The average power delivered by the 500 ∠ 0° voltage source

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The average power delivered by the 500 ∠ 0° voltage source and dependent source for the circuit shown below—

1. 5·432 kW, – 2·833 kW
2. 54·32 kW, – 2·833 kW
3. 28·33 kW, 5·432 kW
4. 2·833 kW, – 5·432 kW

Correct Option: A

The given circuit:

From given circuit 5
00 0° – 4I – 7I – 3V_x = 0 . . . . …(i)
and
V_x = 4I . . . . …(ii)
From equation (i) and (ii) 500 0° – 11I – 3V_x = 0
or 500 – 11I – 3 × 4 I= 0
or

I =	500 0°	= 21·73 0°
	23

Now, the power delivered by the 500 0° voltage source =
1/2 VI

=	1	× 500 0°× 21·73
	2

= 5·432kW
The power delivered by the independent voltage source

=	1	3V_x × I =	1	× 3 × 4I × I
	2		2

= 6I² = 6 (21·73)²
= 2·833 kW
Since the current I entering therefore power delivered will be negative i.e. – 2·833 kW.
Hence alternative (A) is the correct choice.