if drain terminal of an nmos is open and source terminal is connected to 1v+ and the gate is also biased to a positive voltage, the Id=0 and Vds=0. can anyone explain? especially "Vds=0" I thought when a terminal is open it means we don't know its voltage, how does it say: Vds=0??
If the gate and source are both at 1V the MOSFET is off, but there is still a bit of leakage that will quickly discharge any charge that might be on the drain. If the gate is at a high enough potential to turn the MOSFET on, of course the discharge will be much faster.