Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
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 O h. 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 O h. 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)
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 2.7P: Let a 100V sinusoidal source be connected to a series combination of a 3 resistor, an 8 inductor,...
See similar textbooks
icon
Related questions
Question
Circuit theory Power Analysis I want the solution within twenty minutes
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 O h. 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
Transcribed Image Text: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 O h. 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
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Power System Analysis and Design (MindTap Course …
Power System Analysis and Design (MindTap Course …
Electrical Engineering
ISBN:
9781305632134
Author:
J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS