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The fractional slip of an induction motor is to ratio
- a)rotor Cu loss / rotor input
- b)stator Cu loss / stator input
- c)rotor Cu loss / rotor output
- d)rotor Cu loss / stator Cu loss
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The fractional slip of an induction motor is to ratioa)rotor Cu loss /...
In an induction motor, the rotor copper losses are given as I2r2, while the rotor input and rotor output are expressed as
Most Upvoted Answer
The fractional slip of an induction motor is to ratioa)rotor Cu loss /...
Explanation:
The fractional slip of an induction motor is a measure of the difference in speed between the rotating magnetic field produced by the stator and the actual speed of the rotor. It is denoted by the symbol 's' and is expressed as a fraction or a percentage.
Formula:
The fractional slip is calculated using the following formula:
s = (Ns - Nr) / Ns
Where:
- s is the fractional slip
- Ns is the synchronous speed of the rotating magnetic field produced by the stator
- Nr is the actual speed of the rotor
Interpretation of the options:
a) rotor Cu loss / rotor input:
This option does not directly relate to the fractional slip. It is related to the losses in the rotor and the input power to the rotor.
b) stator Cu loss / stator input:
This option also does not directly relate to the fractional slip. It is related to the losses in the stator and the input power to the stator.
c) rotor Cu loss / rotor output:
This option does not directly relate to the fractional slip. It is related to the losses in the rotor and the output power of the rotor.
d) rotor Cu loss / stator Cu loss:
This option does not directly relate to the fractional slip. It is related to the losses in the rotor and the losses in the stator.
Correct answer:
The correct answer is option 'A' which is "rotor Cu loss / rotor input".
Explanation of the correct answer:
The fractional slip can be determined by comparing the rotor copper losses with the rotor input power. The rotor copper loss is the power loss due to the resistance of the rotor windings, and the rotor input power is the power supplied to the rotor by the stator. By comparing these two quantities, we can determine the fractional slip.
The rotor copper loss can be calculated using the formula:
Rotor Cu loss = I2^2 * R2
Where:
- I2 is the current in the rotor windings
- R2 is the resistance of the rotor windings
The rotor input power can be calculated using the formula:
Rotor input = I2^2 * R2 + I2^2 * X2
Where:
- X2 is the reactance of the rotor windings
By dividing the rotor copper loss by the rotor input power, we can obtain the fractional slip. This ratio represents the fraction of power lost in the rotor compared to the power supplied to the rotor. It is a measure of how efficiently the rotor converts electrical energy into mechanical energy.
Therefore, option 'A' is the correct answer.
The fractional slip of an induction motor is a measure of the difference in speed between the rotating magnetic field produced by the stator and the actual speed of the rotor. It is denoted by the symbol 's' and is expressed as a fraction or a percentage.
Formula:
The fractional slip is calculated using the following formula:
s = (Ns - Nr) / Ns
Where:
- s is the fractional slip
- Ns is the synchronous speed of the rotating magnetic field produced by the stator
- Nr is the actual speed of the rotor
Interpretation of the options:
a) rotor Cu loss / rotor input:
This option does not directly relate to the fractional slip. It is related to the losses in the rotor and the input power to the rotor.
b) stator Cu loss / stator input:
This option also does not directly relate to the fractional slip. It is related to the losses in the stator and the input power to the stator.
c) rotor Cu loss / rotor output:
This option does not directly relate to the fractional slip. It is related to the losses in the rotor and the output power of the rotor.
d) rotor Cu loss / stator Cu loss:
This option does not directly relate to the fractional slip. It is related to the losses in the rotor and the losses in the stator.
Correct answer:
The correct answer is option 'A' which is "rotor Cu loss / rotor input".
Explanation of the correct answer:
The fractional slip can be determined by comparing the rotor copper losses with the rotor input power. The rotor copper loss is the power loss due to the resistance of the rotor windings, and the rotor input power is the power supplied to the rotor by the stator. By comparing these two quantities, we can determine the fractional slip.
The rotor copper loss can be calculated using the formula:
Rotor Cu loss = I2^2 * R2
Where:
- I2 is the current in the rotor windings
- R2 is the resistance of the rotor windings
The rotor input power can be calculated using the formula:
Rotor input = I2^2 * R2 + I2^2 * X2
Where:
- X2 is the reactance of the rotor windings
By dividing the rotor copper loss by the rotor input power, we can obtain the fractional slip. This ratio represents the fraction of power lost in the rotor compared to the power supplied to the rotor. It is a measure of how efficiently the rotor converts electrical energy into mechanical energy.
Therefore, option 'A' is the correct answer.
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Community Answer
The fractional slip of an induction motor is to ratioa)rotor Cu loss /...
In an induction motor, the rotor copper losses are given as I2r2, while the rotor input and rotor output are expressed as
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The fractional slip of an induction motor is to ratioa)rotor Cu loss / rotor inputb)stator Cu loss / stator inputc)rotor Cu loss / rotor outputd)rotor Cu loss / stator Cu lossCorrect answer is option 'A'. Can you explain this answer?
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