Correct Option: A

Considering the given circuit for first half cycle in order to determine the peak value of the current through the load resistor.
From above figure

I =	V
	Req

Where, Req = 2 || (2 + 2) =	4	kΩ.
	3

Now, I =		(10 – 0.7) V	=	9.3 x 3	mA
		4 / 3 kΩ		4

= 6.975mA

and IL(peak) = I ×		2	= 6.975 x	2	= 2.325 mA.
		2 + 4		6