{
  "id": "kva",
  "version": "54d0b1aa1d04",
  "status": "published",
  "name": "KVA Calculator",
  "question": "What is the kVA?",
  "summary": "Works out apparent power in kVA from volts and amps for single-phase or three-phase circuits, or the current from kVA and volts, with real power in kW from the power factor.",
  "category": "physics",
  "subcategory": "electricity",
  "url": "https://www.acalculator.org/physics/kva-calculator",
  "markdown": "https://www.acalculator.org/physics/kva-calculator.md",
  "kind": "function",
  "method": "Single-phase: kVA = V × I ÷ 1,000, I = kVA × 1,000 ÷ V. Three-phase: kVA = √3 × V × I ÷ 1,000, I = kVA × 1,000 ÷ (√3 × V). kW = kVA × pf; kVAR = kVA × √(1 − pf²).",
  "assumptions": [
    "RMS values on a sinusoidal supply; for three-phase, a balanced load with the line-to-line voltage and the line current.",
    "The power factor only changes kW and kVAR; kVA and amps do not depend on it."
  ],
  "inputs": {
    "$schema": "https://json-schema.org/draft/2020-12/schema",
    "type": "object",
    "properties": {
      "ph": {
        "title": "Supply",
        "description": "Single-phase, or balanced three-phase with the line-to-line voltage.",
        "type": "string",
        "enum": [
          "single",
          "three"
        ]
      },
      "find": {
        "title": "Find",
        "description": "Work out kVA from volts and amps, or amps from kVA and volts.",
        "type": "string",
        "enum": [
          "kva",
          "amps"
        ]
      },
      "v": {
        "title": "Voltage",
        "description": "The RMS voltage; for three-phase, the line-to-line voltage.",
        "type": [
          "number",
          "string"
        ],
        "x-quantity": "voltage",
        "minimum": 0.001,
        "maximum": 1000000
      },
      "i": {
        "title": "Current",
        "description": "The RMS current in each line.",
        "type": [
          "number",
          "string"
        ],
        "x-quantity": "current",
        "minimum": 0.000001,
        "maximum": 1000000
      },
      "kva": {
        "title": "Apparent power (kVA)",
        "description": "The apparent power in kilovolt-amperes, such as a transformer or generator rating.",
        "type": "number",
        "minimum": 0.000001,
        "maximum": 10000000
      },
      "pf": {
        "title": "Power factor",
        "description": "The share of the apparent power that does work, from 0 to 1: 1 for heaters, about 0.8 to 0.95 for motors.",
        "type": "number",
        "minimum": 0,
        "maximum": 1
      }
    }
  },
  "outputs": {
    "kva": {
      "label": "Apparent power (kVA)",
      "description": "Volts × amps (× √3 for three-phase) ÷ 1,000.",
      "format": "number"
    },
    "amps": {
      "label": "Current (A)",
      "description": "The current in each line, in amperes.",
      "format": "number"
    },
    "kw": {
      "label": "Real power (kW)",
      "description": "kVA × power factor: the power that does work.",
      "format": "number"
    },
    "kvar": {
      "label": "Reactive power (kVAR)",
      "description": "kVA × √(1 − power factor²).",
      "format": "number"
    },
    "va": {
      "label": "Apparent power (VA)",
      "description": "The apparent power in volt-amperes.",
      "format": "number"
    }
  },
  "defaultAnswer": {
    "inputs": {
      "ph": "single",
      "find": "kva",
      "v": "240 V",
      "i": "50 A",
      "kva": 75,
      "pf": 0.8
    },
    "outputs": {
      "kva": 12,
      "amps": 50,
      "kw": 9.6,
      "kvar": 7.2,
      "va": 12000
    },
    "text": "240 V at 50 A is 12 kVA (9.6 kW at the power factor given)."
  },
  "examples": [
    {
      "given": {
        "ph": "single",
        "find": "kva",
        "v": 240,
        "i": 50,
        "pf": 0.8
      },
      "expect": {
        "kva": 12,
        "kw": 9.6,
        "kvar": 7.2,
        "va": 12000
      },
      "source": "OpenStax, University Physics Volume 2, §15.4 Power in an AC Circuit (average power = I_rms V_rms cos φ; power factor cos φ), https://openstax.org/books/university-physics-volume-2/pages/15-4-power-in-an-ac-circuit (retrieved 2026-10-05); hand calculation in content.mdx: 240 × 50 = 12,000 VA = 12 kVA; × 0.8 = 9.6 kW; 12 × 0.6 = 7.2 kVAR",
      "tolerance": 1e-12
    },
    {
      "given": {
        "ph": "three",
        "find": "kva",
        "v": 480,
        "i": 100,
        "pf": 1
      },
      "expect": {
        "kva": 83.13843876330611,
        "kw": 83.13843876330611,
        "kvar": 0
      },
      "source": "Wikipedia, \"Three-phase electric power\" (balanced load: apparent power = √3 × line voltage × line current), https://en.wikipedia.org/wiki/Three-phase_electric_power (retrieved 2026-10-05); hand calculation in content.mdx: √3 × 480 × 100 ÷ 1,000 = 83.138 kVA",
      "tolerance": 1e-12
    },
    {
      "given": {
        "ph": "three",
        "find": "amps",
        "v": 208,
        "kva": 75,
        "pf": 0.9
      },
      "expect": {
        "i": 208.17918360202853,
        "kw": 67.5
      },
      "source": "Wikipedia, \"Three-phase electric power\" (balanced load: apparent power = √3 × line voltage × line current), https://en.wikipedia.org/wiki/Three-phase_electric_power (retrieved 2026-10-05); hand calculation in content.mdx: 75,000 ÷ (√3 × 208) = 208.179 A; 75 × 0.9 = 67.5 kW",
      "tolerance": 1e-12
    },
    {
      "given": {
        "ph": "single",
        "find": "amps",
        "v": 120,
        "kva": 1.8,
        "pf": 1
      },
      "expect": {
        "i": 15
      },
      "source": "OpenStax, University Physics Volume 2, §15.4 Power in an AC Circuit (average power = I_rms V_rms cos φ; power factor cos φ), https://openstax.org/books/university-physics-volume-2/pages/15-4-power-in-an-ac-circuit (retrieved 2026-10-05); hand calculation in content.mdx: 1,800 ÷ 120 = 15 A"
    }
  ],
  "sources": [
    "OpenStax, University Physics Volume 2, §15.4 Power in an AC Circuit: average power = I_rms V_rms cos φ, and the power factor cos φ. https://openstax.org/books/university-physics-volume-2/pages/15-4-power-in-an-ac-circuit (retrieved 2026-10-05)",
    "Wikipedia, \"Three-phase electric power\": for a balanced load, power = √3 × line voltage × line current. https://en.wikipedia.org/wiki/Three-phase_electric_power (retrieved 2026-10-05)"
  ],
  "related": [
    "amp",
    "power-factor",
    "transformer"
  ],
  "changelog": []
}