Motor Current Rating Chart for Full Load Amps The information in this chart was derived from Table 50.1 of UL standard 508A. The voltages listed are rated motor voltages. The currents listed shall be permitted for system voltage ranges of 110-120, 220-240, 380-415, 440-480 and 550-600 volts.

MOTOR CURRENT SINGLE PHASE MOTORS 110V ac 220V ac 240V ac 0.07 kW 1/12 2.4 1.2 1.1 0.1 kW 1/8 3.3 1.6 1.5 0.12 kW 1/6 3.8 1.9 1.7 0.18 kW 1/4 4.5 2.3 2.1 lower current than that shown in the table; while motors of lower speed usually take higher current. Wide variations from these figures can arise, especially on single phase motors, and

cuit protection. If a motor has a long starting time, high starting current profile or is cycled frequently, it may be necessary to use Column 5 or 6. The LPS-RK_SP, LPN-RK_SP, FRS-R and FRN-R fuses sized per this column pro-vide short circuit and ground-fault protection for motor branch circuits (430.52) as discussed in the previous paragraph.

Table 1 shows that for motors of similar rating, the realisable torque can cover a wide range under reduced voltage (or reduced current) starting conditions.

An induction motor draws 5-10 times its rated current during start. In order to reduce the starting current, we need a starter. If we don't use a starter, inrush currents may heat up the motor winding and the life of motor may reduce. Comparison of three phase induction motor starters

Motor terminal voltage During the starting, th- e motor terminal voltage shoube ld maintained at approximately 80% of the rated voltage for type B motors having a standard 150% starting torque at full voltage with a constant torque load applied. A 81.6% rated voltage will develop a torque T = 0.8162 x 150% = 100%.

Use tables 430.247 thru 430.250 to determine the motor FLC. (Note we do not use the motor nameplate information). From table 430.52 we find the correct maximum setting value for standard short-circuit protection We multiply the motor FLC by the value in table 430.52 We round up the nearest standard rating available, found in Table 240.6 (A).

PDF | Induction motors (IMs) are widely used in industry including it be an electrical or not. However during starting period, 

Its voltage is 220V, while the power factor is 0.8 with an efficiency of 90%. With these values in mind, we'll arrive at the following equation: I = (25 x 746) / (√3 x 220 x 0.8 x 0.9) The total pullout current is 67.98 amperes. 2. Get the exact ampere rating relative to the type of circuit breaker you're using.

Induction motors at starting draw larger current than normal operation current. This starting current is represented as 

To determine the starting current use the following formula: A Code letter table and corresponding factors can be found in Na tional Electrical Manufacturer s Association (NEMA) Motor and Generators Standards MG1, beginning in Section 2 Part 12.