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A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______
  • a)
    1.125 Ω    
  • b)
    1.0 Ω 
  • c)
    2.9 Ω    
  • d)
    3.0 Ω 
Correct answer is option 'D'. Can you explain this answer?
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A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency ...

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A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency ...
Given data:
Voltage, V = 500 V
Power, P = 40 kW
Speed, N = 1000 rpm
Efficiency, η = 80%
Armature circuit resistance, Ra = 0.125 ohm
Field current, If = 5 A

To find: Resistance to be inserted in the armature circuit to limit starting current to 1.65 times the full load current.

Solution:
1. Calculation of full load current (Ifl):
Power, P = VIfl
Ifl = P/V = 40000/500 = 80 A

2. Calculation of starting current (Ist):
Ist = 1.65 x Ifl = 1.65 x 80 = 132 A

3. Calculation of armature circuit resistance required to limit starting current:
Let R be the resistance to be inserted in the armature circuit.
Voltage drop across armature circuit, Vra = Ist x R
At starting, the voltage drop across armature and field circuits is equal to the supply voltage, i.e., V = Vra + If x Ra
Therefore, R = (V - If x Ra)/Ist
Substituting the given values, we get:
R = (500 - 5 x 0.125)/132 = 3 ohm

Therefore, the resistance to be inserted in the armature circuit to limit starting current to 1.65 times the full load current is 3 ohm (Option D).
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A 500 V, 40 kW, 1000 rpm d.c. shunt motor has on full load efficiency of 80%. The armature circuit resistance is 0.125 Ω and the field current is 5A. The resistance to be inserted in armature to limit starting current to 1.65 times the full load current, is ______a)1.125 Ω b)1.0 Ωc)2.9 Ω d)3.0 ΩCorrect answer is option 'D'. Can you explain this answer?
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