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A separately exited dc motor has an armature resistance of 0.5W. It runs off a 250V dcsupply drawing an armature current of 20A at 1500 rpm. The torque developed for an armature current of 10A, for the same fieldcurrent, will be
- a)15.28 Nm
- b)15.92 Nm
- c)15.6 Nm
- d)16.55 Nm
Correct answer is option 'A'. Can you explain this answer?
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Given Data:
- Armature resistance, Ra = 0.5 Ω
- DC supply voltage, V = 250 V
- Armature current at 1500 rpm, I1 = 20 A
- Armature current at unknown speed, I2 = 10 A
Formula Used:
- Torque developed by a DC motor is given by the formula: T = (k × φ × I) / Ra
where T is the torque, k is a constant, φ is the flux, I is the armature current, and Ra is the armature resistance.
Explanation:
The torque developed by a DC motor is directly proportional to the armature current. Therefore, we can use the following equation to find the torque at 10 A:
T1 = (k × φ × I1) / Ra
To find the torque at 10 A, we need to know the ratio of the armature currents at 10 A and 20 A. Since the field current remains the same, the flux is constant. Therefore, we can write:
T2 = (k × φ × I2) / Ra
Taking the ratio of the two equations:
T2 / T1 = (I2 / I1)
Substituting the given values:
T2 / T1 = (10 A / 20 A) = 0.5
Therefore, the torque at 10 A is half of the torque at 20 A.
Now, we can calculate the torque at 20 A using the given data:
T1 = (k × φ × I1) / Ra
T1 = (k × φ × 20 A) / 0.5 Ω
T1 = 40 × (k × φ) Nm
Since the torque at 10 A is half of the torque at 20 A, we have:
T2 = (T1 / 2) = 20 × (k × φ) Nm
Therefore, the torque developed for an armature current of 10 A is 20 × (k × φ) Nm, which is equal to 15.28 Nm (approximately).
Hence, the correct answer is option 'A' (15.28 Nm).
- Armature resistance, Ra = 0.5 Ω
- DC supply voltage, V = 250 V
- Armature current at 1500 rpm, I1 = 20 A
- Armature current at unknown speed, I2 = 10 A
Formula Used:
- Torque developed by a DC motor is given by the formula: T = (k × φ × I) / Ra
where T is the torque, k is a constant, φ is the flux, I is the armature current, and Ra is the armature resistance.
Explanation:
The torque developed by a DC motor is directly proportional to the armature current. Therefore, we can use the following equation to find the torque at 10 A:
T1 = (k × φ × I1) / Ra
To find the torque at 10 A, we need to know the ratio of the armature currents at 10 A and 20 A. Since the field current remains the same, the flux is constant. Therefore, we can write:
T2 = (k × φ × I2) / Ra
Taking the ratio of the two equations:
T2 / T1 = (I2 / I1)
Substituting the given values:
T2 / T1 = (10 A / 20 A) = 0.5
Therefore, the torque at 10 A is half of the torque at 20 A.
Now, we can calculate the torque at 20 A using the given data:
T1 = (k × φ × I1) / Ra
T1 = (k × φ × 20 A) / 0.5 Ω
T1 = 40 × (k × φ) Nm
Since the torque at 10 A is half of the torque at 20 A, we have:
T2 = (T1 / 2) = 20 × (k × φ) Nm
Therefore, the torque developed for an armature current of 10 A is 20 × (k × φ) Nm, which is equal to 15.28 Nm (approximately).
Hence, the correct answer is option 'A' (15.28 Nm).
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A separately exited dc motor has an armature resistance of 0.5W. It runs off a 250V dcsupply drawing an armature current of 20A at 1500 rpm. The torque developed for an armature current of 10A, for the same fieldcurrent, will bea)15.28 Nmb)15.92 Nmc)15.6 Nmd)16.55 NmCorrect answer is option 'A'. Can you explain this answer?
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