Section 5: Choose the correct AnswerWhat happens if field winding of the synchronous motor is short* circuitedFirst, starts as induction motor then run as synchronous motorNot startMotor will burn outRun as induction motori da - When the field of a synchronous motor is over excited, the power factorwill beLeading OLaggingZeroUnityThe synchronous motors are not self-starting becauseThe direction of torque on the rotor reverses after every half cycle. OSlip is not present in synchronous machineStarting winding is not present in the synchronous machineDC excitation is used

Answered: Image /qna-images/answer/3ebc4035-3348-4cb4-8952-411996db8e0b.jpg

Section 5: Choose the correct Answer لیس ان What happens if field winding of the synchronous motor is short * circuited First, starts as induction motor then run as synchronous motor Not start Motor will burn out Run as induction motor e- When the field of a synchronous motor is over excited, the power factor will be Leading Lagging O Zero Unity ليس هد The synchronous motors are not self-starting because

Section 5: Choose the correct Answer لیس ان What happens if field winding of the synchronous motor is short * circuited First, starts as induction motor then run as synchronous motor Not start Motor will burn out Run as induction motor e- When the field of a synchronous motor is over excited, the power factor will be Leading Lagging O Zero Unity ليس هد The synchronous motors are not self-starting because

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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There are many methods used for starting induction motors and synchronous machines. For the following three methods of starting three-phase motors, explain how the starting method works and state the type of motor the starter is designed for. Support your answer with illustrations: Star-Delta Rotor Resistance Induction starting (with damper windings).
The 3-phase 3730W, 220V, 2-pole and 50 Hz induction motor has a slip of 0.025 at full load. The future of friction (friction)+reducing losses and other (stray load) losses is 5%. Find the induced (electromagnetic) moment at full load. Answer guide group 12.7863 Nm 84.6395 Nm 70.7351 Nm 25.5725 Nm 40.2582 Nm 38.3588Nm 51,1451 Nm 63.3692Nm
DEGREE: BS ELECTRICAL ENGINEERING COURSE: AC MACHINES Note: Please solve in this way. Given>Missing>Illustration>Solution>FinalAnswer. Please include symbols, units, formulas, step by step solution, please don't cut the solution. Thank you! (#13) The name-plate speed of a 25-cycle induction motor is 720 RPM. If the speed at no load is 745 RPM, calculate: (A) the slip, and (B) percent regulation. [ Answer: 4% and 3.4/%, respectively ]
DEGREE: BS ELECTRICAL ENGINEERING COURSE: AC MACHINES Note: Please solve in this way. Given>Missing>Illustration>Solution>FinalAnswer. Please include symbols, units, formulas, step by step solution, please don't cut the solution. Thank you! (#12) A three-phase,6-pole,220-volt,60-cycle wound-rotor induction motor Is tested with the rotor circuit open. With full voltage applied to the stator, and the rotor at standstill, the voltage between slip rings is 150 volts. Determine rotor voltage and frequency at (A) Synchronous speed, and (B) 1250 RPM. [Answer: 0,0,6.25V,2.5 Hz]
Please explain your answers step by step how did you solve or why u needed those formulas etc. Thanks in advance. You know better than me but I'm mentioning just for clarification this question is on the electromechanical energy conversion. The no-load and locked-rotor test results of a three-phase 10 HP, 220 V, 60 Hz, 6 poles star connected induction motor are; The no-load test: VNL=220 V, PNL=340 W, INL=6.2 A The locked-rotor test: VSC=54 V, PSC =430 W, ISC =15.2 A The voltage across the stator windings is 2 V when nominal DC current flows through it. Coefficient for DC resistance is 1,25. a) Calculate the equivalent circuit parameters and draw the complete single-phase equivalent circuit of the induction motor. b) Calculate efficiency of the motor for s=0.06 by using the approximate equivalent circuit (RFe core resistance is not ignored). c) Shaft torque for s=0,06d) Slip value at the maximum (breakdown) torque, e) Maximum (breakdown) torque of the induction motor. f) Speed of…
3-phase 3730W, 220V, 6-pole and 50 Hz induction motor operates at full load with a slip of 0.025. The sum of the rotational losses including friction + windage losses and other (stray load) losses is 5% of the output power. Find the induced (electromagnetic) moment at full load
• T or F, modify the wrong part if False, please underline twice your answer if True or False. A cumulatively-compound DC machine has its flus for the shunt and series field winding flow in opposite directions.
(b) A 415V, 50Hz, delta-connected, 4-pole, squirrel cage induction motor has a perphase equivalent circuit as shown in Figure Q4(b). The circuit parameter values (in Ω/phase) for the stator is as follows: R1 = 0.095 Ω ; R2 = 0.06 Ω; X1 = 0.68 Ω ; X2 = 0.67 Ω; Xm = 18 Ω The motor is running at 1400 rpm at full-load. The core loss of the motor is assumed to be zero. The total friction and windage losses is 600 W, independent of load. Determine the efficiency of the motor.
Evaluate the suitability for using as a belt conveyor drive of a three‐phase, wound rotor induction motor by estimating the motor breakdown and start‐up torque. The motor particulars as per manufacturer’s specifications are: 380‐V, 6‐pole, star connected, 50‐Hz. Machine’s parameters on per‐phase basis, referred to the stator are:R1 = 0.12 Ω, R2 = 0.0.3 Ω, X1 = 0.6 Ω, X2 = 0.15 Ω, XM = 42.2 Ω, Rc is neglected
The armature of a 400V, 6-pole, 25 Hz single-phase series motor is lap-wounded with 504 conductors Motor field winding has effective resistance and leakage reactionance are 0.06 22 and 0.05 12, respectively and has 18 turns. However, the armature winding has a leakage reactance of 0.04 12 and negligible resistance. At rated motor current of 300A, the flux per pole is 0.05 Wb, the short circuit current is 50A and the equivalent current to iron losses is 40A. Find motor speed , torque and efficiency at rated current .
Here it is the equivalent circuit of and induction motor and it is per phase circuit my question is name for me the variable for example R1 is the winding resistance and so an the all variable and please show me a pic of the real machine with the variable for example for R1 happens because of the winding i need to see the winding in the real machine, please help me to understand well
6pole, 3phase, 50HZ induction motor has arotor resistance of 0.02ohm/phase and stand still leakage reactance is 0.1 ohm/phase determine the speed to give max torque, what exrernal resistance per phase is necessary in the rotor circuit to give half maximun torque at starting?