Q.The DC base current of the transistor is 10$\mu$A. Calculate the AC resistance of the emitter diode of Silicon transistor. Given $\beta$=100

In a working amplifier the DC bias current and the AC signal current flow through the same components at once, so the circuit is analysed in two independent parts.

DC equivalent circuit — reduce every AC source to zero and open-circuit every capacitor (a capacitor blocks DC). What remains is the bias network. The supply divides across $R_1, R_2$, giving $V_2 = V_{CC}\,R_2/(R_1+R_2)$; the base-emitter loop then gives the emitter current $I_E = (V_2 - V_{BE})/R_E$, and the collector loop gives $V_{CC} = V_{CE} + I_E R_E + I_C R_C$. This fixes the operating point.

AC equivalent (re' model) — short every DC source and every capacitor, then replace the transistor by its small-signal model: the emitter-base junction becomes an AC resistance and the collector-base junction a constant-current source. The key quantity is the AC emitter-junction resistance $r_e' = V_T/I_E$, with the thermal voltage $V_T = 26$ mV for silicon and $52$ mV for germanium.

From this model the CE-amplifier results follow: input impedance $Z_{in} = R_1 \parallel R_2 \parallel \beta r_e'$; output impedance $Z_o = R_C$ (or $R_C \parallel R_L$ with a load); voltage gain $A_v = -R_C/r_e'$ (the minus sign is the 180 phase reversal); current gain $A_i = \beta$; and power gain $A_p = A_v \times A_i$.