Correct Option: C

If we consider the circuit carefully we see that there are two voltage source present in the circuit and if we have to calculate the net voltage in a particular resistance it can be easily calculated by using superposition theorem.
Case I. When V_s = 8u (t) = 8 V source is treated, the equivalent circuit becomes.
Apply voltage divides method, we get

I =		V	=	8		8	=	16	amp
		Req		1 + 1 || 1		1 + (1 / 2)		3

I 1 = 16 3 × 1 2 = 8 3 amp
_{Case II.} When | V₀ | = 5 V source is treated, the equivalent circuit becomes.

I′ 2 =		V	=	5
		Req		1 + 1 || 1

=		5	=	10	amp.
		1 + (1 / 2)		3

I′ 1 = I′2 ×		1	=	10	x	1
		1 + 1		3		2

I′ 1 =	5	amp.
	3

The net current in 1 Ω resistance

= I1 + I′ 1 =		8	+	5	=	13	amp.
		3		3		3

Voltage drop across 1 Ω =		13	x 1 =	13	V
		3		3