Mohammed Kamel - Engineer Boards
Engineer Boards   # Mohammed Kamel

Jr Member

3

2 Neutral

• Rank
Intern

## Previous Fields

• Engineering Field
Power systems analysis and distribution
PE
• Calculator
Casio
• Discipline
Electrical

## Profile Information

• Gender
Male
• Location
Makkah, KSA
1. IbB is equal to IaA but lagged by angle 120 degree, hence, IbB = 70 * (Cos(-140) + j Sin(-140)) = -53.623-j45.
2. This problem can be solved easily by applying KVL for loop (a-A-B-b) as it is without converting it into single phase values where, IaA = 65.78-j23.94, IbB = -53.62-j45 ( IbBI = 70, angle = -140), Vab = (65.78-j23.94)*(5+j10) + 12500 - (-53.62-j45)*(5+j10) Vab = 12886.47 + j1299.3 IVabI = 12952 volt
3. I think this question may issued considering the permissions under article 440.41 (A) :"In case the motor controller is rated in horsepower but is without one or both of the foregoing current ratings, equivalent currents shall be determined from the ratings as follows. Table 430.248, Table 430.249, and Table 430.250 shall be used to determine the equivalent full-load current rating. Table 430.251(A) and Table 430.251(B) shall be used to determine the equivalent locked-rotor current ratings.", and article 430.83(D).
×
×
• Create New...