For the circuit shown below each diode has Vr = 0.6 V and

For the circuit shown below each diode has V_r = 0.6 V and r_f = 0. Both diode will be ON if:

V_A = V_S.	500	=	V_S
	500 + 5000		11

and given cut-in voltage i.e., V_γ = 0.6 for each diode, for V_S > 0.6 V, D₁ is ON., while for V_A > 0.6 V, both the diode will be ON.
KCL at node A.

V_A – V_S + 0.6	+	V_A – V_S	+	V_A	+	V_A – 0.6	= 0
5 kΩ		5 kΩ		500 kΩ		500 kΩ

V_A =	2V_S + 5.4	> 0.6 V
	22

⇒ V_S > 3.9 V
Hence alternative (A) is the correct choice.