The following excerpt from a book explains the functioning of the circuit below:
When V_in < V_th (device threshold voltage), the supply voltage (V_dd) is measured at the outlet. When V_in is increased above V_th, the NMOS turns on and V_dd is now dropped across the load resistor; V_out is now in common with ground, and the signal at V_out is inverted relative to V_in.
Question: Why is V_out = V_dd when the NMOS transistor is off (i.e. when V_in < V_th)? With the transistor off, it seems we should effectively be able to ignore that part of the circuit and compute V_out using Ohm's law to predict the drop of V_dd across the resistor. Why is this not the case?