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A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ___ . (round off to nearest integer).
Correct answer is '483'. Can you explain this answer?
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A 280 V, separately excited DC motor with armature resistance of 1 &Om...
Ohm and field resistance of 50 Ohm is connected to a 250 V supply. The motor has a rated power of 10 kW and operates at a rated speed of 1500 rpm. Determine the rated current, rated field current, and rated torque of the motor.
To determine the rated current, we can use the formula:
I = P / V
Where I is the current, P is the power, and V is the voltage. Plugging in the given values:
I = 10,000 W / 250 V = 40 A
So, the rated current of the motor is 40 A.
The rated field current can be calculated using the formula:
If = Vf / Rf
Where If is the field current, Vf is the field voltage, and Rf is the field resistance. Plugging in the given values:
If = 250 V / 50 Ohm = 5 A
So, the rated field current of the motor is 5 A.
The rated torque can be calculated using the formula:
T = (P * 60) / (2 * pi * N)
Where T is the torque, P is the power, N is the speed in revolutions per minute (rpm), and pi is a constant. Plugging in the given values:
T = (10,000 W * 60) / (2 * 3.14 * 1500 rpm) = 38.28 Nm
So, the rated torque of the motor is 38.28 Nm.
To determine the rated current, we can use the formula:
I = P / V
Where I is the current, P is the power, and V is the voltage. Plugging in the given values:
I = 10,000 W / 250 V = 40 A
So, the rated current of the motor is 40 A.
The rated field current can be calculated using the formula:
If = Vf / Rf
Where If is the field current, Vf is the field voltage, and Rf is the field resistance. Plugging in the given values:
If = 250 V / 50 Ohm = 5 A
So, the rated field current of the motor is 5 A.
The rated torque can be calculated using the formula:
T = (P * 60) / (2 * pi * N)
Where T is the torque, P is the power, N is the speed in revolutions per minute (rpm), and pi is a constant. Plugging in the given values:
T = (10,000 W * 60) / (2 * 3.14 * 1500 rpm) = 38.28 Nm
So, the rated torque of the motor is 38.28 Nm.
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Community Answer
A 280 V, separately excited DC motor with armature resistance of 1 &Om...
Given: V = 280, Sep. excited motor, Ra = 1 Ω, φ : Constant, Ta ∝ Ia = 30 A, N = 1000
rpm, Rext = 10 Ω
T ∝ φIa
∴ Flux constant, T ∝ Ia
T2/T1 = Ia2/Ia1
Given: T ∝ N
∴ T2/T1 = N2/N1
∴ Ia2/Ia1 = N2/N1
Let, Ia1 = 30 A, N1 = 1000 rpm
250N2 = 280 x 1000 - 330 N2
N2 = 482.76 rpm
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A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ___ . (round off to nearest integer).Correct answer is '483'. Can you explain this answer?
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A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ___ . (round off to nearest integer).Correct answer is '483'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ___ . (round off to nearest integer).Correct answer is '483'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ___ . (round off to nearest integer).Correct answer is '483'. Can you explain this answer?.
A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ___ . (round off to nearest integer).Correct answer is '483'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ___ . (round off to nearest integer).Correct answer is '483'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ___ . (round off to nearest integer).Correct answer is '483'. Can you explain this answer?.
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A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ___ . (round off to nearest integer).Correct answer is '483'. Can you explain this answer?, a detailed solution for A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ___ . (round off to nearest integer).Correct answer is '483'. Can you explain this answer? has been provided alongside types of A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ___ . (round off to nearest integer).Correct answer is '483'. Can you explain this answer? theory, EduRev gives you an
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