicate the source voltage or the current through the resistor, respectively.

Ow., I see. my purpose for the simulation is just to determine if by chance that the circuit is also possible in actual set up like doing it using real voltage. Basing from what you said i totally agree because if you try a simple mesh it would totally have no sense. I think putting a Voltmeter in between the hanging resistor and ground would simply mean nothing. I think Voltmeter must be shunted to a particular component to work as intended for it. What could you say?
By the way thanks =)

bq). I think Voltmeter must be shunted to a particular component to work as intended for it.

If you do that for the resistor it will indicate zero volts because the resistor's other end is hanging. You will have to connect it to ground and then it will simply indicate the source voltage. Also between the resistor's hanging end and ground the voltmeter will still indicate the source voltage as long as the voltmeter's impedance is more than 10 times the resistor's value. Otherwise it will indicate a voltage based on the following formula: Vmeter = (Rmeter * Vsrc)/(Rmeter + R). If Rmeter >> R, then Vmeter approaches Vsrc.

I forgot to mention that what i mean earlier is that its already connected to ground. But for the resistor in series with the voltmeter to ground that was my first time to hear about it. I hadn't had the knowledge about that. One more thing, so if the voltmeter could be in series with the resistor with the condition of impedance greater than the resistor. Could it be possible that the ammeter can be in shunted with the resistor also?