kVA is kilo-volt-ampere. kVA is a unit of apparent power, which is electrical power unit.

1 kilo-volt-ampere is equal to 1000 volt-ampere:

1kVA = 1000VA

1 kilo-volt-ampere is equal to 1000 times 1 volt times 1 ampere:

1kVA = 1000⋅1V⋅1A

So The apparent power S in volt-amps (VA) is equal to 1000 times the apparent power S in kilovolt-amps (kVA).

*S*_{(VA)} = 1000 × *S*_{(kVA)}

So The real power P in kilowatts (kW) is equal to the apparent power S in kilovolt-amps (kVA), times the power factor [PF]

*P*_{(kW)} = * S*_{(kVA)}* * × *PF*

What is the real power in kilowatts when the apparent power is 8 kVA and the power factor is 0.8?

Solution:

*P* = 8kVA × 0.8 = 6.4kW

What is the real power in kilowatts when the apparent power is 35 kVA and the power factor is 0.8?

Solution:

*P* = 35kVA × 0.8 = 28kW

So The real power P in watts (W) is equal to 1000 times the apparent power S in kilovolt-amps (kVA), times the power factor PF.

*P*_{(W)} = 1000 × *S*_{(kVA)}* * × *PF*

What is the real power in watts when the apparent power is 7 kVA and the power factor is 0.8?

Solution:

*P* = 1000 × 7kVA × 0.8 = 5600W

What is the real power in watts when the apparent power is 16 kVA and the power factor is 0.8?

Solution:

*P* = 1000 × 16kVA × 0.8 = 12800W

The current I in amps is equal to 1000 times the apparent power S in kilovolt-amps, divided by the voltage V in volts:

*I*_{(A)} = 1000 × *S*_{(kVA)}* */ *V*_{(V)}

Question: What is the phase current in amps when the apparent power is 6 kVA and the RMS voltage supply is 110 volts?

Solution:

*I* = 1000 × 6kVA / 110V = 54.545A

Question: What is the phase current in amps when the apparent power is 6 kVA and the RMS voltage supply is 120 volts?

Solution:

*I* = 1000 × 6kVA / 120V = 50A

The phase current I in amps (with balanced loads) is equal to 1000 times the apparent power S in kilovolt-amps, divided by the square root of 3 times the line to line RMS voltage V_{L-L} in volts:

*I*_{(A)} = 1000 × *S*_{(kVA)}* */ (*√*3 × *V*_{L-L(V)}* *)

Question: What is the phase current in amps when the apparent power is 3 kVA and the line to line RMS voltage supply is 180 volts?

Solution:

*I* = 1000 × 3kVA / (*√*3 × 180V) = 9.623A

Question: What is the phase current in amps when the apparent power is 4 kVA and the line to line RMS voltage supply is 180 volts?

Solution:

*I* = 1000 × 4kVA / (*√*3 × 180V) = 12.83A

So The phase current I in amps (with balanced loads) is equal to 1000 times the apparent power S in kilovolt-amps, divided by 3 times the line to neutral RMS voltage V_{L-N} in volts:

*I*_{(A)} = 1000 × *S*_{(kVA)}* */ (3 × *V*_{L-N(V)}* *)

Question: What is the phase current in amps when the apparent power is 5 kVA and the line to neutral RMS voltage supply is 120 volts?

Solution:

*I* = 1000 × 5kVA / (3 × 120V) = 13.889A

Question: What is the phase current in amps when the apparent power is 5 kVA and the line to neutral RMS voltage supply is 180 volts?

Solution:

*I* = 1000 × 5kVA / (3 × 180V) = 9.259A

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- Ampere (A)
- dB-milliwatt (dBm)
- dB-watt (dBW)
- Decibel (dB)
- Farad (F)
- Kilovolt-amp (kVA)
- Kilowatt (kW)
- Kilowatt-hour (kWh)
- Ohm (Ω)
- Volt (V)
- Watt (W)