In the circuit shown VCC = 10, RC = 2.7 K, RF = 200 K, β

In the circuit shown V_CC = 10, R_C = 2.7 K, R_F = 200 K, β = 99, V_BE = 0.6V. The operating point V_CE, I_C are given by—

1. 4.6 V and 1.98 mA
2. 3.18 V and 2.5 mA
3. 5.4 V and 1.56 mA
4. 4.2 V and 2.1 mA

From given figure

V_CC – I_e R_C – I_b R_f – V_BE = 0
V_CC – I_C R_C – I_b (R_f + R_c) – V_BE = 0
V_CC – I_C R_C – I_C β (R_f + R_C) – V_BE = 0

10 – I_C		R_C +	(R_f + R_C)		– 0.6 = 0
			β

or I_C =	9.4³
	2.7 × 10³ + { (200 + 2.7) 10³ / 990}

or I_C =	9.4³
	2700 + 2047.47

or I_C = 1.98 mA
or V_CE = V_CC – I_C R_C = V_CC – (I_C + I_b) R_C

or V_CE = 10 –		1.98 +	1.98		× 10^–3 × 2.7 × 10³
			99

or V_CE = 10 – 5.4 = 4.6 V
Hence alternative (A) is the correct answer.