How do you calculate voltage drop for starting motor current
All you can do is get in the ballpark knowing resistance of windings and the supply voltage. Current = Voltage divided by resistance. Wattage = voltage x current x power factor. For a motor the power factor is between zero and sone number less than one, with one being just a resistive load. So if you calculate the current and use a PF = 1 you can get worse case wattage.
The first statement is true, the motor needs a power source to operate.
Motors overheat due to excessive current, not necessarily voltage. Normal voltage can cause a motor to overheat if it is stuck (not spinning). The problem is not usually the voltage, but whatever is causing excessive current flow (usually because the motor is not spinning like it is supposed to).
Yes. If voltage leads the current, the impedance is inductive (this would be the case if the load is a motor). If current leads the voltage, the impedance is capacitive (this would be the case for a CFL light bulb).
You need to specify the rating , voltage and phases of the motor to answer this question.
To calculate the starting current of a motor, the run current must be stated. The voltage is only associated with this calculation in as the higher the motor's voltage is the lower the run current is. The start current can be as high as 300 to 600 percent of the run current. Also taken into account, without the specific make and model of the A/C, the run current is hard to guess as there is quite a variety of amperage drawn by these units.
This configuration is used to reduce the starting current. Utility companies do not like large motor loads starting across the line. It dips the voltage level of the line. By reducing the starting current to a lower level also reduces the voltage dip in the supply lines.
When a motor is stationary, it is not generating a back-mmf which would otherwise act to oppose the applied voltage and, thus, reduce the supply current. However, as the motor runs up to speed, it generates an increasing back-emf, and the supply current falls.
All you can do is get in the ballpark knowing resistance of windings and the supply voltage. Current = Voltage divided by resistance. Wattage = voltage x current x power factor. For a motor the power factor is between zero and sone number less than one, with one being just a resistive load. So if you calculate the current and use a PF = 1 you can get worse case wattage.
One need to know the voltage, the power factor and starting torque, to derive the starting current. The information provided in the question is not sufficient.
When an induction motor starts up, it draws a large current, 6-8 times for direct on line starting and around 3 times for star delta or soft starting. This results in the voltage dipping. VSD controlled induction motors which start from zero speed do not result in voltage dips.
Only when you are Starting the car to turn the starter motor is the battery drawing current. Once the car starts, the alternator delivers the current and the voltage regulator regulates the voltage.
starter is used to limit the starting current to save the motor. for star-delta first star mode is enabled so the applied voltage is reduced to safe value after the motor catches the rated speed the mode is changed to the delta mode. hence the full supply voltage is applied to the motor.
To answer this question the voltage of the motor needs to be stated.
Motors with same horse-powers have different full load amps when operating. To calculate the size of wire to supply the motor feeder the voltage or current of the motor has to be known.
The first statement is true, the motor needs a power source to operate.
Motor starting current is typically 5-7 times the rated current of the motor. (For three phase induction motors)