a) Find the useful power output of an elevator motor that lifts a 3675-kg load a height of 44 m in 18s, if it also increases the speed from rest to 6 m/s. Note that the total mass of the counterbalanced system is 16888 kg, so that only 3675 kg is raised in height, but the full 16888 kg is accelerated. 104924.67 W b) What does it cost, if electricity is $0.0546 per kW.h? 0.5728 dollars/h c) If the motor operates from a 600V DC supply, what is the average current draw by the motor? (assume the power of the motor is given by P=IV) 174.87 A 12 V -11x 2 A + 2 A + + 2 24 V 3 V3 36 V (+ Ix 1 Given that I 2, determine the following quantities: = a. The power disipation in the 36V source is: 72 b. The power dispation in element 1 is: 24 W W c. The power dispation in element 2 is: 48 W d. The power dispation in the dependent source is: 4 W e. Determine the voltage V3 22
a) Find the useful power output of an elevator motor that lifts a 3675-kg load a height of 44 m in 18s, if it also increases the speed from rest to 6 m/s. Note that the total mass of the counterbalanced system is 16888 kg, so that only 3675 kg is raised in height, but the full 16888 kg is accelerated. 104924.67 W b) What does it cost, if electricity is $0.0546 per kW.h? 0.5728 dollars/h c) If the motor operates from a 600V DC supply, what is the average current draw by the motor? (assume the power of the motor is given by P=IV) 174.87 A 12 V -11x 2 A + 2 A + + 2 24 V 3 V3 36 V (+ Ix 1 Given that I 2, determine the following quantities: = a. The power disipation in the 36V source is: 72 b. The power dispation in element 1 is: 24 W W c. The power dispation in element 2 is: 48 W d. The power dispation in the dependent source is: 4 W e. Determine the voltage V3 22
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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