P, the power in kW (kilowatts) is related to V (the supply voltage) and I (the current in amps) by the Power Law:
P = V x I
So, to get the current:
I = P / V
If you have something that uses 5 kW (which is the same as 5000 watts) and the source voltage is 120 V,
then the current drawn in amps = 5000 / 120 = 41.67 amps.
Wiki User
∙ 12y agoWiki User
∙ 13y agoA watt IS a volt-ampere.
For reasons wattage and volt-amperes are sometimes specified differently see 'Volt-ampere on Wikipedia' in related links below.
The area required would depend primarily on the efficiency of the system and on its location. A photovoltaic system works better in a desert in the southwest than in Seattle, Washington.
motor connection to inverter
You need to know the power factor, the phase angle, or the KVA in order to determine KVAR from KW.Power factor is the cosine of the phase angle between voltage and current, and is also the ratio of KW over KVA.Once you know one of those other things, you simply consider that reactive power factor is the sine of the phase angle between voltage and current, and is also the ratio of KVAR or KVA.It is simple arithmetic from there. You don't even need trigonometry, as I've already given you the ratios.
FORMULA1. Input Power =Wattmeter reading × Multiplication factor in WattsWhere, Multiplication factor =2. Output power = VSY × ISY × cosf in Watts.Where VSY - Secondary Voltage in Volts.ISY- Secondary current in Amps.3.Percentage of Efficiency = × 100 %4.Percentage of Regulation = × 100 %Where, VO - No Load Voltage in VoltsVL - Load Voltage in VoltsIMPORTANT!No Load Condition should be observed at the time of startingMeters are checked for proper Type and rating.PROCEDUREConnections are given as per the circuit diagram.The SPST Switch on the Primary side is closed and the DPST Switch on the Secondary side is opened.The Autotransformer is adjusted to Energize the transformer with rated Primary VoltageThe Volt meters and Ammeters Readings are noted and tabulated at No load conditionThe DPST switch on the secondary side is closed.The transformer is loaded upto 130% of the Rated Load, corresponding Ammeters, Voltmeters and Wattmeters readings are noted and tabulated.After the observation of all the readings the load is released gradually to its initial position.The Autotransformer is brought to its initial positionThe Supply is switched off.To find load test on SF transformer you will need:AMMETER - Range (0-10)A - (0-5) A Type MIVOLTMETER (0-150)V - (0-300) V Type MIWattmeter (300V, 5A) - (150V, 5A) Type UPFAuto Transformer 1f, (0-260)VResistive load 5KW, 230VConnecting wires 2.5sq.mm Type Copper
One gallon of #2 fuel oil gives approximately 41 kWh
5000w / 120 volts = 41.66 amps P=E*I transposed to I=P/E
Power is obtained using the equation: Power (Watts) = Voltage (Volts) x Current (Amps) To convert to Watts to Kilowatts (kW), simply divide by 1000 (or shift the DP left by 3 places). eg. 5000W = 5kW Bringing these together: Kilowatts = ( Volts x Amps ) / 1000
5000 (watts) /120 (volts) = 41.6 amps , but to be safe you should allow 1000 watts margin for peaks, so 4000/120 is 33.3amps, although peaks might sometimes get higher than 5kw
At 120 Volts you would draw about 42 amps. At 240 Volts it would be about 21 amps. For 120 Volts you would need 6 AWG and for 240 Volts you would need 10 AWG.
This is a voltage drop question. To answer is question the voltage must be stated.
5kw = 6.25 kva becoz kva = kw/ pf if we take pf is o.8
The area required would depend primarily on the efficiency of the system and on its location. A photovoltaic system works better in a desert in the southwest than in Seattle, Washington.
120 15 amp service ? 210 7 amp service ?
1kw electric power
No! It doesn't matter if you use two legs or three. The output is 5000 watts
5KW
If the heater is RATED at 5kw and is DESIGNED for 240v, it will draw 20.8 amps if 240v is applied.I say it this way because you probably will not have exactly 240v. You may have as little as 220v or as much as 250v and still successfully operate this heater. But the difference in voltages would give you different currents.For USA, Canada and countries running a 60 Hz supply service...In general practice you would connect this heater to a 30amp 2-pole breaker using #10 gauge wire.As always, if you are in doubt about what to do, the best advice anyone should give you is to call a licensed electrician to advise what work is needed.Before you do any work yourself,on electrical circuits, equipment or appliances,always use a test meter to ensure the circuit is, in fact, de-energized.IF YOU ARE NOT ALREADY SURE YOU CAN DO THIS JOBSAFELY AND COMPETENTLYREFER THIS WORK TO QUALIFIED PROFESSIONALS.