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 …