If the transistors in figure have high values of β and

If the transistors in figure have high values of β and a V_BE of 0.65V, the current I flowing through the 2 kΩ resistance will be—

For DC current
I_E2 ≈ I_E1
or I_C2 ≈ I_C1
Neglecting I_B1 which will be very small, β being very high, the base voltage of Q₁ is

V_B1 =	V_CC	(by using potential divider method)
	(1.65 kΩ) 6.5 kΩ + 1.85 kΩ + 1.65 kΩ

= 10 ×	1.65	= 1.65 V I_E1
	10

=		V_B1 – V_BE	=	= (1.65 – .65) V	mA.
		1 kΩ		1 kΩ

The current flowing through the 2 kΩ resistor is 1 mA.
Since I_C2 ≈ I_C1.
Hecne alternative (A) is the correct answer.