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1000 or Eg2 135 82 = 0.8 x = 90 V 1200 Example 8.20(a). A4-pole, d.c. shunt generator with a shunt field resistance of 100 $2 and an armature resistance of 1 92 has 378 wave-connected conductors in its armature. The flux per pole is 0.02 Wb. Ifa load resistance of 10 2 is connected across the armature terminals and the generator is driven at 1000 r.p.m., calculate the power absorbed by the load. (Elect. Technology, Hyderabad Univ. 1991) Solution. Induced e.m.f. in the generator is E = OZN 60 ~ (²) volt 0.02 × 378 × 1000 (4) = 252 volt 1Ω 60 Now, let Vbe the terminal voltage i.e. the voltage available across the load as well as the shunt resistance (Fig. 8.59). Load current = V10 A and Shunt current = V100 A V V 11 V + 100 = Armature current= 2 Fig. 8.59 10 1,00 Now, V=E-armature dron ? 11 V V = 252 -1x :: V = 227 volt 100 Load current = 227/10 = 22.7 A, Power absorbed by the load is = 227 x 22.7 = 5,153 W Example 8.20(b). A four-pole, lap-wound shunt generator has 300 armature-conductors and a flux/pole of 0.1 Wb. It runs at 1000 r.p.m. The armature and field-resistances are 0.2 ohm and 125 Calculate the terminal voltage when it is loaded to take a load current of 90 A Nagpur University, April 1999) elle 100 Ω www VOI

1000 or Eg2 135 82 = 0.8 x = 90 V 1200 Example 8.20(a). A4-pole, d.c. shunt generator with a shunt field resistance of 100 $2 and an armature resistance of 1 92 has 378 wave-connected conductors in its armature. The flux per pole is 0.02 Wb. Ifa load resistance of 10 2 is connected across the armature terminals and the generator is driven at 1000 r.p.m., calculate the power absorbed by the load. (Elect. Technology, Hyderabad Univ. 1991) Solution. Induced e.m.f. in the generator is E = OZN 60 ~ (²) volt 0.02 × 378 × 1000 (4) = 252 volt 1Ω 60 Now, let Vbe the terminal voltage i.e. the voltage available across the load as well as the shunt resistance (Fig. 8.59). Load current = V10 A and Shunt current = V100 A V V 11 V + 100 = Armature current= 2 Fig. 8.59 10 1,00 Now, V=E-armature dron ? 11 V V = 252 -1x :: V = 227 volt 100 Load current = 227/10 = 22.7 A, Power absorbed by the load is = 227 x 22.7 = 5,153 W Example 8.20(b). A four-pole, lap-wound shunt generator has 300 armature-conductors and a flux/pole of 0.1 Wb. It runs at 1000 r.p.m. The armature and field-resistances are 0.2 ohm and 125 Calculate the terminal voltage when it is loaded to take a load current of 90 A Nagpur University, April 1999) elle 100 Ω www VOI

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...
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Egi ΦΜ Eg2 or = 0.8 x 1000 1200 = 90 V 135 Example 8.20(a). A4-pole, d.c. shunt generator with a shunt field resistance of 100 $2 and an armature resistance of 1 9 has 378 wave-connected conductors in its armature. The flux per pole is 0.02 Wb. If a load resistance of 10 2 is connected across the armature terminals and the generator is driven at 1000 r.p.m., calculate the power absorbed by the load. (Elect. Technology, Hyderabad Univ. 1991) Solution. Induced e.m.f. in the generator is E = OZN 60 (²) volt 0.02 × 378 x 1000 (²) = 252 volt 192 60 Now, let Vbe the terminal voltage i.e. the voltage available across the load as well as the shunt resistance (Fig. 8.59). Load current = V10 A and Shunt current = V100 A 11 V = V V + 10 100 100 2.7 - Fig. 8.59 Armature current Now, V= E-armature dron 11 V V = 252-1 x :: V = 227 volt 100 Load current = 227/10 = 22.7 A, Power absorbed by the load is = 227 x 22.7-5,153 W Example 8.20(b). A four-pole, lap-wound shunt generator has 300 armature-conductors and a flux/pole of 0.1 Wb. It runs at 1000 r.p.m. The armature and field-resistances are 0.2 ohm and 125 inly Calculate the terminal voltage when it is loaded to take a load current of 90 A Nogpur University, April 1999) elle 100 92 www VOI
Transcribed Image Text:Egi ΦΜ Eg2 or = 0.8 x 1000 1200 = 90 V 135 Example 8.20(a). A4-pole, d.c. shunt generator with a shunt field resistance of 100 $2 and an armature resistance of 1 9 has 378 wave-connected conductors in its armature. The flux per pole is 0.02 Wb. If a load resistance of 10 2 is connected across the armature terminals and the generator is driven at 1000 r.p.m., calculate the power absorbed by the load. (Elect. Technology, Hyderabad Univ. 1991) Solution. Induced e.m.f. in the generator is E = OZN 60 (²) volt 0.02 × 378 x 1000 (²) = 252 volt 192 60 Now, let Vbe the terminal voltage i.e. the voltage available across the load as well as the shunt resistance (Fig. 8.59). Load current = V10 A and Shunt current = V100 A 11 V = V V + 10 100 100 2.7 - Fig. 8.59 Armature current Now, V= E-armature dron 11 V V = 252-1 x :: V = 227 volt 100 Load current = 227/10 = 22.7 A, Power absorbed by the load is = 227 x 22.7-5,153 W Example 8.20(b). A four-pole, lap-wound shunt generator has 300 armature-conductors and a flux/pole of 0.1 Wb. It runs at 1000 r.p.m. The armature and field-resistances are 0.2 ohm and 125 inly Calculate the terminal voltage when it is loaded to take a load current of 90 A Nogpur University, April 1999) elle 100 92 www VOI
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