For the circuit shown below. Power is delivered by

1. Dependent source of 192 W
2. Dependent source of 368 W
3. Independent source of 192 W
4. Independent source of 436 W

The given circuit → Circuit correction

KCL at node A.

	V_A – 40	+	V_A – 0	= 0 ...................(i)
	500		600

V_A = 2i_x ........................(ii)
Again from given circuit

i_x =	40 – V_A
	500

V_A		1	+	1		=	40
		500		600			500

or V_A			1100		=	40
			500 × 600			500

or V_A =	40 × 600
	1100

or V_A = 21.82 V

l_x =	V_A	=	21.82	= 10.90 A
	2		2

Since lx is positive. So power is delivered by independent source i.e. P_delivered = 40 × l_x
or = 40 × 10.90 = 436 W