Performing DC analysis of the circuit

RS is the combination of two indivigual resistances

$R_S=330+100=430\ \Omega$

Here, $V_G=0$

$V_{GS}=V_G-V_S$

$V_{GS}=-V_S$

and $V_S=I_SR_S$

But, $I_S=I_D$

$\therefore V_{GS}=-I_DR_S$

$\therefore V_{GS}=-430I_D\cdots(1)$

To find drain current, we use the Shockely equation

$I_D=I_{DSS}\bigg[1-\dfrac{V_{GS}}{V_P}\bigg]^2$

$I_D=10\times10^{-3}\bigg[1-\dfrac{-430I_D}{-3.5}\bigg]^2$

$I_D=10\times10^{-3}\bigg[1-\dfrac{430I_D}{3.5}\bigg]^2$

$I_D=\dfrac{10\times10^{-3}}{12.25}\big(3.5-430I_D\big)^2$

$I_D=0.816\times10^{-3}(12.25-3010I_D+184900{I_D}^2\big)$

$I_D=9.996\times10^{-3}-2.456I_D+150.87{I_D}^2$

$150.87{I_D}^2-3.456I_D+9.996\times10^{-3}=0$

$I_D=19.5\ mA\ \&\ 3.3957\ mA$

But, (I_D\not>I_{DSS})

$\therefore \underline{I_D=3.3957\ mA}$

Substituting the value of ID in 1

$V_{GS}=-430\times3.396\times10^{-3}$

$\therefore V_{GS}=-1.46\ V$

Applying KVL to the Drain - Source Loop

$V_{DD}-I_DR_D-V_{DS}-I_SR_S=0$

But, $I_S=I_D$

$V_{DD}-I_D\big(R_D+R_S\big)=V_{DS}$

$\therefore V_{DS}=15-3.396\times10^{-3}\big(8000+430\big)$

$\therefore \underline{V_{DS}=-13.628\ V}$

Therefore, $\underline{\underline{\text{Q - Point}=\big(-13.628\ V,3.396\ \text{mA}\big)}}$

Replacing the FET by the small signal equivalent model