Open circuit voltage gain of JFET transistor circuit
If you look at the AC small signal model of a MOSFET, you will see this:



Therefore, you need to convert your MOSFET circuit into this form and analyze it. Beware of memorizing and thinking with formulas. This is circuit analysis, not circuit memorization.

First since this is AC, we can just short out the capacitors so \$V_G=V_{in}\$ and \$V_D=V_{out}\$ (Assuming capacitors are big enough). Then, we can just ignore the \$R_G\$ as it is huge making it essentially an open circuit in the system. We can also ground the \$+V_{DD}\$ DC source as we are looking at this in AC domain. So now gate of the MOSFET goes directly to \$V_{in}\$, drain goes to the ground through RD and source again goes to ground through RS as you can see below.







simulate this circuit &ndash; Schematic created using CircuitLab

Now, onto the analysis:

We are trying to find $$A_v=\frac{V_{out}}{V_{in}}=\frac{V_{GS}}{V_{in}}\frac{V_{out}}{V_{GS}}$$


  For \$\frac{V_{GS}}{V_{in}}\$, \$V_{in}=V_G\$ and \$V_S=g_mV_{GS}R_S\$. So \$V_{GS}=V_G-V_S=V_{in}-g_mV_{GS}R_S\$. Therefore, \$\frac{V_{GS}}{V_{in}}=\frac{1}{1+g_mR_S}\$.


This is our first expression.


  For \$\frac{V_{out}}{V_{GS}}\$, since all the current, \$g_mV_{GS}\$, is flowing through \$R_D\$, \$V_{out}=-g_mV_{GS}R_D\$. So \$\frac{V_{out}}{V_{GS}}=-g_mR_D\$


This is our second expression.

Finally, if we put both expressions into \$A_v\$,

$$A_v=\frac{V_{out}}{V_{in}}=\frac{-g_mR_D}{1+g_mR_S}$$.

When you plug in these, the answer you get is -4.17 V/V. As you can see with this, \$R_S\$ provides negative feedback into this amplifier, lowering the gain.