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A 10 kW, 400 V, 3-phase induction motor draw a current of 5 times its full-load current at rated voltage and at standstill developing a starting toque of 1.5 times its full-load torque. An autotransformer is installed to reduce the starting current and to give full-load torque at starting. Neglecting exciting current of auto-transformer, the voltage applied to the motor terminals at the time of starting with auto transformer is
- a)220 volts
- b)241.8 volts
- c)326.6 volts
- d)215.9 volts
Correct answer is option 'C'. Can you explain this answer?
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A 10 kW, 400 V, 3-phase induction motor draw a current of 5 times its ...
Given data:
- Power rating of the motor (P) = 10 kW
- Rated voltage (V) = 400 V
- Full-load current (Ifl)
- Starting current (Ist) = 5 * Ifl
- Full-load torque (Tfl)
- Starting torque (Tst) = 1.5 * Tfl
To reduce the starting current and provide full-load torque at starting, an autotransformer is used.
Working Principle of Autotransformer:
An autotransformer is a type of transformer where a part of the winding is common to both the primary and secondary windings. It allows voltage transformation and current reduction.
Step 1: Finding Full-load Current (Ifl)
The full-load current can be calculated using the formula:
P = √3 * V * Ifl * cos(ϕ)
where ϕ is the power factor.
Since the power factor is not given, we'll assume it to be 0.8 (typical value for an induction motor).
So, Ifl = P / (√3 * V * cos(ϕ))
Ifl = 10,000 / (√3 * 400 * 0.8)
Ifl ≈ 14.56 A
Step 2: Finding Starting Current (Ist)
Since Ist = 5 * Ifl
Ist = 5 * 14.56
Ist ≈ 72.8 A
Step 3: Finding Full-load Torque (Tfl)
The full-load torque can be calculated using the formula:
Tfl = P / (2 * π * n)
where n is the synchronous speed of the motor.
Since the synchronous speed is not given, we'll assume it to be 1800 RPM (typical value for a 3-phase induction motor).
So, Tfl = 10,000 / (2 * π * (1800/60))
Tfl ≈ 8.86 Nm
Step 4: Finding Starting Torque (Tst)
Since Tst = 1.5 * Tfl
Tst = 1.5 * 8.86
Tst ≈ 13.29 Nm
Step 5: Finding the Voltage Applied at Starting with Autotransformer
The autotransformer is used to reduce the starting current and provide full-load torque at starting. Assuming the autotransformer reduces the starting current to the full-load current (Ifl) and provides the full-load torque (Tfl) at starting, we can use the torque equation to find the applied voltage.
T = (3 * V^2 * R2) / (ωs * (R1 + R2)^2)
where T is the torque, V is the voltage applied, R1 is the stator resistance, R2 is the rotor resistance, and ωs is the synchronous speed.
Since the motor is at standstill, we can assume the rotor resistance (R2) is zero.
Also, we can assume the stator resistance (R1) is negligible for simplification.
T = (3 * V^2 * R2) / (ωs * (R1 + R2)^2)
Tst = (
- Power rating of the motor (P) = 10 kW
- Rated voltage (V) = 400 V
- Full-load current (Ifl)
- Starting current (Ist) = 5 * Ifl
- Full-load torque (Tfl)
- Starting torque (Tst) = 1.5 * Tfl
To reduce the starting current and provide full-load torque at starting, an autotransformer is used.
Working Principle of Autotransformer:
An autotransformer is a type of transformer where a part of the winding is common to both the primary and secondary windings. It allows voltage transformation and current reduction.
Step 1: Finding Full-load Current (Ifl)
The full-load current can be calculated using the formula:
P = √3 * V * Ifl * cos(ϕ)
where ϕ is the power factor.
Since the power factor is not given, we'll assume it to be 0.8 (typical value for an induction motor).
So, Ifl = P / (√3 * V * cos(ϕ))
Ifl = 10,000 / (√3 * 400 * 0.8)
Ifl ≈ 14.56 A
Step 2: Finding Starting Current (Ist)
Since Ist = 5 * Ifl
Ist = 5 * 14.56
Ist ≈ 72.8 A
Step 3: Finding Full-load Torque (Tfl)
The full-load torque can be calculated using the formula:
Tfl = P / (2 * π * n)
where n is the synchronous speed of the motor.
Since the synchronous speed is not given, we'll assume it to be 1800 RPM (typical value for a 3-phase induction motor).
So, Tfl = 10,000 / (2 * π * (1800/60))
Tfl ≈ 8.86 Nm
Step 4: Finding Starting Torque (Tst)
Since Tst = 1.5 * Tfl
Tst = 1.5 * 8.86
Tst ≈ 13.29 Nm
Step 5: Finding the Voltage Applied at Starting with Autotransformer
The autotransformer is used to reduce the starting current and provide full-load torque at starting. Assuming the autotransformer reduces the starting current to the full-load current (Ifl) and provides the full-load torque (Tfl) at starting, we can use the torque equation to find the applied voltage.
T = (3 * V^2 * R2) / (ωs * (R1 + R2)^2)
where T is the torque, V is the voltage applied, R1 is the stator resistance, R2 is the rotor resistance, and ωs is the synchronous speed.
Since the motor is at standstill, we can assume the rotor resistance (R2) is zero.
Also, we can assume the stator resistance (R1) is negligible for simplification.
T = (3 * V^2 * R2) / (ωs * (R1 + R2)^2)
Tst = (
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A 10 kW, 400 V, 3-phase induction motor draw a current of 5 times its full-load current at rated voltage and at standstill developing a starting toque of 1.5 times its full-load torque. An autotransformer is installed to reduce the starting current and to give full-load torque at starting. Neglecting exciting current of auto-transformer, the voltage applied to the motor terminals at the time of starting with auto transformer isa)220 volts b)241.8 voltsc)326.6 voltsd)215.9 voltsCorrect answer is option 'C'. Can you explain this answer?
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A 10 kW, 400 V, 3-phase induction motor draw a current of 5 times its full-load current at rated voltage and at standstill developing a starting toque of 1.5 times its full-load torque. An autotransformer is installed to reduce the starting current and to give full-load torque at starting. Neglecting exciting current of auto-transformer, the voltage applied to the motor terminals at the time of starting with auto transformer isa)220 volts b)241.8 voltsc)326.6 voltsd)215.9 voltsCorrect answer is option 'C'. 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 10 kW, 400 V, 3-phase induction motor draw a current of 5 times its full-load current at rated voltage and at standstill developing a starting toque of 1.5 times its full-load torque. An autotransformer is installed to reduce the starting current and to give full-load torque at starting. Neglecting exciting current of auto-transformer, the voltage applied to the motor terminals at the time of starting with auto transformer isa)220 volts b)241.8 voltsc)326.6 voltsd)215.9 voltsCorrect answer is option 'C'. 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 10 kW, 400 V, 3-phase induction motor draw a current of 5 times its full-load current at rated voltage and at standstill developing a starting toque of 1.5 times its full-load torque. An autotransformer is installed to reduce the starting current and to give full-load torque at starting. Neglecting exciting current of auto-transformer, the voltage applied to the motor terminals at the time of starting with auto transformer isa)220 volts b)241.8 voltsc)326.6 voltsd)215.9 voltsCorrect answer is option 'C'. Can you explain this answer?.
A 10 kW, 400 V, 3-phase induction motor draw a current of 5 times its full-load current at rated voltage and at standstill developing a starting toque of 1.5 times its full-load torque. An autotransformer is installed to reduce the starting current and to give full-load torque at starting. Neglecting exciting current of auto-transformer, the voltage applied to the motor terminals at the time of starting with auto transformer isa)220 volts b)241.8 voltsc)326.6 voltsd)215.9 voltsCorrect answer is option 'C'. 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 10 kW, 400 V, 3-phase induction motor draw a current of 5 times its full-load current at rated voltage and at standstill developing a starting toque of 1.5 times its full-load torque. An autotransformer is installed to reduce the starting current and to give full-load torque at starting. Neglecting exciting current of auto-transformer, the voltage applied to the motor terminals at the time of starting with auto transformer isa)220 volts b)241.8 voltsc)326.6 voltsd)215.9 voltsCorrect answer is option 'C'. 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 10 kW, 400 V, 3-phase induction motor draw a current of 5 times its full-load current at rated voltage and at standstill developing a starting toque of 1.5 times its full-load torque. An autotransformer is installed to reduce the starting current and to give full-load torque at starting. Neglecting exciting current of auto-transformer, the voltage applied to the motor terminals at the time of starting with auto transformer isa)220 volts b)241.8 voltsc)326.6 voltsd)215.9 voltsCorrect answer is option 'C'. Can you explain this answer?.
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