Please question 4 1. A DC motor develops a torque of 30 N-m. Determine the torque when the armaturecurrent is increased by 50% and the flux is reduced by 10%.2. A 10-hp 230-V DC series motor has a line current of 37 A and a rated speed of 1200 rpm.The armature and series field resistances are 0.4 22 and 0.2 2 respectively. The total brushvoltage drop is 2 V. Calculate the following:a. Speed at a line current of 20 A.b. No-load speed when the line current is 1 A.c. Speed at 150% of the rated load when the line current is 60 A and the series field flux is125% of the full load flux.3. A 240-V DC shunt DC motor has an armature resistance of 0.25 2 and a field resistanceof 120 2. At full load the armature draws a current of 40 A and the speed is 1100 rpm.(i)Find the developed torque(ii)The field rheostat is adjusted so that the field resistance is 150 2. Find the newoperating speed if the torque and the armature current remain constant.4. Compute the efficiency of 100 kW, 250-V, 6-pole, 1000 rpm long shunt compoundgenerator. The generator has the following parametersArmature resistance R₁ = 0.015Series field resistance R, = 0.0102Shunt field resistance R₁ = 100.0 2Total rotational losses = 4000 WAssume stray losses to be 1% of the output and a brush drop of 2 volts.5. A 100 V de series motor has an armature resistance of 0.2 2 and series field resistance of0.250. When its pulley exerts a torque of 27.58 N.m it runs at a speed of 600 rpm, ironand friction losses at this speed are 300 W. Determinelost torquecopper losses and(i)(ii)(iii) efficiency.

Rating of the generator 100 kWTerminal voltage Number of poles Speed Long shunt compound generator…

4. Compute the efficiency of 100 kW, 250-V, 6-pole, 1000 rpm long shunt compound generator. The generator has the following parameters Armature resistance R₂ = 0.015 Series field resistance R,=0.0102 Shunt field resistance Rr = 100.0 2 Total rotational losses = 4000 W Assume stray losses to be 1% of the output and a brush drop of 2 volts.

4. Compute the efficiency of 100 kW, 250-V, 6-pole, 1000 rpm long shunt compound generator. The generator has the following parameters Armature resistance R₂ = 0.015 Series field resistance R,=0.0102 Shunt field resistance Rr = 100.0 2 Total rotational losses = 4000 W Assume stray losses to be 1% of the output and a brush drop of 2 volts.

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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1. Calculate the no load speed of a DC motor with specs 12V Vdc KE=0.1 and flux of 2Wb 2. For a series motor, what is the effect with the torque if a. 2x the current, b. 2 times the speed, c. 2 times the voltage?
A 500 V shunt motor has an armature resistance of 0.1 ohm shunt field resistance of 125 ohms. Find the resistance of the shunt field regulator to increase the speed from 750 r.p.m, to 1000 r.p.m if the current taken by the motor is 500 A. Assume the magnetization characteristic to be a straight line. (Ans. 41.7 ohm)
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