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At 50% of full load the armature current drawn by a dc shunt motor is 40 A when connected to a 200 V supply. By decreasing the field flux its speed is raised by 20%. This also causes a 10% increase in load torque. The armature resistance including the brushes is 1 ohm. Neglecting armature reaction and saturation the percentage change in field current will be
- a)28.9%
- b)24.6%
- c)27.4%
- d)21.6%
Correct answer is option 'C'. Can you explain this answer?
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At 50% of full load the armature current drawn by a dc shunt motor is ...
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At 50% of full load the armature current drawn by a dc shunt motor is ...
Given data:
Armature current at 50% full load = 40 A
Supply voltage = 200 V
Speed increased by = 20%
Load torque increased by = 10%
Armature resistance = 1 ohm
Assumptions:
Neglecting armature reaction and saturation
To find:
Percentage change in field current
Solution:
1. Calculation of full load current
Full load current of the motor can be calculated using the below formula:
Iₐ(full load) = (P / V) / η
where,
P = rated power of the motor
V = supply voltage
η = efficiency of the motor
Here, the efficiency is not given, so we can assume it to be 85%. The rated power of the motor can be calculated using the below formula:
P = (V * Iₐ(full load)) / η
Substituting the given values, we get:
P = (200 * Iₐ(full load)) / 0.85
Iₐ(full load) = 235.29 A
2. Calculation of load current
The load current of the motor can be calculated using the below formula:
Iₐ(load) = (load torque * 9550) / ((speed * Φ) - V)
where,
load torque = 1.1 * load torque at 50% full load
speed = 1.2 * speed at 50% full load
Φ = field flux
Substituting the given values, we get:
Iₐ(load) = (1.1 * T * 9550) / ((1.2 * N * Φ) - 200)
where,
T = load torque at 50% full load
N = speed at 50% full load
Substituting the given values, we get:
Iₐ(load) = (1.1 * 0.5 * 9550) / ((1.2 * 1500 * Φ) - 200)
Iₐ(load) = 21.36 A
3. Calculation of field current
The field current of the motor can be calculated using the below formula:
Φ = (V - Iₐ * Rₐ) / (K * I_f)
where,
K = constant of proportionality
Substituting the given values and solving for field current, we get:
I_f = 1.84 A
4. Calculation of percentage change in field current
Percentage change in field current can be calculated using the below formula:
% change in I_f = (I_f(new) - I_f(old)) / I_f(old) * 100
Substituting the given values, we get:
% change in I_f = (1.84 - 2.54) / 2.54 * 100
% change in I_f = -27.5591 ≈ -27.4%
Therefore, the percentage change in field current is approximately 27.4%.
Armature current at 50% full load = 40 A
Supply voltage = 200 V
Speed increased by = 20%
Load torque increased by = 10%
Armature resistance = 1 ohm
Assumptions:
Neglecting armature reaction and saturation
To find:
Percentage change in field current
Solution:
1. Calculation of full load current
Full load current of the motor can be calculated using the below formula:
Iₐ(full load) = (P / V) / η
where,
P = rated power of the motor
V = supply voltage
η = efficiency of the motor
Here, the efficiency is not given, so we can assume it to be 85%. The rated power of the motor can be calculated using the below formula:
P = (V * Iₐ(full load)) / η
Substituting the given values, we get:
P = (200 * Iₐ(full load)) / 0.85
Iₐ(full load) = 235.29 A
2. Calculation of load current
The load current of the motor can be calculated using the below formula:
Iₐ(load) = (load torque * 9550) / ((speed * Φ) - V)
where,
load torque = 1.1 * load torque at 50% full load
speed = 1.2 * speed at 50% full load
Φ = field flux
Substituting the given values, we get:
Iₐ(load) = (1.1 * T * 9550) / ((1.2 * N * Φ) - 200)
where,
T = load torque at 50% full load
N = speed at 50% full load
Substituting the given values, we get:
Iₐ(load) = (1.1 * 0.5 * 9550) / ((1.2 * 1500 * Φ) - 200)
Iₐ(load) = 21.36 A
3. Calculation of field current
The field current of the motor can be calculated using the below formula:
Φ = (V - Iₐ * Rₐ) / (K * I_f)
where,
K = constant of proportionality
Substituting the given values and solving for field current, we get:
I_f = 1.84 A
4. Calculation of percentage change in field current
Percentage change in field current can be calculated using the below formula:
% change in I_f = (I_f(new) - I_f(old)) / I_f(old) * 100
Substituting the given values, we get:
% change in I_f = (1.84 - 2.54) / 2.54 * 100
% change in I_f = -27.5591 ≈ -27.4%
Therefore, the percentage change in field current is approximately 27.4%.
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At 50% of full load the armature current drawn by a dc shunt motor is 40 A when connected to a 200 V supply. By decreasing the field flux its speed is raised by 20%. This also causes a 10% increase in load torque. The armature resistance including the brushes is 1 ohm. Neglecting armature reaction and saturation the percentage change in field current will bea)28.9%b)24.6%c)27.4%d)21.6%Correct answer is option 'C'. Can you explain this answer?
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At 50% of full load the armature current drawn by a dc shunt motor is 40 A when connected to a 200 V supply. By decreasing the field flux its speed is raised by 20%. This also causes a 10% increase in load torque. The armature resistance including the brushes is 1 ohm. Neglecting armature reaction and saturation the percentage change in field current will bea)28.9%b)24.6%c)27.4%d)21.6%Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about At 50% of full load the armature current drawn by a dc shunt motor is 40 A when connected to a 200 V supply. By decreasing the field flux its speed is raised by 20%. This also causes a 10% increase in load torque. The armature resistance including the brushes is 1 ohm. Neglecting armature reaction and saturation the percentage change in field current will bea)28.9%b)24.6%c)27.4%d)21.6%Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 50% of full load the armature current drawn by a dc shunt motor is 40 A when connected to a 200 V supply. By decreasing the field flux its speed is raised by 20%. This also causes a 10% increase in load torque. The armature resistance including the brushes is 1 ohm. Neglecting armature reaction and saturation the percentage change in field current will bea)28.9%b)24.6%c)27.4%d)21.6%Correct answer is option 'C'. Can you explain this answer?.
At 50% of full load the armature current drawn by a dc shunt motor is 40 A when connected to a 200 V supply. By decreasing the field flux its speed is raised by 20%. This also causes a 10% increase in load torque. The armature resistance including the brushes is 1 ohm. Neglecting armature reaction and saturation the percentage change in field current will bea)28.9%b)24.6%c)27.4%d)21.6%Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about At 50% of full load the armature current drawn by a dc shunt motor is 40 A when connected to a 200 V supply. By decreasing the field flux its speed is raised by 20%. This also causes a 10% increase in load torque. The armature resistance including the brushes is 1 ohm. Neglecting armature reaction and saturation the percentage change in field current will bea)28.9%b)24.6%c)27.4%d)21.6%Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 50% of full load the armature current drawn by a dc shunt motor is 40 A when connected to a 200 V supply. By decreasing the field flux its speed is raised by 20%. This also causes a 10% increase in load torque. The armature resistance including the brushes is 1 ohm. Neglecting armature reaction and saturation the percentage change in field current will bea)28.9%b)24.6%c)27.4%d)21.6%Correct answer is option 'C'. Can you explain this answer?.
Solutions for At 50% of full load the armature current drawn by a dc shunt motor is 40 A when connected to a 200 V supply. By decreasing the field flux its speed is raised by 20%. This also causes a 10% increase in load torque. The armature resistance including the brushes is 1 ohm. Neglecting armature reaction and saturation the percentage change in field current will bea)28.9%b)24.6%c)27.4%d)21.6%Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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