Correct Option: D

The given circuit:

Applying KCL at node A, we get

VA + 80	+	VA – 40	+	VA – 100	= 0
2		3		4

VA		1	+	1	+	1		= -	80	+	40	+	100
		2		3		4			2		3		4

or

VA		6 + 4 + 3	+		=	- 80 × 6 × + 40 ×4 +100 ×3
		12				12

or

I4Ω =	100 + 20	=	120
	4		4

= 30A
Power absorbed by 4Ω resistor = I²_4Ω R = 30² × 4 = 3600W
Hence alternative (D) is the correct choice.