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Electric Circuits, 9th Edition P41. Designed for use in a one or two-semester Introductory Circuit Analysis or Circuit Theory Course taught in Electrical or Computer Engineering Departments. Electric Circuits 9/e is the most widely used introductory circuits textbook of the past 25 years. As this book has evolved over the years to meet the changing learning styles of students, importantly, the underlying teaching approaches and philosophies remain unchanged. | 376 Sinusoidal Steady-State Power Calculations The average power lost in the line results from the current flowing through the line resistance Fline I 2K 89.44 2 0.05 400 W Note that the power supplied totals 20 000 400 20 400 W even though the loads require a total of only 20 000 W. c As we can see from the power triangle in Fig. 10.15 c we can correct the power factor to 1 if we place a capacitor in parallel with the existing loads such that the capacitor supplies 10 kVAR of magnetizing reactive power. The value of the capacitor is calculated as follows. First find the capacitive reactance from Eq. 10.37 Y IK-nl2 X Q _ 250 2 -10 000 -6.25 O. Recall that the reactive impedance of a capacitor is 1 wC and w 2ti 60 376.99 rad s if the source frequency is 60 Hz. Thus C - 424.4 lcF. o X 376.99 6.25 The addition of the capacitor as the third load is represented in geometric form as the sum of the two power triangles shown in Fig. 10.16. When the power factor is 1 the apparent power and the average power are the same as seen from the power triangle in Fig. 10.16 c . Therefore the apparent power once the power factor has been corrected is 5 P 20 kVA. The magnitude of the current that supplies this apparent power is 20 000 w r 80A- The average power lost in the line is thus reduced to Plinc I. 80 2 0.05 320 W. Now the power supplied totals 20 000 320 20 320 W. Note that the addition of the capacitor has reduced the line loss from 400 W to 320 W. -lOkVAR b 20 kW c Figure 10.16 a The sum of the power triangles for loads 1 and 2. b The power triangle for a 424.4 iF capacitor at 60 Hz. c The sum of the power triangles in a and b . Example 10.7 Balancing Power Delivered with Power Absorbed in an ac Circuit a Calculate the total average and reactive power delivered to each impedance in the circuit shown in Fig. 10.17. b Calculate the average and reactive powers associated with each source in the circuit. V5 150 0 V Vi 78 - 104 V V2 72 104 V V3 150 - 130 V 11 -26 - 52 A I -2 6 A
