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220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.?
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220 V shunt motor has armature and field resistances of 0.2 Ω and 220 ...
Solution:
Given data:
- Voltage supply (V) = 220 V
- Armature resistance (Ra) = 0.2 Ω
- Field resistance (Rf) = 220 Ω
- Load torque (TL) ∝ n2
- Initial speed (n1) = 1000 rpm
- Initial current (I1) = 10 A
- External armature resistance (Re) = 5 Ω
Calculations:
1. Calculate the initial armature current (Ia1) using Ohm's Law:
Ia1 = V / (Ra + Re) = 220 / (0.2 + 5) = 37.61 A
2. Calculate the initial torque (T1) using the load torque equation:
T1 = k * TL = k * n1^2
Since TL ∝ n2, we can write TL = k * n^2. Therefore, T1 = k * n1^2
3. Calculate the initial value of k using the given data:
T1 = k * n1^2
10 = k * 1000^2
k = 10 / 1000000 = 0.00001
4. Calculate the initial value of back emf (Eb1) using the motor equation:
Eb1 = V - I1 * Ra - Ia1 * Re
Eb1 = 220 - 10 * 0.2 - 37.61 * 5 = 10.95 V
5. Calculate the initial value of the armature current (Ia1) using the motor equation:
Eb1 = k * n1 * Φ
where Φ is the magnetic flux in the motor.
Φ = Eb1 / (k * n1)
Φ = 10.95 / (0.00001 * 1000) = 1095000 Wb
Ia1 = (V - Eb1) / (Ra + Re)
Ia1 = (220 - 10.95) / (0.2 + 5) = 37.61 A
6. Calculate the new value of back emf (Eb2) and armature current (Ia2) when the external armature resistance (Re) is added:
Eb2 = k * n2 * Φ
Since the torque is proportional to the square of the speed, we can write:
(TL2 / TL1) = (n2 / n1)^2
n2 = √(TL2 / TL1) * n1
We are not given any value of TL2, so let's assume that it remains the same as TL1.
n2 = √(1) * 1000 = 1000 rpm
Eb2 = k * n2 * Φ
Eb2 = 0.00001 * 1000 * 1095000 = 109.5 V
Ia2 = (V - Eb2) / (Ra + Re)
Ia2 = (220 - 109.5) / (0.2 + 5) = 24.25 A
7. Verify that the new torque (T2) is the same as the initial torque (T1):
T2 = k * n2^2
T2 = 0.00001 *
Given data:
- Voltage supply (V) = 220 V
- Armature resistance (Ra) = 0.2 Ω
- Field resistance (Rf) = 220 Ω
- Load torque (TL) ∝ n2
- Initial speed (n1) = 1000 rpm
- Initial current (I1) = 10 A
- External armature resistance (Re) = 5 Ω
Calculations:
1. Calculate the initial armature current (Ia1) using Ohm's Law:
Ia1 = V / (Ra + Re) = 220 / (0.2 + 5) = 37.61 A
2. Calculate the initial torque (T1) using the load torque equation:
T1 = k * TL = k * n1^2
Since TL ∝ n2, we can write TL = k * n^2. Therefore, T1 = k * n1^2
3. Calculate the initial value of k using the given data:
T1 = k * n1^2
10 = k * 1000^2
k = 10 / 1000000 = 0.00001
4. Calculate the initial value of back emf (Eb1) using the motor equation:
Eb1 = V - I1 * Ra - Ia1 * Re
Eb1 = 220 - 10 * 0.2 - 37.61 * 5 = 10.95 V
5. Calculate the initial value of the armature current (Ia1) using the motor equation:
Eb1 = k * n1 * Φ
where Φ is the magnetic flux in the motor.
Φ = Eb1 / (k * n1)
Φ = 10.95 / (0.00001 * 1000) = 1095000 Wb
Ia1 = (V - Eb1) / (Ra + Re)
Ia1 = (220 - 10.95) / (0.2 + 5) = 37.61 A
6. Calculate the new value of back emf (Eb2) and armature current (Ia2) when the external armature resistance (Re) is added:
Eb2 = k * n2 * Φ
Since the torque is proportional to the square of the speed, we can write:
(TL2 / TL1) = (n2 / n1)^2
n2 = √(TL2 / TL1) * n1
We are not given any value of TL2, so let's assume that it remains the same as TL1.
n2 = √(1) * 1000 = 1000 rpm
Eb2 = k * n2 * Φ
Eb2 = 0.00001 * 1000 * 1095000 = 109.5 V
Ia2 = (V - Eb2) / (Ra + Re)
Ia2 = (220 - 109.5) / (0.2 + 5) = 24.25 A
7. Verify that the new torque (T2) is the same as the initial torque (T1):
T2 = k * n2^2
T2 = 0.00001 *
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220 V shunt motor has armature and field resistances of 0.2 Ω and 220 ...
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220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.?
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220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about 220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.?.
220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about 220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.?.
Solutions for 220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.? in English & in Hindi are available as part of our courses for GATE.
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220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.?, a detailed solution for 220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.? has been provided alongside types of 220 V shunt motor has armature and field resistances of 0.2 Ω and 220 Ω respectively. The motor is driving load torque, TL ∝ n2 and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation.? theory, EduRev gives you an
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