Of a shunt excited DA generator; Excitation winding internal resistance R = 2.5 e Armature winding internal resistance Ra = 0.5 N A rheostat has been placed in the excitation circuit to keep the end voltage of the machine constant at Va = 30 V. Machine www Voltage magnetization induced in the armature while operating at n = 1400 rpm at If = 7 A excitation current It is determined as Ea = 30 V from its characteristic. According to this; %3D wwmu www a) End voltage Va = 30 V, excitation current If = 7 A, when the number of revolutions is n = 1400 d/d, What is the resistance of the rheostat with the %3D ww mw power supplied to the load? b) Spindle revolutions are reduced to n = 1000 d/d. In this case Va = 30 V and Ea = 37 V constant to what value should the rheostat be set to remain? Calculate the power given to the load in this new case. www ww m ww w mh m wmw m w

Of a shunt excited DA generator; Excitation winding internal resistance R = 2.5 e Armature winding internal resistance Ra = 0.5 N A rheostat has been placed in the excitation circuit to keep the end voltage of the machine constant at Va = 30 V. Machine www Voltage magnetization induced in the armature while operating at n = 1400 rpm at If = 7 A excitation current It is determined as Ea = 30 V from its characteristic. According to this; %3D wwmu www a) End voltage Va = 30 V, excitation current If = 7 A, when the number of revolutions is n = 1400 d/d, What is the resistance of the rheostat with the %3D ww mw power supplied to the load? b) Spindle revolutions are reduced to n = 1000 d/d. In this case Va = 30 V and Ea = 37 V constant to what value should the rheostat be set to remain? Calculate the power given to the load in this new case. www ww m ww w mh m wmw m w

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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Q23) A shunt generator supplies a 20KW load at 200V through cables of resistance, R = 100mΩ. If the field winding resistance ,Rf=50Ω and the armature resistance, Ra=40mΩ, i) Estimate terminal voltage across the load. ii) Calculate the emf generated in the armature.
Q#1(a): Illustrate the EMF equation of generator. Q#1(b): An 8-pole, lap-wound armature rotated at 350 r.p.m is required to generate 260V. The useful magnetic flux per pole is 0.07Wb. If the armature has 110 slots. Use appropriate equation to calculate the number of conductors per slot.
The flux density distribution in the airgap of a 60Hz,4-pole, salient pole machine is sinusoidal, having an amplitude of 0.8 T. Calculate the instantaneous and rms values of the voltage induced in 1150-turn, If the axial length of the armature and its inner diameter are both 100 mm.
A shunt generator supplies a 20kW load at 200V through cables of resistance, R= 100mΩ. If the field winding resistance, Rf =50Ω and the armature resistance, Ra =40mΩ, determine (a) the terminal voltage, and (b) the e.m.f. generated in the armature
A short shunt compound generator supplies a load of 10 kW at 200 V through a pair of feeder and series field resistance 0.10 ohm. Armature resistance, shunt field resistance and series field resistance are 0.01 ohm, 52.2 ohms and 0.10 ohm, respectively. If the core friction and windage losses amount to 631 watts, the BHP of the driving engine is nearest to
A DC shunt generator supplies a load of 27 kW at 210 V through the cables of resistance, R = 126 mΩ. If the field winding resistance, Rf = 62 Ω and the armature resistance, Ra = 35 mΩ, Determine (a) the terminal voltage, and (b) the E.m.f. generated in the armature. The terminal voltage is = The E.m.f generated in armature is =
. When a self-excited compound DC generator supplies load, the terminal voltage is not the same as the generated emE Why? a) voltage drop across the armature resistance will reduce the voltage available at the load terminals b) where there is a series field, voltage drop across the series field resistance will reduce the voltage available at the load terminals c) both "a" and "b" d) neither "a" nor "b"
A 250 V shunt generator has a full load armature current of 40 A, under these conditions the losses are: FW = 200 watts, shunt field = 100 watts, core = 260 watts, brush contact = 120 watts, armature copper = 400 watts. For operation at maximum efficiency, when the constant losses are equal to those losses that vary as the square of the load, calculate the maximum efficiency.
111/ The armature reaction of the DC generator reduces the generated EMF. Select one: True False
4. A short shunt compound generator supplies a load of 10 kW at 200 V through a pair of feeder resistance of 0.10 ohm. Armature, shunt field resistance and series field resistance are 0.10 ohm, 52.5 ohms and 0.10 ohm, respectively. If the core, friction, and windage losses amount to 631 watts, determine (a) HP of driving engine (b) commercial, electrical and mechanical efficiencies.
The armature winding resistance of a long shunt connected DC generator is 0.4ohm, the resistances of the shunt and serial field windings are 125 ohms and 0.1 ohms, respectively. While the generator supplies 8KW of power to the load it feeds, which is the voltage it produces since the output voltage is 250V?
Q/ dc shunt generator 4 pole with 500 condcutor , wave running at speed 1000 r.p.m the resistance of armature and field are 0.02 , 55 ohm and delivering load 43.12kwatt at terminal voltage 220 volt, the iron losses is 750 watt 1-Torque developed by the prime mover .........and The torque developed by the armature ........... when running at 1000 r.p.m * 2-The load current ,,,,,,,,,,,,when the speed drops to 800 r.p.m. if If is unchanged ? * 3-Output of the prime motor…….kwatt * 4-Mechanical , electrical and overall efficiency …….. * 5-Flux per pole is......... wb *