• anonymous
This question in my book says "Assume the motor in Q.35 is moved some distance from the supply. Determine the maximum allowable total line resistance for proper motor operation.(hint: find the maximum allowable voltage drop assume 25-A current, and apply Ohm's law)" Q.35=A 220-V, 25-A motor is to be directly connected to a 230-V (+-5%) supply. Assume the motor has a voltage tolerance rating of (+-10%). I don't know what to do!
  • Stacey Warren - Expert
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  • schrodinger
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  • anonymous
As the motor is moved away from the source the resistance of the extension cable comes into effect as their will be voltage due to the added resistance of the extension cord. You have to work out the volt drop per lenth of cable to retriceertain what length must not be exceeded causing the the motor supply to drop ideal operating voltage. so as your motor has a +-10% rating at 220v, therefore the maximum allowed drop in voltage will be 220-22=190V The supply is 230V+-5%, allowing for -5% at worst your supply will be 230-11.5=218.5V The max loss you can operate with is 218.5-190=28.5V. All cabling have a loss factor ie.; x volts per meter, therefore from the cable cheracteristics you can safely calculate how many meters of cable can be used to use up the 28.5V acceptable loss. Hope this helps. :)

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