Basic_Electrical_Engineering_4th_edition

A-C CIRCUITS	63
Power	P = 120 x 5.37 cos 63.4 = 288 watts
Power can also be calculated using the relation
	P = Real [VI*]
In Cartesian co-ordinates
	V= l20 +j0
	I= 2.4 -j4.8
Hence	I* = 2.4 +j4.8
and	P = Re [(120 + j0)(2.4 + j4.8)]
	= 120 x 2.4
	= 288 watts

Example 1.3. A 15 V source is applied to a capacitive circuit that has an impedance of (10-j25) Determine the current and the power in the circuit.

Solution:		as reference and using j-notation method we have
Q.	Taking voltage	(15 + jO)(lO + j25)
	I=	(10 - j25)(10			j25)
		150	j375	6		jl5
	-	102	+ 252	+	29
	= 0.2 ++j0.517 A				+

This means the current leads the voltage

I - 0.55 tan-1 ---0.517 o0.552L68.85

which means the current leads the supply voltage by 68.85° which, in this case is also the angle of the impedance (10 -j25).

- 25
tan E l =W'	E l =- 68.2°
This being capacitive circuit the impedance triangle is as shown
here in Fig. El.3.

The power loss Re [VI*]

= {15[0.2 -j0.517]}

= 3 watts	Fig. E1 .3
This can also be obtained using the formula
P= VI cos El

=15 x 0.55 cos 68.8

=3 watts

Since it is a series circuit and the current through the circuit is 0.55 A and resistance is 10 Q, the power consumed by the circuit is I2R

0.552 x 10 = 3 watts

The reactive power Q = VIsin E l =15 x 0.55 sin (- 68.8) = - 7.7 Var