Correct Option: A

The given circuit can be redrawn by applying source transformation

KCL at node A

Ix +	VA	= 1.6 Ix ..............(i)
	8

Also,	VA – 10 90°	= Ix
	(5 + j2)

or V_A = I_x (5 + j2) + 10 90º ..........................(ii)
from equations (i) and (ii)

Ix +	Ix (5 + j2) + 10 90°	= 1.6 Ix
	8

or 8I_x + I_x (5 + j2) – 12.8 I_x = – 10 90º
or I_x (0.2 + j2) = – 10 90º

or Ix =	–10 90°	=	–10 90°
	0.2 + j2		2.01 84.23°

or I_x = – 4.975 5.77°
V_A = (1.6 I_x – I_x) × 8
or V_A = .6 I_x × 8
or V_A = 4.8 I_x
or V_A = – 23.88 V

Now, average power =	1	VA. Ix
	2

=	1	(– 23.88) × (– 4.975)
	2

= 59.40 W
Positive power means supplied by the dependent source.