# 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:
 P.single phase [W] S.single phase [VA] P.single phase [W] S.single phase [VA] I [A] U = 115V U = 230V 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:
 P.three phase [W] S.three phase [VA] P.three phase [W] S.three phase [VA] I [A] U = 230V U = 400V 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 [VA
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.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= Pthree phase/ cos φ