Q1.a)b)A 200V DC series motor has armature resistance of 0.1 Q and fieldresistance of 0.15 Q. The motor runs at a speed of 600 rev/min when theshaft torque is 28 Nm. Friction and windage losses at this speed are 251.3W. Calculate armature current, copper loss and efficiency.A DC shunt motor has an armature resistance of 0.22, a field resistanceof 200 and is connected to a 200 V supply.i) Draw the equivalent electrical circuit of the motorii) If the motor runs at 1500 rpm and takes a current of 11 A from thesupply, calculate the output torque of the motoriii) If the supply voltage is kept constant but the load torque is changedso that the supply current decreased to 6 A, determine the motorspeed and the output torque.

Step 1:Explanation of part (a): Step 2:Explanation for part (b): (ii) Output Torque:Field Current:…

Answered: Q1. a) b) A 200V DC series motor has…

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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A 10 kW, 250 V Shunt motor has an armature resistance of 0.5 ohm and a field resistance of 200 ohm. At no load and rated voltage, the speed is 1200 rpm and the armature current is 3 A. At full load and rated voltage, the line current is 47 A and because of armature reaction, the flux is 4 % less than it no load value. a.) What is the full load speed? b.) What is the developed torque at full load?
Please question 4
A 240-volt series motor has an armature resistance of 0.42 ohm and a series-field resistance of 0.18 ohm. If the speed is 500 rpm when the current is 36 A, what will be the motor speed when the load reduces the line current to 21 A? (Assume a 3-volt brush drop and that of the flux is proportional to the current.). Ans.: 1435 rpm show diagram and step by step process
QUESTION 2 a) Draw the equivalent circuit and torque-speed characteristic curve for a series DC motor. b) Briefly explain the induced torque occurs in a series DC motor. e) A 20-hp 240-V 76-A 900 rimin series motor has a field winding of 33 turns per pole. Its armature resistance is 0.09 2, and its field resistance is 0.06 2. The magnetization curve expressed in terms of magnetomotive force versus Ex at 900 rimin is given by the following table: Table Q2 c): The magnetomotive force table for a de motor taken at 900r'min E. V 95 150 188 212 229 243 F.A tuus 1000 1500 2000 2500 500 3000 If the motor running at 33 percent of the full-koad armature current, determine: ) The motor's power converted from electrical to mechanical form, Prene i) The motor's torque, Tnit ii) The motor's speed, n
Please question 5 solution
A 250 V series motor has an armature resistance of 0.2 ohm and a series-fieldresistance of 0.18 ohm. If the speed is 6oo RPM when the current is 32 amps.What will be the motor speed when the load is reduced so that the current is18 amps, under which condition the percent flux change is 50 % of the percentcurrent change.
A 220 V dc shunt motor develops a total torque of 100 N-M and takes 29A at 600 rpm. The armature and shunt field resistance are 0.3 ohm and 220 ohms respectively. If the speed is to be increased to 800 rpm determine the percentage reduction of the field and additional resistance to be inserted in the field circuit. The torque develop at 800 rpm is 70 N-M.
show circuit diagram and step by step solution plz
(b) A 2 pole 220 V, 100A, and 850 rpm separately excited DC motor has an armature resistance of 0.15 0. Two field coils are connected in parallel and used to drive a load with the torque of TL = 400 – 0.2N, Nm Where N is the motor speed in rpm. Armature voltage control and two field windings connected in series are used in speed control below and above the rated, respectively. (i) Calculate the motor armature current and speed when the armature voltage is reduced to 110VV. (ii) Examine the motor speed and current when field coils are connected in series.
A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 and 0.1 respectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is
2. A 2-pole, 50 Hz induction motor supplies 15 kW to a load at a speed of 2950 rpm. a. What is the motor's slip? b. What is the induced torque in the motor in N-m under these conditions? c. What will be the operating speed of the motor be if its torque is doubled? d. How much power will be supplied by the motor when the torque is doubled?
200 +21. If maximum torque of an induction motor is kg-m at a slip of 12%, the torque at 6% slip would be...... kg-m. 1 (a) 100 (b) 160 (c) 50 (d) 40 B, 1993)

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