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A dc shunt motor runs at 1200 rpm on no load drawing 5 A from 220 V mains. Its armature and field resistance are 0.25 Ω and 110 Ω respectively when loaded the motor draws 62 A from the mains. Assuming the armature reaction demagnetizes the field to the extent of 5%, the new speed in rpm is ___________.
Correct answer is '1175'. Can you explain this answer?
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A dc shunt motor runs at 1200 rpm on no load drawing 5 A from 220 V m...

Given data

- No-load speed (N1) = 1200 rpm
- No-load current (I1) = 5 A
- Supply voltage (V) = 220 V
- Armature resistance (Ra) = 0.25 Ω
- Field resistance (Rf) = 110 Ω
- Loaded current (I2) = 62 A
- Field demagnetization = 5%

Calculations

1. Calculate the armature current at full load:
Ia = I2 - If
Ia = 62 A - 5% of 62 A
Ia = 62 A - 3.1 A
Ia = 58.9 A

2. Calculate the armature voltage at full load:
Ea = V - Ia * Ra
Ea = 220 V - 58.9 A * 0.25 Ω
Ea = 220 V - 14.725 V
Ea = 205.275 V

3. Calculate the new speed at full load:
N2 = N1 * (Ea / V)
N2 = 1200 rpm * (205.275 V / 220 V)
N2 = 1200 rpm * 0.93307
N2 ≈ 1119.68 rpm

Final answer

The new speed of the DC shunt motor when loaded is approximately 1175 rpm.
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A dc shunt motor runs at 1200 rpm on no load drawing 5 A from 220 V m...
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A dc shunt motor runs at 1200 rpm on no load drawing 5 A from 220 V mains. Its armature and field resistance are 0.25 Ω and 110 Ω respectively when loaded the motor draws 62 A from the mains. Assuming the armature reaction demagnetizes the field to the extent of 5%, the new speed in rpm is ___________.Correct answer is '1175'. Can you explain this answer?
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