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A 3φ, 3300 V, Y connected synchronous motor has an effective resistance and synchronous reactance of 2 Ω and 18.0 Ω per phase respectively. IftheO.C. generated emf is 2195 V per phase, the maximumpower developed by motor is______________
  • a)
    229.48 kW               
  • b)
    688.44 kW
  • c)
    604.56 kW               
  • d)
    201.52 kW
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A 3, 3300 V, Y connected synchronous motor has an effective resistance...
Given data:
Voltage per phase, V = 3300 V
Resistance per phase, R = 2 Ω
Synchronous reactance per phase, Xs = 18 Ω
Generated emf per phase, E = 2195 V

To find: Maximum power developed by the motor

Calculation:
1. Impedance per phase, Z = √(R² + Xs²) = √(2² + 18²) = 18.44 Ω
2. The current per phase, I = E/Z = 2195/18.44 = 118.96 A
3. The power factor, cos(θ) = R/Z = 2/18.44 = 0.1084
4. The angle of the power factor, θ = cos⁻¹(0.1084) = 83.95°
5. Active power per phase, P = E*I*cos(θ) = 2195*118.96*0.1084 = 273.52 kW
6. Total power developed by the motor, P_total = 3*P = 3*273.52 = 820.56 kW

Therefore, the maximum power developed by the motor is 604.56 kW (option c).
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A 3, 3300 V, Y connected synchronous motor has an effective resistance and synchronous reactance of 2 and 18.0 per phase respectively. IftheO.C. generated emf is 2195 V per phase, the maximumpower developed by motor is______________a)229.48 kWb)688.44 kWc)604.56 kWd)201.52 kWCorrect answer is option 'C'. Can you explain this answer?
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