Electrical Engineering: Principles & Applications (6th Edition)

Published by Prentice Hall
ISBN 10: 0133116646
ISBN 13: 978-0-13311-664-9

Chapter 1 - 1.3 - Problems - Power and Enery - Page 36: P1.25

Answer

Power - 694.44 W Current - 5.787 A Energy - reduced by 8.64%

Work Step by Step

Total energy consumed = Total bill / Bill per unit energy = 60/0.12 = 500 kWh Power = Total energy / Total time = $\frac{500 \times 1000 Wh}{30 \times 24h}$ = 694.44 W If voltage = 120 V, current = power/voltage = 694.44/120 = 5.787 A If 60 W light is turned off, power would be reduced by 60. Hence reduced percentage = $60 \times 100 / 694.44$ = 8.64%
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