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A 50 kW. 240 shunt motor takes 15A when running light at 1400 rpm. The resistance of the armature and field are 0.25 Ω and 100 Ω respectively. If mechanical losses to be negligible, the efficiency (in%) of the motor when taking 150 A is
Correct answer is between '74,76'. Can you explain this answer?
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Verified Answer
A 50 kW. 240 shunt motor takes 15A when running light at 1400 rpm. The...
Shunt resistance = 100 Ω
Shunt Current
Armature current (Ia) = IL - Ish
= 15 – 2.4 = 12.6 A
At no load, Eb = 240 – (12.6 × 0.25) = 236.86 V
No load losses = (236.85) (12.6) + (12.6)2 (0.25) + (240) (2.4) = 3600 W
At load,
Ia = 150 – 2.4 = 147.6 A
P losses = (147.6)2 (0.25) + 3600
= 9046.44 W
Input power = 240 × 150 = 36 kW
Efficiency 
Most Upvoted Answer
A 50 kW. 240 shunt motor takes 15A when running light at 1400 rpm. The...
To find the resistance of the armature and field, we can use the formula:
R = V / I
Where R is the resistance, V is the voltage, and I is the current.
Given that the motor takes 15A when running light, we can use this value to find the resistance.
For the armature resistance:
R_armature = V_armature / I_armature
Since the motor is running light, there is no load on the motor, so the voltage across the armature is the same as the supply voltage.
Given that the motor is a shunt motor, the supply voltage is 240V.
R_armature = 240V / 15A
R_armature = 16 Ω
For the field resistance:
R_field = V_field / I_field
In this case, the voltage across the field is not given. However, we can assume that the voltage across the field is the same as the supply voltage, since the motor is running light.
R_field = 240V / I_field
To find the field current, we can use the power equation:
P = VI
Given that the motor takes 50 kW, we can rearrange the equation to solve for the current:
I_field = P / V_field
I_field = 50,000W / 240V
I_field = 208.33 A
Now we can find the field resistance:
R_field = 240V / 208.33 A
R_field = 1.15 Ω
Therefore, the resistance of the armature is 16 Ω and the resistance of the field is 1.15 Ω.
R = V / I
Where R is the resistance, V is the voltage, and I is the current.
Given that the motor takes 15A when running light, we can use this value to find the resistance.
For the armature resistance:
R_armature = V_armature / I_armature
Since the motor is running light, there is no load on the motor, so the voltage across the armature is the same as the supply voltage.
Given that the motor is a shunt motor, the supply voltage is 240V.
R_armature = 240V / 15A
R_armature = 16 Ω
For the field resistance:
R_field = V_field / I_field
In this case, the voltage across the field is not given. However, we can assume that the voltage across the field is the same as the supply voltage, since the motor is running light.
R_field = 240V / I_field
To find the field current, we can use the power equation:
P = VI
Given that the motor takes 50 kW, we can rearrange the equation to solve for the current:
I_field = P / V_field
I_field = 50,000W / 240V
I_field = 208.33 A
Now we can find the field resistance:
R_field = 240V / 208.33 A
R_field = 1.15 Ω
Therefore, the resistance of the armature is 16 Ω and the resistance of the field is 1.15 Ω.
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A 50 kW. 240 shunt motor takes 15A when running light at 1400 rpm. The resistance of the armature and field are 0.25 Ω and 100 Ω respectively. If mechanical losses to be negligible, the efficiency (in%) of the motor when taking 150 A isCorrect answer is between '74,76'. Can you explain this answer?
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A 50 kW. 240 shunt motor takes 15A when running light at 1400 rpm. The resistance of the armature and field are 0.25 Ω and 100 Ω respectively. If mechanical losses to be negligible, the efficiency (in%) of the motor when taking 150 A isCorrect answer is between '74,76'. 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 50 kW. 240 shunt motor takes 15A when running light at 1400 rpm. The resistance of the armature and field are 0.25 Ω and 100 Ω respectively. If mechanical losses to be negligible, the efficiency (in%) of the motor when taking 150 A isCorrect answer is between '74,76'. 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 50 kW. 240 shunt motor takes 15A when running light at 1400 rpm. The resistance of the armature and field are 0.25 Ω and 100 Ω respectively. If mechanical losses to be negligible, the efficiency (in%) of the motor when taking 150 A isCorrect answer is between '74,76'. Can you explain this answer?.
A 50 kW. 240 shunt motor takes 15A when running light at 1400 rpm. The resistance of the armature and field are 0.25 Ω and 100 Ω respectively. If mechanical losses to be negligible, the efficiency (in%) of the motor when taking 150 A isCorrect answer is between '74,76'. 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 50 kW. 240 shunt motor takes 15A when running light at 1400 rpm. The resistance of the armature and field are 0.25 Ω and 100 Ω respectively. If mechanical losses to be negligible, the efficiency (in%) of the motor when taking 150 A isCorrect answer is between '74,76'. 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 50 kW. 240 shunt motor takes 15A when running light at 1400 rpm. The resistance of the armature and field are 0.25 Ω and 100 Ω respectively. If mechanical losses to be negligible, the efficiency (in%) of the motor when taking 150 A isCorrect answer is between '74,76'. Can you explain this answer?.
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