Correct Option: D

Given V₂ = 1 – e^{– 2t}

I R =		V2	=	3(1 – e–2t)
		2 / 3		2

I c =		d	V2 = 1	d	(1 – e– 2t)
		dt		dt

= – (– 2) e^{– 2t} = 2e^{– 2t}
current across 2Ω resistance

(I) = IR + IC =	3	(1 – e– 2t) + 2e– 2t
	2

=		3	+	1	e– 2t
		2		2

V₁ = I.2 + V₂

=			3	+	1	e– 2t		. 2 + 1 – e– 2t
			2		2

= 3 + e^{– 2t} + 1 – e^{– 2t}
= 4