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The value of 4.9 kW is equal to 4900 watts. Use the following equation to find the amperage for your question. I = W/E, Amps = Watts / Volts.

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11y ago

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You may need to step up the voltage to minimize the power loss / voltage dip at the end of the 800 feet depending on: 1. the type of cable used (the higher resistance, the large cables would be required to not do this) 2. the amperage that you will need to draw. The voltage drop in the wires is due to the I^2*R losses in the cable. If your cable has a resistance of 1 ohm over 800 feet, and you draw 5 amps (so your load resistance is R = V/R = 240 / 5 = 48 ohms), the actual current delivered will be (I = V / R = 240 / 49) 4.9 amps, and the voltage at the end of the 800 feet will be (240 - 4.9*1) 235 volts. voltage is probably fine for whatever equipment you're using. If used continuously, the 800 ft of cable would waste 210kWh of energy/yr (15-35 dollars/year). If you need 20 amps, the load resistance would be (240 / 20) 12 ohms, current delivered to the load would be (240/13) = 18.5, and the voltage at the end of the 800 feet would be (240 - 18.5*1) 222 volts. This might be too low, depending on the equipment you're using and if you're area has a naturally low voltage to begin with. If used continuously, the 800 ft of cable would waste ~3,000kWh of energy/yr (180-450 dollars/year). As your question is written, a straight yes or no cannot be given. I hope the above explains enough to give you the knowledge you need to determine this on your own.


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Without more information you cannot.


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