Performing DC analysis by open circuiting all capacitors

We know that, $I_C\approx I_E$

$\therefore I_C=1\ mA$

Applying KVL to Base - Collector Loop

$V_C=V_{CC}-I_CR_C$

$V_C=5-1\times10^{-3}\times2\times10^3$

$\therefore V_C=3\ V$

$V_{BE}=V_B-V_E$

But, $V_B=0$ and $V_{BE}=0.7$

$\therefore V_E=-0.7\ V$

$V_{CEQ}=V_C-V_E$

$V_{CEQ}=3-(-0.7)$

$\therefore \underline{\underline{V_{CEQ}=3.7\ V}}$

Replacing the BJT by it's small signal equivalent model

As you may observe, the current source has vanished, that is because during ac analysis we neglect any DC sources and since the current source is DC, it gets kicked out

By observation

$V_{in}=-V_\pi$

$V_o=-g_mV_\pi(R_C||R_L)$

Voltage gain is given as

$A_V=\dfrac{V_o}{V_{in}}$

$A_V=\dfrac{-g_mV_\pi(R_C||R_L)}{-V_\pi}$

$\therefore A_V=g_m(R_C||R_L)$

where, $g_m=\dfrac{I_{C}}{V_T}=\dfrac{1\times10^{-3}}{26\times10^{-3}}=38.46\ mA/V$

$\therefore A_V=38.46\times10^{-3}(2\times10^3||10\times10^3)$

$\therefore \underline{\underline{A_V=64.1}}$

Output resistance is simply $R_C||R_L$

$R_O=\dfrac{2\times10^3+10\times10^3}{2\times10^3+10\times10^3}$

$\therefore \underline{\underline{R_O=1.66\ K\Omega}}$