# How do you convert watts into kVA

In order to convert the current strength (in amperes [A]) into real power (in watts [W]) and apparent power (in volt-amps [VA], you need to know the voltage of the network.

The tables below provide a quick amperage conversion.

Attention: we use cos φ = 0.8 here. If you are dealing with a different cos φ, you can use the formula for calculating the active power.

Single phase:

Used formulas:

Three-phase:

Used formulas:

When we need to determine the performance of a load, we first see whether this load is single-phase or three-phase.

The real power P (in watts) is equal to the voltage U (in volts) times the current I (in amperes) and the cos φ.

P.

The apparent power S (in volt amperes) is equal to the voltage U (in volts) times the current I (in amperes).

S.

The effective power P (in watts) is equal to the voltage U (in volts) times the current I (in amperes), the cos φ and the concatenation factor √3.

P.

The apparent power S (in volt amperes) is equal to the voltage U (in volts) times the current I (in amperes) and the concatenation factor √3.

S.

If we now want to convert the active power P in watts (W) into apparent power S in volt amperes (VA), we apply the following formula:

S.

The tables below provide a quick amperage conversion.

Attention: we use cos φ = 0.8 here. If you are dealing with a different cos φ, you can use the formula for calculating the active power.

Single phase:

I [A] | U = 115V | U = 230V | ||

P._{single phase} [W] | S._{single phase} [VA] | P._{single phase} [W] | S._{single phase} [VA] | |
---|---|---|---|---|

0.1 | 9.2 | 11.5 | 18.4 | 23 |

0.25 | 23 | 28.75 | 46 | 57.5 |

0.5 | 46 | 57.5 | 92 | 115 |

0.75 | 69 | 86.25 | 138 | 172.5 |

1 | 92 | 115 | 184 | 230 |

1.25 | 115 | 143.75 | 230 | 287.5 |

1.5 | 138 | 172.5 | 276 | 345 |

1.75 | 161 | 201.25 | 322 | 402.5 |

2 | 184 | 230 | 368 | 460 |

2.25 | 207 | 258.75 | 414 | 517.5 |

2.5 | 230 | 287.5 | 460 | 575 |

2.75 | 253 | 316.25 | 506 | 632.5 |

3 | 276 | 345 | 552 | 690 |

3.25 | 299 | 373.75 | 598 | 747.5 |

3.5 | 322 | 402.5 | 644 | 805 |

3.75 | 345 | 431.25 | 690 | 862.5 |

4 | 368 | 460 | 736 | 920 |

Used formulas:

- P.
_{single phase }= U * I * cos φ - S.
_{single phase}= U * I

Three-phase:

I [A] | U = 230V | U = 400V | ||

P._{three phase} [W] | S._{three phase} [VA] | P._{three phase} [W] | S._{three phase} [VA] | |
---|---|---|---|---|

0.1 | 31.87 | 39.84 | 55.43 | 69.28 |

0.25 | 79.67 | 99.59 | 138.56 | 173.21 |

0.5 | 159.35 | 199.19 | 277.13 | 346.41 |

0.75 | 239.02 | 298.78 | 415.69 | 519.62 |

1 | 318.70 | 398.37 | 554.26 | 692.82 |

1.25 | 398.37 | 497.96 | 692.82 | 866.03 |

1.5 | 478.05 | 597.56 | 831.38 | 1039.23 |

1.75 | 557.72 | 697.15 | 969.95 | 1212.44 |

2 | 637.39 | 796.74 | 1108.51 | 1385.64 |

2.25 | 717.07 | 896.34 | 1247.08 | 1558.85 |

2.5 | 796.74 | 995.93 | 1385.64 | 1732.05 |

2.75 | 876.42 | 1095.52 | 1524.20 | 1905.26 |

3 | 956.09 | 1195.11 | 1662.77 | 2078.46 |

3.25 | 1035.77 | 1294.71 | 1801.33 | 2251.67 |

3.5 | 1115.44 | 1394.30 | 1939.90 | 2424.87 |

3.75 | 1195.11 | 1493.89 | 2078.46 | 2598.08 |

4 | 1274.79 | 1593.49 | 2217.02 | 2771.28 |

Used formulas:

- P.
_{three phase }= U * I * cos φ * √3 - S.
_{three phase}= U * I * √3

When we need to determine the performance of a load, we first see whether this load is single-phase or three-phase.

The real power P (in watts) is equal to the voltage U (in volts) times the current I (in amperes) and the cos φ.

P.

_{single-phase}> = U * I * cos φ [W]The apparent power S (in volt amperes) is equal to the voltage U (in volts) times the current I (in amperes).

S.

_{single-phase}= U * I [VAThe effective power P (in watts) is equal to the voltage U (in volts) times the current I (in amperes), the cos φ and the concatenation factor √3.

P.

_{three-phase}= U * I * cos φ * √3 [W]The apparent power S (in volt amperes) is equal to the voltage U (in volts) times the current I (in amperes) and the concatenation factor √3.

S.

_{three-phase}= U * I * √3 [VA]If we now want to convert the active power P in watts (W) into apparent power S in volt amperes (VA), we apply the following formula:

S.

_{three phase}= P_{three phase}/ cos φ- How do mebendazole and albendazole differ
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