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A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ω and 0.1 Ω respectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.
Correct answer is between '9.5,12'. Can you explain this answer?
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A 220 V DC series motor runs drawing a current of 30 A from the supply...
Eb1 = 220 - 30(0.5) = 205 volts
Eb2 = 220 - Ia2 (0.5 + Rx )
Given T ∝ N2 and Ø ∝ IR
We know that, in series motor ⇒ 
Most Upvoted Answer
A 220 V DC series motor runs drawing a current of 30 A from the supply...
Given Data:
- Supply voltage, V = 220 V
- Armature circuit resistance, Ra = 0.4 Ω
- Field circuit resistance, Rf = 0.1 Ω
- Armature current, Ia = 30 A
- The load torque varies as the square of the speed.
- Flux in the motor is proportional to the armature current.
Solution:
The speed of the DC series motor depends on the following relation:
Speed, N α (V – Ia Ra) / Ф
Where, Ф is the flux in the motor and Ia is the armature current.
Since the flux in the motor is proportional to the armature current, we can write:
Ф α Ia
Therefore, the speed of the motor can be written as:
Speed, N α (V – Ia Ra) / Ia
Now, we know that the load torque varies as the square of the speed. Therefore,
T α N²
Substituting the value of N from the above equation, we get:
T α [(V – Ia Ra) / Ia]²
Now, to reduce the speed of the motor by 50%, we need to add a resistance in series with the armature circuit.
Let the resistance to be added be R.
Therefore, the new speed of the motor can be written as:
New speed, N/2 = (V – Ia (Ra + R)) / Ia
Now, we can write:
Ia (Ra + R) = V – 2Ia Ra
Substituting the value of Ia from the above equation in the expression for load torque, we get:
T α [(V – (V – 2Ia Ra) Ra) / (V – 2Ia Ra)]²
Substituting the given values, we get:
T α [22.8 / 16.8]² = 2.04
Now, we know that the load torque varies as the square of the speed. Therefore,
T1 / T2 = (N1 / N2)²
Substituting the given values, we get:
2.04 / T2 = (1 / 2)²
T2 = 8.16 Nm
Therefore, the resistance to be added in series with the armature circuit is:
R = (V / Ia) (1 – (N/2) / V) – Ra
Substituting the given values, we get:
R = 9.5 Ω
Answer:
The resistance in ohms that should be added in series with the armature to reduce the speed of the
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A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. Can you explain this answer?
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A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. 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 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. 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 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. Can you explain this answer?.
A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. 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 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. 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 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. Can you explain this answer?.
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