The common emitter forward current gain of the transistor

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The common emitter forward current gain of the transistor shown is β_F = 100.

The transistor is operating in

1. Saturation region
2. Cut-off region
3. Reverse active region
4. Forward active region

Correct Option: C

Assume V_BE = 0.7 and since β >> 1, I_C ≈ I_E
using KVL in base-emitter circuit,
10 – V_BE – I _ER_E – I_BR_B = 0
⇒ 10 – 0.7 = (β + 1) I_B. R_E + R_B. I_B

⇒ I_B =	9.3	mA = 0.025 mA
	101 + 270

Now, using KVL in Collector-emitter circuit,
10 – V_CE – (I_E + I_C)1 = 0 [Assume V_CE = 0.2 V]

⇒	9.8	= I_C ⇒ I_C = 4.9 mA
	2

(I_B)_min =	I_C	=	4.9	= 0.049 mA
	β		100

As (I_B)_min < I_B = 0.025 mA, therefore transistor operates in active region.