A DC, separately excited motor is used to drive a constant-torque load. The field circuit is excited by a full-wave, AC/DC SCRconverter. The armature circuit of the motor is connected to a constant dc voltage source of 160 V. The inductance of thefield circuit is large, and the field current is continuous. The field AC voltage (input to the converter) is 127 V (rms), and thefield resistance is 110 N. The armature resistance is 2.2 N. When the triggering angle of the SCRS is adjusted to zero, themotor speed is 1300 rpm, and the armature current is 13 A.A1• Calculate the average current and dc power of the field circuit when the triggering angle isequal to 15°.LaRaIf1.04AVaMP¡ =110.89WA2If the motor speed is to be increased to 1700 rpm, what is the triggering angle?(a) Circuita =

Answered: Image /qna-images/answer/c04689f5-8629-48de-a47f-a26ccee7af93.jpg

A DC, separately excited motor is used to drive a constant-torque load. The field circuit is excited by a full-wave, AC/DC SCR converter. The armature circuit of the motor is connected to a constant dc voltage source of 160 V. The inductance of the field circuit is large, and the field current is continuous. The field AC voltage (input to the converter) is 127 V (rms), and the field resistance is 110 2. The armature resistance is 2.2 2. When the triggering angle of the SCRS is adjusted to zero, the motor speed is 1300 rpm, and the armature current is 13 A. • Calculate the average current and dc power of the field circuit when the triggering angle is i, A equal to 15°. La Ra Va I = 1.04 Lf Pf = 110.89 W M F2 • If the motor speed is to be increased to 1700 rpm, what is the triggering angle? A2 (a) Circuit a =

A DC, separately excited motor is used to drive a constant-torque load. The field circuit is excited by a full-wave, AC/DC SCR converter. The armature circuit of the motor is connected to a constant dc voltage source of 160 V. The inductance of the field circuit is large, and the field current is continuous. The field AC voltage (input to the converter) is 127 V (rms), and the field resistance is 110 2. The armature resistance is 2.2 2. When the triggering angle of the SCRS is adjusted to zero, the motor speed is 1300 rpm, and the armature current is 13 A. • Calculate the average current and dc power of the field circuit when the triggering angle is i, A equal to 15°. La Ra Va I = 1.04 Lf Pf = 110.89 W M F2 • If the motor speed is to be increased to 1700 rpm, what is the triggering angle? A2 (a) Circuit a =

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...

See similar textbooks

Similar questions

Colve asap
Solve it fast fast fast plz
what are the disadvantages of slip ring induction motor?
Q4) What is the effect in series motor if you assume V = Eb , with low resistance fielding winding ? a) b) The given information about series motor on full – load (with that the flux is proportional to the field current) is a ) takes ( 78A) and runs at (850 r.p.m) , the series winding resistance is (0.04 2) , with diverter in parallel with the field winding is a (0.18 2) and the armature resistance is (0.09 S2). ( 220 V Neglected the iron, friction and windage losses. Determine the efficiency , speed when developing full load torque ,the maximum efficiency .
Please all subparts
For a 1.8 degree, 2 phase bipolar stepper motor, steeping rate is 100 steps/seconds. The rotational speed of the motor in rpm is
The supply voltage to a DC motor is 160V. If the back emf Ez=155V and the armature resistance A6 is 0.250, the armature current is a) 5A b) 10A c) 15A d) 20A 1800 ro/min The frequency
In order to get higher speed of Dc motor the ..... ....... voltage supply most be low back emf (Eb) most be high O frequency most be constant
A de machine is controlled by the step-down de-de converter shown below. The armature inductance La =0.2 mH and the armature resistance can be neglected. The armature current is 5A. The Buck converter is operating at the switching frequency is 30 kHz and the duty ratio is 0.8. L₁ + Answer the following questions: 1) Calculate the input voltage (Va) if the output voltage (ea = 200 V). 2) Find the ripple in the armature current (a). 3) Calculate the maximum and minimum values of the armature current (la). 4) Sketch the armature current (ia (t)) and the supply current (id (t)).
Reducing the supply voltage so that Vnew = Vold / V3 reduces the maximum torque of the motor so that Tmaxnew= ...Tmaxold . * .... ...