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There will be a significant reduction in the mechanical power output available from the motor.

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10y ago

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What is percent of power loss at 415 volts against 11000 volts?

Power = voltage times current, and the power loss is the loss in the line, I^2 * R. At 11,000 volts, the current will be (11,000 / 415 = ) 3.77% of what it is at 415 volts. So the power loss in the line at 11,000 volts will be (3.77% ^2 = ) .14% of what it is at 415 volts.


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reactive power depends on motor winding losses and the reason for increasing of temperature. At running condition the motor power reduces by copper loss.


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Depends on the motor and the load on that motor. If the motor is loaded to its capacity, it will draw the same amount of power as it would on 690 volts - which will result in ( 690/480 = ) 144% of normal current, which will thermally damage the motor, or will trip overload protection.


Calculate the power loss when the voltage is 120 volts and has a resistance of 230 ohms?

By Ohm's law, 120 volts across 230 ohms is 0.522 amps. By the power law, that translates to 62.6 watts. You ask about power loss. In order to answer that, you need to provide more information, such as some change in configuration, or some other component. Please review and restate your question.


What is the power loss in watts of a conductor that carries 24Ampere and has voltage drop of 7.2Volts?

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How do you get fire to the wiper motor on 1985 s-10 blazer?

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What is the efficiency of a 12 volt electric motor?

Efficiency of any electrical machine is maximum when the load on that machine is such that the variable loss ( copper loss) is equal to constant loss (eddy current loss, hysteresis etc).the same applies to dc machines too.


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How many amps draw is there on a half hp 3 phase electric motor?

5 hp equals 3730 watts, and on a 3 phase 480 v system the line voltage is 277, so the current times 277 times 3 equals 3730. The answer in theory is 4.5 amps. But you have to allow for 10% power-loss in the motor, and also the power factor, which could be 0.8. Therefore the current is probably 6 to 6½ amps. Maybe 10 amps on starting up. <<>> The formula you are looking for when amperage is desired when horsepower is shown is - I = HP x 746 / 1.73 x volts x % efficiency x power factor. A standard motor's efficiency between 5 to 100 HP is .84 to .91. A standard motor's power factor between 10 to 100 HP is .86 to .92. Amps = 5 x 746 = 3730/1.73 x 480 = 3730/830 x .84 x .84 = 3730/586 = 6.37 amps. Starting current will be 300% of the motors run current.


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What is motor efficiency?

Motor efficiency refers to the ratio of the mechanical power output of a motor to the electrical power input. It indicates how effectively a motor converts electrical energy into mechanical energy, with higher efficiency values indicating less energy loss during operation. Efficient motors help reduce energy consumption and operating costs.