Since (in North America, at least), the horsepower of a motor describes its rated output power, you need to know its efficiency in order to determine its input power, then its rated voltage, to determine its full-load current.
The ratio is between 30% to 40 %
Full load starting current is typically in the region of 5or 6 times the full load motor current;.
90/3=30ma
The first thing you have to do is find the full load amps of the motor. The wire size feeding the motor has to be 125% of the full load current. The breaker is usually 250% of the full load current. If the voltage and amperage had been added to the question the exact breaker size could have been calculated.
whenever the load increases,the current drawn by the motor to do or to fulfill the required energy to the load. so the current will increase generally.Increase in load will cause the full utilization of motor,so speed of rotor will decrease.
The current flowing through a transformer's secondary is the current drawn by the load, so it will be exactly the same as the current flowing through your induction motor -assuming that is the load. Don't really understand the point of your question!
The no-load current of a motor, such as a 90 kW motor operating at 440V and 60Hz, can vary based on its design and efficiency. Typically, the no-load current for such motors ranges from 10% to 30% of the full-load current. To estimate the no-load current, you can use the formula: No-load current ≈ Full load current × (no-load current percentage). The full-load current can be calculated using the formula: Full Load Current (A) = Power (W) / (Voltage (V) × √3 × Power Factor).
The full load torque of an induction motor is the maximum torque it can produce at its rated speed when operating at full load conditions. This torque is necessary to drive the load at the desired speed under normal operating conditions. It is an important parameter to consider when selecting a motor for a specific application.
starting current of 3 phase 75 KW induction motor
480V about 20A 240V about 40A look at the nameplate duty cycle and efficiencies vary
The starting current of a motor is usually much higher than the normal current of that motor's full load condition. So for the protection purpose, in case of selecting fuse, the fuse has to be rated 2-3 times the full load of the motor so it does not blow on start. An induction motor will have a starting current of around 6 times FLC.
1. Induction Motor has an air-gap but transformer has no air gap rather it is mutually link. 2. Induction Motor has high no-load current than transformer. 3. Induction Motor is a dynamic device. 4. Induction Motor has high power factor.