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Current in Voltage drop Equation

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    Current in Voltage drop Equation

    There is a voltage drop from the panel to the final load (say motor). Generally, the full load ampere is determined by using the rated voltage. If the full load ampere is determined by using the voltage at the terminals (Voltage at the panel minus the voltage drop in the cables), then the FLA would be bit higher than the FLA at rated voltage at the terminals.


    Which current to be used in the Voltage drop equation to determine the voltage drop?

    #2
    It all depends on how accurate you need to be.

    In general using the rated full load current is close enough. But the exact voltage drop depends on the actual current flow.

    If you need to be that accurate then you also need to consider things such as the temperature coefficient of resistance of the conductors, the magnetic interaction of the conduit, the actual load on the motor, etc.

    But if you are being this detailed, then maybe you also want to look at things like voltage drop during starting, which blows any of the above details out of the water.

    Jon

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      #3
      If the purpose of the calculation is to size conductors, FLA is close enough, as Winnie states.

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        #4
        Rated currents for motors from the tables in 430 already take this into account. The current rating for a 120V motor is at 115V, a 208V motor is at 200 Volts, etc....

        Unless you exceed the assumed voltage drop, and have less than the rated voltage at the motor. Then the current would be higher, and the voltage drop would also be higher. But that can be hard on motors.

        But on the flip side, utility voltages always seem like they run a little high. And I think a 1 HP motor is almost never connected to a 1 HP load. So both of those help to reduce the actual current, and the voltage drop.

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