If kVA is known for a three-phase system, which expression gives current?

If kVA is known for a three-phase system, which expression gives current?

• A. I = kVA × 1000 / Volts × PF × 1.73
• B. I = kVA × 1000 / Volts × PF
• C. I = kVA × 1000 / (Volts × PF × 1.73)
• D. I = kVA × 1000 / (Volts × PF)

Answer: (C)

In a three-phase system, the apparent power S is related to line voltage and line current by the equation S = √3 × V_L × I_L. When S is given in kilovolt-amperes (kVA), convert it to volt-amperes by multiplying by 1000, so I_L = S/(√3 × V_L). Plugging S = kVA × 1000 gives I_L = (kVA × 1000) / (√3 × V_L). Since √3 is about 1.73, this can be written as I_L ≈ (kVA × 1000) / (V_L × 1.73). PF (the power factor) does not affect this relationship for current when you’re starting from apparent power, because P and S are linked by P = S × PF, and PF cancels out in the derivation of current from S. So the key idea is using the three-phase power relation with the √3 factor, yielding I = (kVA × 1000) / (Volts × 1.73).

In a three-phase system, the apparent power S is related to line voltage and line current by the equation S = √3 × V_L × I_L. When S is given in kilovolt-amperes (kVA), convert it to volt-amperes by multiplying by 1000, so I_L = S/(√3 × V_L).

Plugging S = kVA × 1000 gives I_L = (kVA × 1000) / (√3 × V_L). Since √3 is about 1.73, this can be written as I_L ≈ (kVA × 1000) / (V_L × 1.73). PF (the power factor) does not affect this relationship for current when you’re starting from apparent power, because P and S are linked by P = S × PF, and PF cancels out in the derivation of current from S.

So the key idea is using the three-phase power relation with the √3 factor, yielding I = (kVA × 1000) / (Volts × 1.73).