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d.c. series motor has a wave wound armature with 180 conductors. The armature resistance and the field winding resistance are 0.10 ohm and0.2 ohm respectively. If the motor is drawing 40 A current at 0.015 Wb flux per pole, then the speed under this condition is O 2533.3 rpm O 2644.4 rpm O 2756.9 rpm O 2987.6 rpm

d.c. series motor has a wave wound armature with 180 conductors. The armature resistance and the field winding resistance are 0.10 ohm and0.2 ohm respectively. If the motor is drawing 40 A current at 0.015 Wb flux per pole, then the speed under this condition is O 2533.3 rpm O 2644.4 rpm O 2756.9 rpm O 2987.6 rpm

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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d.c. series motor has a wave wound armature with 180 conductors. The armature current of 20 A has 360 If Eb is the back emf of d.c. motor and V is armature resistance and the field winding conductors. If the flux per pole is 0.015 Wb the terminal voltage, then the condition for maximum power is a. * resistance are 0.10 ohm ando.2 ohm then the gross torque developed by the respectively. If the motor is drawing 40 A current at 0.015 Wb flux per pole, then the armature of motor is * speed under this condition is Eb = V 15.56 N-m O 2533.3 rpm O 2644.4 rpm Eb = (V/ 2) 10.23 N-m O 2756.9 rpm Eb = 2V O 19.08 N-m O 2987.6 rpm O 17.17 N-m Eb = Va 2 dc machine 6 pole wave winding is d. The number of parallel paths are? * The back emf of a dc motor Adds to the supply voltage 1 regulates its armature voltage 4 helps in energy conversion 2 usually exceeds the supply voltage 20 V dc machine has an armature stance of 1ohm if the full load current is A series motor runs at (800O r.p.m) when the voltage is (40O V) and the current is (25 A). A, the difference of induced voltage, en the machine is running as a motor and erator, is * The armature resistance is (0.4 Q) and the series field resistance is (0.2 Q). Determine 20 the resistance to be connected in series to reduce the speed to (600 r.p.m) with same 40 current 30 3.85 60 3 4.25 2.85
Transcribed Image Text:d.c. series motor has a wave wound armature with 180 conductors. The armature current of 20 A has 360 If Eb is the back emf of d.c. motor and V is armature resistance and the field winding conductors. If the flux per pole is 0.015 Wb the terminal voltage, then the condition for maximum power is a. * resistance are 0.10 ohm ando.2 ohm then the gross torque developed by the respectively. If the motor is drawing 40 A current at 0.015 Wb flux per pole, then the armature of motor is * speed under this condition is Eb = V 15.56 N-m O 2533.3 rpm O 2644.4 rpm Eb = (V/ 2) 10.23 N-m O 2756.9 rpm Eb = 2V O 19.08 N-m O 2987.6 rpm O 17.17 N-m Eb = Va 2 dc machine 6 pole wave winding is d. The number of parallel paths are? * The back emf of a dc motor Adds to the supply voltage 1 regulates its armature voltage 4 helps in energy conversion 2 usually exceeds the supply voltage 20 V dc machine has an armature stance of 1ohm if the full load current is A series motor runs at (800O r.p.m) when the voltage is (40O V) and the current is (25 A). A, the difference of induced voltage, en the machine is running as a motor and erator, is * The armature resistance is (0.4 Q) and the series field resistance is (0.2 Q). Determine 20 the resistance to be connected in series to reduce the speed to (600 r.p.m) with same 40 current 30 3.85 60 3 4.25 2.85
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