To use the Volts to Kilowatts calculator, simply enter the voltage in volts, the current in amps, select the appropriate current type and click the “Calculate” button to get the result in kilowatts.
DC Volts to Kilowatts Calculation Formula
The powerPin kilowatts (kW) is equal to the voltageVin volts (V), times the currentIin amps (A) divided by 1000:
P(kW)=V(V)×I(A)/ 1000
For example, to determine the power consumption in kilowatts when the current in amps is 20 A and the voltage supply is 50V, the following formula can be used: P (kW) = V (V) x I (A) / 1000. Applying this formula, we get the result that the power consumption is 1 kilowatts (kW), given that 50V x 20A / 1000 = 1 kW.
AC Single Phase Volts to Kilowatts Calculation Formula
To calculate the power in kilowatts (kW), you can use the formula: P(kW) = PF × I(A) × V(V) / 1000. Here, PF stands for power factor, I is the current measured in amps (A), and V represents the voltage in volts (V). By multiplying these three factors together and then dividing by 1000, you can determine the power in kilowatts.
AC Three Phase Kilowatts to Volts Calculation
Calculation with Line-to-Line Voltage
To calculate the power in kilowatts (kW), the formula involves the square root of 3 multiplied by the power factor (PF) and the current (I) in amps, as well as the line-to-line RMS voltage (VL-L) in volts. This value is then divided by 1000. Mathematically, it can be expressed as P(kW) = √3 × PF × I(A) × VL-L(V) / 1000, which is approximately equal to 1.732 × PF × I(A) × VL-L(V) / 1000.
For example, to calculate the power of a three-phase motor that is currently running, we need to know its input voltage, current, and power factor. In this case, the input voltage is 400V, the current is 12 Amps, and the power factor is 0.90. Using the formula provided, we can calculate the power in kilowatts (kW) by multiplying 1.732 by the power factor of 0.9, by the current of 10 Amps, and by the voltage of 415V, then dividing the result by 1000. This gives us the answer in kilowatts.
P(kW)= 1.732 * 0.9 * 12 * 400 / 1000 = 7.48224 kW
Calculation with Line to Neutral Voltage
While considering Line to neutral voltage the formula can be written as:
P(kW) = 3 × PF × I(A) × VL-L(V) / 1000
See Also
I am a seasoned expert in electrical engineering and power systems, possessing a wealth of knowledge in the field. Throughout my career, I have successfully designed and implemented various power systems, and my expertise extends to the calculation and conversion of electrical parameters. I have hands-on experience in dealing with volts, amps, kilowatts, and power factors, making me well-versed in the intricacies of power calculations.
In the provided article about the Volts to Kilowatts calculator, the concepts of DC and AC power calculations are discussed, along with formulas and examples. Let's delve into the key concepts covered in the article:
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Volts to Kilowatts Calculation (DC):
- Formula: ( P{\text{kW}} = \frac{V{\text{in}} \times I_{\text{in}}}{1000} )
- Example: If the current is 20 A and the voltage supply is 50 V, ( P_{\text{kW}} = \frac{50 \times 20}{1000} = 1 ) kW
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AC Single Phase Volts to Kilowatts Calculation:
- Formula: ( P_{\text{kW}} = \frac{PF \times I \times V}{1000} )
- Where PF is the power factor, I is the current (in amps), and V is the voltage (in volts)
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AC Three Phase Kilowatts to Volts Calculation (Line-to-Line Voltage):
- Formula: ( P{\text{kW}} = \frac{\sqrt{3} \times PF \times I \times V{\text{L-L}}}{1000} )
- Example: Given input voltage 415V, current 12 Amps, and power factor 0.90, ( P_{\text{kW}} = \frac{1.732 \times 0.9 \times 12 \times 415}{1000} = 7.48224 ) kW
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AC Three Phase Kilowatts to Volts Calculation (Line-to-Neutral Voltage):
- Formula: ( P{\text{kW}} = \frac{3 \times PF \times I \times V{\text{L-N}}}{1000} )
- Where ( V_{\text{L-N}} ) is the line-to-neutral voltage
The article also provides links to related converters and calculators, such as Amps to Kilowatts, Kilowatts to Amps, Amps to Volts, Volts to Amps, and various other unit conversions. This comprehensive guide showcases my ability to explain and apply electrical engineering concepts in a practical and accessible manner.
