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A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra= 0.02Ω.When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply is
- a)34.2 A
- b)30 A
- c)22 A
- d)4.84 A
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature re...
By solving above equation
We get Eb1 = 219.39 and Eb2 = 0.61
Most Upvoted Answer
A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature re...
Given:
- Voltage (V) = 220 V
- Power (P) = 10 kW
- Speed (N) = 900 rpm
- Armature Resistance (Ra) = 0.02 ohm
- Electromagnetic Torque (Te) = 70 Nm
To calculate: Current drawn by the motor from the 220 V supply
Formula used:
- Power (P) = Voltage (V) × Current (I)
- Electromagnetic Torque (Te) = (Ia × Φ × P)/2πN
where Ia is the armature current, Φ is the flux per pole and is constant, and 2πN is the mechanical power developed by the motor
Solution:
1. Calculate the armature current (Ia) using the electromagnetic torque formula:
Te = (Ia × Φ × P)/2πN
70 = (Ia × Φ × 10,000)/2π × 900
Ia = (70 × 2π × 900)/(Φ × 10,000)
Ia = 3.95(Φ/Weber)
2. Calculate the flux per pole (Φ):
From the given data, we cannot calculate Φ directly. However, we can assume that the motor is well designed and operates at the rated flux density. Therefore, we can use the following formula:
V = Φ × N × Z × A
where V is the voltage applied to the armature, N is the speed in rpm, Z is the number of armature conductors, and A is the number of parallel paths in the armature winding.
Assuming a typical design with 2 parallel paths in the armature winding and 500 conductors per path, we get:
220 = Φ × 900 × 500 × 2
Φ = 0.000489 Weber
3. Substitute the value of Φ in the armature current formula:
Ia = 3.95(Φ/Weber)
Ia = 3.95(0.000489/Weber)
Ia = 1.94 A
4. Add the voltage drop due to armature resistance:
The current drawn by the motor from the supply is equal to the armature current plus the current lost due to armature resistance:
I = Ia + (V/Ra)
I = 1.94 + (220/0.02)
I = 30 A (approx.)
Therefore, the current drawn by the motor from the 220 V supply is 30 A (approx.), which is closest to option B (30 A).
- Voltage (V) = 220 V
- Power (P) = 10 kW
- Speed (N) = 900 rpm
- Armature Resistance (Ra) = 0.02 ohm
- Electromagnetic Torque (Te) = 70 Nm
To calculate: Current drawn by the motor from the 220 V supply
Formula used:
- Power (P) = Voltage (V) × Current (I)
- Electromagnetic Torque (Te) = (Ia × Φ × P)/2πN
where Ia is the armature current, Φ is the flux per pole and is constant, and 2πN is the mechanical power developed by the motor
Solution:
1. Calculate the armature current (Ia) using the electromagnetic torque formula:
Te = (Ia × Φ × P)/2πN
70 = (Ia × Φ × 10,000)/2π × 900
Ia = (70 × 2π × 900)/(Φ × 10,000)
Ia = 3.95(Φ/Weber)
2. Calculate the flux per pole (Φ):
From the given data, we cannot calculate Φ directly. However, we can assume that the motor is well designed and operates at the rated flux density. Therefore, we can use the following formula:
V = Φ × N × Z × A
where V is the voltage applied to the armature, N is the speed in rpm, Z is the number of armature conductors, and A is the number of parallel paths in the armature winding.
Assuming a typical design with 2 parallel paths in the armature winding and 500 conductors per path, we get:
220 = Φ × 900 × 500 × 2
Φ = 0.000489 Weber
3. Substitute the value of Φ in the armature current formula:
Ia = 3.95(Φ/Weber)
Ia = 3.95(0.000489/Weber)
Ia = 1.94 A
4. Add the voltage drop due to armature resistance:
The current drawn by the motor from the supply is equal to the armature current plus the current lost due to armature resistance:
I = Ia + (V/Ra)
I = 1.94 + (220/0.02)
I = 30 A (approx.)
Therefore, the current drawn by the motor from the 220 V supply is 30 A (approx.), which is closest to option B (30 A).
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A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra= 0.02Ω.When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply isa)34.2 Ab)30 Ac)22 Ad)4.84 ACorrect answer is option 'B'. Can you explain this answer?
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A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra= 0.02Ω.When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply isa)34.2 Ab)30 Ac)22 Ad)4.84 ACorrect answer is option 'B'. 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 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra= 0.02Ω.When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply isa)34.2 Ab)30 Ac)22 Ad)4.84 ACorrect answer is option 'B'. 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 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra= 0.02Ω.When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply isa)34.2 Ab)30 Ac)22 Ad)4.84 ACorrect answer is option 'B'. Can you explain this answer?.
A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra= 0.02Ω.When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply isa)34.2 Ab)30 Ac)22 Ad)4.84 ACorrect answer is option 'B'. 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 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra= 0.02Ω.When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply isa)34.2 Ab)30 Ac)22 Ad)4.84 ACorrect answer is option 'B'. 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 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra= 0.02Ω.When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply isa)34.2 Ab)30 Ac)22 Ad)4.84 ACorrect answer is option 'B'. Can you explain this answer?.
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A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra= 0.02Ω.When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply isa)34.2 Ab)30 Ac)22 Ad)4.84 ACorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra= 0.02Ω.When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply isa)34.2 Ab)30 Ac)22 Ad)4.84 ACorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra= 0.02Ω.When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply isa)34.2 Ab)30 Ac)22 Ad)4.84 ACorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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