Mohammad S. answered 11/19/20
An experienced and knowledgeable tutor
Saturation:
Ohmic:
According to above:
for i., vDS > (vGS - vt) or 10 > 3, thus saturation:
I_D = 0.5*(0.5)(mA/V^2)(4 -1)^2 = 2.25 mA
for ii. vDS > (vGS - vt) or 2 > 3 which is not true, thus Ohmic:
I_D = 0.5(mA/V^2)*((3)(2) - 2)= 2.5 mA
for iii. vGS = 0 (not > = Vt), therefore, the transistor is off and I_D = 0.