- An AC Motor with impedance Z₁ = 45+j100 is supplied by a 111 Vrms, 68 Hz source. Find a) Power factor (pf), b) Average power (active power), c) Reactive power, d) Complex power, e) Apperant power, f) Determine the capacitor required to be connected in parallel with motor so that power factor is corrected to 0.92. O a. a) pf=0.410, b) active power = 23.054 W, c) reactive power = 102.462 VAR, d) Complex Power= 23.054 + j102.462 VA, e) Apperant Power= 112.358 VA, f) Capacitor Value 31.481 uF O b. a) pf=0.357, b) active power = 46.108 W, c) reactive power = 51.231 VAR, d) Complex Power= 46.108 + j51.231 VA, e) Apperant Power= 112.358 VA, f) Capacitor Value 15.741 uF O c. a) pf=0.460, b) active power = 23.054 W, c) reactive power = 102.462 VAR, d) Complex Power= 46.108 + j102.462 VA, e) Apperant Power= 56.179 VA, f) Capacitor Value = 31.481 uF O d. a) pf=0.410, b) active power = 46.108 W, c) reactive power = 102.462 VAR, d) Complex Power= 46.108 + j102.462 VA, e) Apperant Power= 112.358 VA, f) Capacitor Value 15.741 uF O e. a) pf=0.460, b) active power = 46.108 W, c) reactive power = 102.462 VAR, d) Complex Power= 46.108 + j102.462 VA, e) Apperant Power= 112.358 VA, f) Capacitor Value = 31.481 uF O f. a) pf=0.357, b) active power = 23.054 W, c) reactive power = 51.231 VAR, d) Complex Power= 23.054+j51.231 VA, e) Apperant Power= 56.179 VA, f) Capacitor Value = 15.741 uF O g. a) pf=0.410, b) active power = 46.108 W, c) reactive power = 51.231 VAR, d) Complex Power= 46.108 + j51.231 VA, e) Apperant Power= 56.179 VA, f) Capacitor Value = 31.481 uF Oh. a) pf=0.357, b) active power = 23.054 W, c) reactive power = 102.462 VAR, d) Complex Power= 23.054+j102.462 VA, e) Apperant Power= 56.179 VA, f) Capacitor Value 15.741 uF
- An AC Motor with impedance Z₁ = 45+j100 is supplied by a 111 Vrms, 68 Hz source. Find a) Power factor (pf), b) Average power (active power), c) Reactive power, d) Complex power, e) Apperant power, f) Determine the capacitor required to be connected in parallel with motor so that power factor is corrected to 0.92. O a. a) pf=0.410, b) active power = 23.054 W, c) reactive power = 102.462 VAR, d) Complex Power= 23.054 + j102.462 VA, e) Apperant Power= 112.358 VA, f) Capacitor Value 31.481 uF O b. a) pf=0.357, b) active power = 46.108 W, c) reactive power = 51.231 VAR, d) Complex Power= 46.108 + j51.231 VA, e) Apperant Power= 112.358 VA, f) Capacitor Value 15.741 uF O c. a) pf=0.460, b) active power = 23.054 W, c) reactive power = 102.462 VAR, d) Complex Power= 46.108 + j102.462 VA, e) Apperant Power= 56.179 VA, f) Capacitor Value = 31.481 uF O d. a) pf=0.410, b) active power = 46.108 W, c) reactive power = 102.462 VAR, d) Complex Power= 46.108 + j102.462 VA, e) Apperant Power= 112.358 VA, f) Capacitor Value 15.741 uF O e. a) pf=0.460, b) active power = 46.108 W, c) reactive power = 102.462 VAR, d) Complex Power= 46.108 + j102.462 VA, e) Apperant Power= 112.358 VA, f) Capacitor Value = 31.481 uF O f. a) pf=0.357, b) active power = 23.054 W, c) reactive power = 51.231 VAR, d) Complex Power= 23.054+j51.231 VA, e) Apperant Power= 56.179 VA, f) Capacitor Value = 15.741 uF O g. a) pf=0.410, b) active power = 46.108 W, c) reactive power = 51.231 VAR, d) Complex Power= 46.108 + j51.231 VA, e) Apperant Power= 56.179 VA, f) Capacitor Value = 31.481 uF Oh. a) pf=0.357, b) active power = 23.054 W, c) reactive power = 102.462 VAR, d) Complex Power= 23.054+j102.462 VA, e) Apperant Power= 56.179 VA, f) Capacitor Value 15.741 uF
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter2: Fundamentals
Section: Chapter Questions
Problem 2.21P: An industrial plant consisting primarily of induction motor loads absorbs 500 kW at 0.6 power factor...
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