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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 Information:
- No-load speed of the DC shunt motor: 1200 rpm
- No-load current drawn by the motor: 5 A
- Supply voltage: 220 V
- Armature resistance: 0.25 Ω
- Field resistance: 110 Ω
- Load current drawn by the motor: 62 A
- Armature reaction demagnetizes the field by 5%
Calculation:
1. Finding the Armature Voltage:
The armature voltage can be calculated using Ohm's Law:
V = I * R
where,
V is the armature voltage,
I is the load current, and
R is the armature resistance.
Substituting the given values, we have:
V = 62 A * 0.25 Ω
V = 15.5 V
2. Finding the Field Voltage:
The field voltage can be calculated by subtracting the armature voltage from the supply voltage:
Field Voltage = Supply Voltage - Armature Voltage
Field Voltage = 220 V - 15.5 V
Field Voltage = 204.5 V
3. Finding the Field Current:
The field current can be calculated using Ohm's Law:
I_field = V_field / R_field
where,
I_field is the field current,
V_field is the field voltage, and
R_field is the field resistance.
Substituting the given values, we have:
I_field = 204.5 V / 110 Ω
I_field ≈ 1.859 A
4. Finding the Field Weakening Effect:
The armature reaction demagnetizes the field to the extent of 5%. This means the effective field current is reduced by 5%.
Field Weakening Effect = 5% of Field Current
Field Weakening Effect = 5/100 * 1.859 A
Field Weakening Effect ≈ 0.093 A
Effective Field Current = Field Current - Field Weakening Effect
Effective Field Current ≈ 1.859 A - 0.093 A
Effective Field Current ≈ 1.766 A
5. Finding the New Speed:
The speed of a DC motor is inversely proportional to the flux produced by the field winding. As the effective field current decreases, the flux and hence the speed increases.
Speed ∝ (1 / Flux)
The new speed can be calculated using the formula:
New Speed = No-load Speed * (Field Current / Effective Field Current)
Substituting the given values, we have:
New Speed = 1200 rpm * (1.859 A / 1.766 A)
New Speed ≈ 1175 rpm
Therefore, the new speed of the DC shunt motor 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 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?
Question Description
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