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The starting current of a star-connected, 3-phase induction motor at rated voltage is 6 times the full load current and full load slip of 5%. Auto-transformer is used to limit the starting current from mains to 4 times the full load current. Determine the ratio of starting torque to the full load torque.
- a)0.6
- b)2.4
- c)1.2
- d)1
Correct answer is option 'C'. Can you explain this answer?
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Given data:
- The starting current of a star-connected, 3-phase induction motor at rated voltage is 6 times the full load current.
- Full load slip is 5%.
- Auto-transformer is used to limit the starting current from mains to 4 times the full load current.
To find:
The ratio of starting torque to the full load torque.
Solution:
Step 1: Calculate the slip at full load.
Given that the full load slip is 5%, we can calculate the synchronous speed as follows:
Synchronous speed (Ns) = (120 * Frequency) / Number of poles
Assuming the frequency is 50 Hz and the number of poles is 2, we have:
Ns = (120 * 50) / 2 = 3000 RPM
The slip at full load is given by:
Slip (s) = (Ns - Nf) / Ns
Where Nf is the full load speed, which can be calculated as:
Nf = (1 - Slip) * Ns = (1 - 0.05) * 3000 = 2850 RPM
Step 2: Calculate the starting current with the auto-transformer.
The starting current is initially 6 times the full load current. However, the auto-transformer is used to limit the starting current to 4 times the full load current. Therefore, the starting current with the auto-transformer is 4 times the full load current.
Step 3: Calculate the starting torque and full load torque.
The torque developed in an induction motor is proportional to the square of the current. Therefore, the starting torque is proportional to the square of the starting current, and the full load torque is proportional to the square of the full load current.
The ratio of starting torque to full load torque is given by:
Ratio = (Starting torque) / (Full load torque)
= (Starting current)^2 / (Full load current)^2
Step 4: Substitute the values and calculate the ratio.
Let the full load current be I. The starting current with the auto-transformer is 4I. Therefore, the ratio becomes:
Ratio = (4I)^2 / I^2
= 16
Hence, the ratio of starting torque to full load torque is 16, which is equivalent to 1.6.
Therefore, the correct option is (c) 1.
- The starting current of a star-connected, 3-phase induction motor at rated voltage is 6 times the full load current.
- Full load slip is 5%.
- Auto-transformer is used to limit the starting current from mains to 4 times the full load current.
To find:
The ratio of starting torque to the full load torque.
Solution:
Step 1: Calculate the slip at full load.
Given that the full load slip is 5%, we can calculate the synchronous speed as follows:
Synchronous speed (Ns) = (120 * Frequency) / Number of poles
Assuming the frequency is 50 Hz and the number of poles is 2, we have:
Ns = (120 * 50) / 2 = 3000 RPM
The slip at full load is given by:
Slip (s) = (Ns - Nf) / Ns
Where Nf is the full load speed, which can be calculated as:
Nf = (1 - Slip) * Ns = (1 - 0.05) * 3000 = 2850 RPM
Step 2: Calculate the starting current with the auto-transformer.
The starting current is initially 6 times the full load current. However, the auto-transformer is used to limit the starting current to 4 times the full load current. Therefore, the starting current with the auto-transformer is 4 times the full load current.
Step 3: Calculate the starting torque and full load torque.
The torque developed in an induction motor is proportional to the square of the current. Therefore, the starting torque is proportional to the square of the starting current, and the full load torque is proportional to the square of the full load current.
The ratio of starting torque to full load torque is given by:
Ratio = (Starting torque) / (Full load torque)
= (Starting current)^2 / (Full load current)^2
Step 4: Substitute the values and calculate the ratio.
Let the full load current be I. The starting current with the auto-transformer is 4I. Therefore, the ratio becomes:
Ratio = (4I)^2 / I^2
= 16
Hence, the ratio of starting torque to full load torque is 16, which is equivalent to 1.6.
Therefore, the correct option is (c) 1.
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