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A 220 v, dc series motor runs at 1000 rpm (clockwise) and takes an armature current of 100 a when driving a load with a constant torque. resistances of the armature and field windings are 0.05 ω each. find the magnitude and direction of motor speed and armature current if the motor terminal voltage is reversed and the number of turns in field winding is reduced to 80 %. assume linear magnetic circuit?
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A 220 v, dc series motor runs at 1000 rpm (clockwise) and takes an arm...
Given data:
- Terminal voltage (V) = 220 V
- Motor speed (N) = 1000 rpm (clockwise)
- Armature current (Ia) = 100 A
- Armature resistance (Ra) = 0.05 Ω
- Field resistance (Rf) = 0.05 Ω
- Field winding turns reduced to 80%

Calculations:
1. Back EMF calculation:
Back EMF (Eb) = V - Ia * Ra
Eb = 220 - 100 * 0.05 = 215 V
2. Torque calculation:
Torque (T) = Eb / (2 * π * N / 60)
T = 215 / (2 * π * 1000 / 60) ≈ 10.88 Nm
3. Motor speed and armature current reversal:
- Reversing the terminal voltage will reverse the direction of the motor.
- Since the field winding turns are reduced to 80%, the field flux will decrease, affecting the motor's speed and armature current.
4. Effects of reversing terminal voltage:
- The motor will run in the opposite direction (anti-clockwise).
- The back EMF will change accordingly.
5. Effects of reducing field winding turns:
- The field flux will decrease, leading to a decrease in the motor speed.
- The armature current may increase due to reduced counter EMF.

Final results:
- Magnitude of motor speed: Decreased due to reduced field flux.
- Direction of motor speed: Reversed (anti-clockwise).
- Magnitude of armature current: May increase due to reduced counter EMF.
In conclusion, reversing the terminal voltage and reducing the field winding turns will significantly affect the motor's speed and armature current. The linear magnetic circuit assumption helps simplify the analysis by assuming a linear relationship between flux and current.
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A 220 v, dc series motor runs at 1000 rpm (clockwise) and takes an armature current of 100 a when driving a load with a constant torque. resistances of the armature and field windings are 0.05 ω each. find the magnitude and direction of motor speed and armature current if the motor terminal voltage is reversed and the number of turns in field winding is reduced to 80 %. assume linear magnetic circuit?
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A 220 v, dc series motor runs at 1000 rpm (clockwise) and takes an armature current of 100 a when driving a load with a constant torque. resistances of the armature and field windings are 0.05 ω each. find the magnitude and direction of motor speed and armature current if the motor terminal voltage is reversed and the number of turns in field winding is reduced to 80 %. assume linear magnetic circuit? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 220 v, dc series motor runs at 1000 rpm (clockwise) and takes an armature current of 100 a when driving a load with a constant torque. resistances of the armature and field windings are 0.05 ω each. find the magnitude and direction of motor speed and armature current if the motor terminal voltage is reversed and the number of turns in field winding is reduced to 80 %. assume linear magnetic circuit? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 220 v, dc series motor runs at 1000 rpm (clockwise) and takes an armature current of 100 a when driving a load with a constant torque. resistances of the armature and field windings are 0.05 ω each. find the magnitude and direction of motor speed and armature current if the motor terminal voltage is reversed and the number of turns in field winding is reduced to 80 %. assume linear magnetic circuit?.
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