Well, a 120V appliance that draws 15A would be using 1800W. (15x120). A killowatt hour is one killowatt used continuously for an hour. If you used that device for one hour straight, it would use 1.8 kWh. (Keep in mind the device may not draw a full 15A)
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To calculate the kWh usage of a 120V 15A appliance, you can use the formula: Power (in watts) = Voltage x Current. For the given appliance, power = 120V x 15A = 1800 watts. To convert watts to kilowatt-hours (kWh), divide by 1000: 1800 watts / 1000 = 1.8 kWh. Therefore, the appliance uses 1.8 kWh of energy per hour.
The number of appliances that can be used in 3.5 kWh depends on the energy consumption of each appliance. To calculate, divide 3.5 kWh by the average energy consumption of each appliance in kWh to determine how many can be used.
To calculate the cost per month for the appliance, you first need to find the power consumption in kilowatts per hour (kWh). For a 110V appliance drawing 12A, the power consumption would be 1.32 kW (110V * 12A = 1320W = 1.32 kW). Multiply this by the number of hours the appliance runs per month to get the total kilowatt-hours (kWh) used, then multiply by the cost per kWh. If the appliance runs for, say, 100 hours a month, the calculation would be: 1.32 kW * 100 hours = 132 kWh * $0.09 = $11.88 per month.
To calculate electricity usage, multiply the power rating of the appliance in kilowatts by the time it was used in hours. Then, multiply this by the electricity rate in kilowatt-hours to determine the cost. Finally, multiply this by the number of days the appliance was used to get the total electricity usage.
To calculate the cost per kWh given 35 kWh of usage and the total cost, divide the total cost by the number of kWh. For example, if the total cost is $70 for 35 kWh, the cost per kWh would be $70 / 35 kWh = $2 per kWh.
To calculate the kWh consumed by a 3-phase motor, you'll need to know both the power factor and operating hours. The formula is: kWh = (√3 x Volts x Amps x Power Factor x Hours) / 1,000. Without the power factor and hours of operation, a precise kWh calculation cannot be provided using just voltage and current.