A CE amplifier has a resistor RF connected between collector

A CE amplifier has a resistor R_F connected between collector and base. R_F = 40 K, R_C (collector load resistance) = 4 K, Given h_fe = 50, r_π = 1 K. The output resistance R₀ is given by:

R₀ = R′_L = R_c \|\| R_F =	4 × 40	=	160	= 3.64 k
	4 + 40		44

output resistance after feedback is given by (since RF is the feedback resistance connected between the collector and base)

R_of =	R₀
	1 + Aβ

(∴ due to voltage shunt feedback)

Aβ =	R_C	. β =	4	. 50 =	50
	R_F + A_C		40 + 4		11

or 1 + Aβ = 1 +	50	=	61
	11		11

Now, R_of =	R₀	=	3·64	= 0.66 kΩ
	1 + Aβ		61/11

Hence alternative (D) is the correct choice.