Performing DC analysis by open circuiting all capacitors

The voltage across resistor R2 can be given by the expression

The current through R1 is

$I_1=\dfrac{V_{DD}-V_G}{R_1}$

Similarly, current through R2 is

$I_2=\dfrac{V_G-V_{SS}}{R_2}$

Since there is no path for these currents to take another loop, these currents have got to be the same which brings us to the foll. equation

$\dfrac{V_{DD}-V_G}{R_1}=\dfrac{V_G-V_{SS}}{R_2}$

$\dfrac{V_{DD}-V_G}{R_1}-\dfrac{V_G-V_{SS}}{R_2}=0$

$\dfrac{V_{DD}R_2-V_GR_2-V_GR_1+V_{SS}R_1}{R_1R_2}=0$

${V_{DD}R_2-V_GR_2-V_GR_1+V_{SS}R_1}=0$

${V_{DD}R_2+V_{SS}R_1}=V_G(R_1+R_2)$

$\therefore V_G=\dfrac{V_{DD}R_2+V_{SS}R_1}{R_1+R_2}$

NOTE: Using the above equation, we can easily get VB but just in case you want a easier equation, we will simplify the obtained equation

$\therefore V_G=\dfrac{V_{DD}R_2+V_{SS}R_2-V_{SS}R_2+V_{SS}R_1}{R_1+R_2}$

$V_G=\dfrac{V_{SS}(R_1+R_2)}{R_1+R_2}+\dfrac{R_2(V_{DD}-V_{SS})}{R_1+R_2}$

$\therefore V_G=V_{SS}+(V_{DD}-V_{SS})\times\dfrac{R_2}{R_1+R_2}$

$V_G=-5+(10)\times\dfrac{30\times10^3}{180\times10^3+30\times10^3}$

$V_G=0.714\ V$

$R_G=R_1||R_2$

$R_G=\dfrac{180\times10^3\times30\times10^3}{180\times10^3+30\times10^3}$

$R_G=27.71\ K\Omega$

$V_S=I_SR_S-V_{SS}$

But, $I_S=I_D$

$\therefore V_S=I_DR_S-V_{SS}$

$V_S=1\times10^3I_D-5$

Applying KVL to Gate - Source Loop

$V_{GS}=V_G-V_S$

$V_{GS}=0.714-(1000I_D-5)$

$V_{GS}=0.714-1000I_D+5$

$V_{GS}=5.714-1000I_D\cdots(1)$

The formula for ID is given by

$I_D=k\big(V_{GS}-V_T\big)^2$

$I_D=1\times10^{-3}\big(5.714-1000I_D-0.8\big)^2$

$I_D=1\times10^{-3}\big(4.914-1000I_D\big)^2$

$I_D=1\times10^{-3}\big(24.14-9828I_D+1\times10^6{I_D}^2\big)$

$I_D=0.024-9.828I_D+1000{I_D}^2$

$1000{I_D}^2-10.828I_D+0.024=0$

$I_D=7.718\ mA \ \&\ 3.1\ mA$

If $I_D=7.718\ mA$, this will result in negative value of VDS. Hence,

$I_D=3.1\ mA$

Putting the value of ID in 1, we get

$V_{GS}=5.714-1000\times3.1\times10^{-3}$

$V_{GS}=2.614\ V$

Replacing the MOSFET by it's small signal equivalent circuit we get