Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
2. A long shunt compound generator has full load output of 100 kW at 250 volts. The armature, series and shunt windings have resistances of 0-05 2,0-03 2 and 55 2 respectively. Find the armature current and generated e.m.f. [404-5 A, 282-3 VJ

2. A long shunt compound generator has full load output of 100 kW at 250 volts. The armature, series and shunt windings have resistances of 0-05 2,0-03 2 and 55 2 respectively. Find the armature current and generated e.m.f. [404-5 A, 282-3 VJ

Introductory Circuit Analysis (13th Edition)
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...
See similar textbooks
icon
Related questions
Question
A long shunt compound generator has full load output of 100 kW at 250 volts. The armature, series and shunt windings have resistances of 0-05 2,0-03 2 and 55 2 respectively. Find the armature current and generated e.m.f.
Principles of Electrical Machines 64 100 A 4.8 A 0.1 Ω5 g0.04 2 120 V 25 Q 0.06 2 (1) 202 . (iii) Fig. 2.52 "1 When a 0.1 N diverter is connected across the series field of long-shunt generator [See Fig. 2.52 (iii)], then by current divider rule, Rdiv Ry + R. - 0-1 0-1 + 0-04 = 74-86 A Current through series winding = I,x = 104-8 x Since the number of series turns is the same, change in series field ampere-turns is the same as the change of current in the series winding. :. Decrease in series field ampere-turns 104-8- 74-86 x 100 = 28.6% ai gonb gsllev 104-8 Example 2.41. A separately excited d.c: generator has armature circuit resistance of 0.12 and the total brush drop is 2V. When running at 1000 r.p.m., it delivers a current of 100A at 250 V to a load of constant resistance. If the generator speed drops to 700 r.p.m. with field current unaltered, find the current delivered to load. Solution. Load resistance, R, '= 250/100 = 2.5 N nihniw are sgeitov gnbniv muf O galov For the first case, E, = V+I; R, + Brush drop = 250 + 100 x 0.1 + 2 = 262 V For the second case, N' = 700 r.p.m. and generated e.m.f. E', is given by; E, N' N' x E. 700 x 262 = 183.4 volts %3D or E = N N 1000 Let the new load current be . Then terminal voltage of the generator is 2:5 I. 2.5 I = E,-1, R,-Brush drop 2.5 T = 183.4 –I, (0.1) – 2 I = 69.77 A Now or TUTORIAL PROBLEMS a0.0 1. A 4-pole shunt generator with wave-wound armature has 41 slots, each having 12 conductors. The / armature resistance is 0-05 2 and shunt field resistance is 200 S. The flux per pole is 25 mWb. If a load resistance of 10Q is connected across the armature terminals, calculate the voltage across the load when generator is driven at 1000 r.p.m. 2. A long shunt compound generator has full load output of 100 kW at 250 volts. The armature, series and shunt windings have resistances of 0-05 2, 0-03 N and 55 N respectively. Find the armature current and [389-5 V] generated e.m.f. [404-5 A, 282-3 V] ele Load
Transcribed Image Text:Principles of Electrical Machines 64 100 A 4.8 A 0.1 Ω5 g0.04 2 120 V 25 Q 0.06 2 (1) 202 . (iii) Fig. 2.52 "1 When a 0.1 N diverter is connected across the series field of long-shunt generator [See Fig. 2.52 (iii)], then by current divider rule, Rdiv Ry + R. - 0-1 0-1 + 0-04 = 74-86 A Current through series winding = I,x = 104-8 x Since the number of series turns is the same, change in series field ampere-turns is the same as the change of current in the series winding. :. Decrease in series field ampere-turns 104-8- 74-86 x 100 = 28.6% ai gonb gsllev 104-8 Example 2.41. A separately excited d.c: generator has armature circuit resistance of 0.12 and the total brush drop is 2V. When running at 1000 r.p.m., it delivers a current of 100A at 250 V to a load of constant resistance. If the generator speed drops to 700 r.p.m. with field current unaltered, find the current delivered to load. Solution. Load resistance, R, '= 250/100 = 2.5 N nihniw are sgeitov gnbniv muf O galov For the first case, E, = V+I; R, + Brush drop = 250 + 100 x 0.1 + 2 = 262 V For the second case, N' = 700 r.p.m. and generated e.m.f. E', is given by; E, N' N' x E. 700 x 262 = 183.4 volts %3D or E = N N 1000 Let the new load current be . Then terminal voltage of the generator is 2:5 I. 2.5 I = E,-1, R,-Brush drop 2.5 T = 183.4 –I, (0.1) – 2 I = 69.77 A Now or TUTORIAL PROBLEMS a0.0 1. A 4-pole shunt generator with wave-wound armature has 41 slots, each having 12 conductors. The / armature resistance is 0-05 2 and shunt field resistance is 200 S. The flux per pole is 25 mWb. If a load resistance of 10Q is connected across the armature terminals, calculate the voltage across the load when generator is driven at 1000 r.p.m. 2. A long shunt compound generator has full load output of 100 kW at 250 volts. The armature, series and shunt windings have resistances of 0-05 2, 0-03 N and 55 N respectively. Find the armature current and [389-5 V] generated e.m.f. [404-5 A, 282-3 V] ele Load
Expert Solution
steps

Step by step

Solved in 4 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
DC Generator
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,
SEE MORE TEXTBOOKS