Since watts = volts times amps times 1.73 times power factor (for a three phase motor): W = E * I * 1.73 * PF Solving for I: I = W / (E * 1.73 * PF) Lets say your motor has a PF of 0.85: I = 10000 / (415 * 1.73 * 0.85) I = 16.39A If this is for your homework (I hope not) and PF is not given, then you will substitute 1.0 for the PF and, of course, arrive at a different answer!
To calculate the current in a Star-Delta configuration for a 30 kW, 415 V, 3-phase induction motor, you first determine the full load current (FLC) in Star configuration using the formula: ( I = \frac{P}{\sqrt{3} \times V \times \text{PF}} ), where ( P ) is power (30,000 W), ( V ) is line voltage (415 V), and PF is the power factor (assumed to be around 0.8 if not specified). For the Delta configuration, the current would be calculated as ( I_{Delta} = I_{Star} \times \sqrt{3} ). In the Star configuration, the current will be approximately 41.6 A, and in Delta, it will be around 72 A.
Sir, what is the cable size for 1.5kw 3phase induction motor 400v?
Ohm's Law - V = IR.
60.4
no load means the motor is acting like a coil
570amps on 3phase 415volts
Dear Sir,I want to know that how much draw starting current of 380 volts 3 phase 50 hertz AC motors as per rating values?
How to calculate a cable size of 3kw motor
Starting capacitors are only required for single-phase induction motors. They are not necessary for three-phase motors.
Pls.answer me of above question
A star-delta starter is a method used to reduce the starting current drawn by a 3-phase induction motor. It involves initially connecting the motor windings in a star configuration for starting, and then switching to a delta configuration for running. This helps to minimize voltage drops and prevents excessive current flow during start-up.
This wouldn't be on a farm would it? 1000' line would need it's own utility pole, and transformer.