MOSFET circuit analysis

Answer

$V_{gat}e =V_{ss}+ (V dd-V ss)* R2/(R1+R2) =V ss+ (10+10) *40/200 =40/10 =-10+4 =-6 V$

$V_{gate}$ is the gate voltage with reference to GND (0 volts)

V(GS) is the gate source voltage

$V gate +V(GS) +Id*Rs =0$

$Id = k*(V(GS)-VT)^2$

$V gate +V(GS)+k*[(VGS) -VT]^2*Rs =0$

$-6 +V(GS) +2 (mA/V)* (V(GS)-1)^2*2 K =0$

$-6 +V(GS) +4(V(GS)^2-2V(GS) +1) =0$

$-6 +Vgs +4Vgs^2 -8Vgs +4 =0$

$4 Vgs^2 -7Vgs -2 =0$

$Vgs = 2V$ (only this root is positive)

$Id=k*(V gs-VT)^2 =2 mA* (2-1)^2 =2 mA$

Drain current is 2 mA.