Overview
The Three-Phase V-I-P Calculator solves for Voltage, Current, or Power in balanced three-phase AC circuits. It is the fundamental building block for power distribution design, feeder sizing, and motor installation verification.
Select what you want to solve for and provide the remaining known values. The calculator returns the solved quantity along with apparent power, reactive power, phase angle, and auto-scaled display units.
Input Parameters
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
solve_for |
Enum: Current / Power / Voltage | Current | Required | Determines which variable to calculate |
voltage |
Number | 415 V | min 0 | Phase-to-phase voltage (line voltage) |
current |
Number | 24.55 A | min 0 | Line current per phase |
power |
Number | 15 | min 0 | Active power (W/kW/MW) or apparent power (VA/kVA/MVA) depending on power_units |
power_units |
Enum: Watts / Kilowatts / Megawatts / VA / KVA / MVA | Kilowatts | — | Unit of power input |
power_factor |
Number | 0.85 | 0 – 1 | \(\cos\phi\) — ratio of active to apparent power |
efficiency |
Number | 100% | 1 – 100% | Motor / load efficiency |
Output Results
| Output | Unit | Description |
|---|---|---|
voltage |
V | Phase-to-phase voltage |
current |
A | Line current |
active_power |
W / kW / MW | Real power consumed (auto-scaled) |
apparent_power |
VA / kVA / MVA | Total power delivered (auto-scaled) |
reactive_power |
VAr / kVAr / MVAr | Reactive component (auto-scaled) |
power_factor |
— (dimensionless) | \(\cos\phi\) |
phase_angle |
degrees | Angle between voltage and current phasors |
power_suffix |
text | Auto-scale prefix (k, M, or empty) |
Formulas
Core Three-Phase Power Equation
Solving for Current
Active power input (W, kW, MW):
Apparent power input (VA, kVA, MVA) — efficiency and PF are not used:
Solving for Voltage
Active power input (W, kW, MW):
Apparent power input (VA, kVA, MVA):
Solving for Power
Apparent Power
Reactive Power
\(S\) is the input apparent power, so this holds for any efficiency: with \(\eta < 100\) % the reported \(P\) is the equipment's output power and the input active power is \(S\cos\phi = P/\eta\) (IEC 60050-131-11-44: \(S^2 = P_{in}^2 + Q^2\)).
Phase Angle
Power Unit Handling
The calculator treats active power units (W, kW, MW) and apparent power units (VA, kVA, MVA) differently:
Active power input — pure scaling to watts:
\( P_{\text{base}} = P_{\text{input}} \times \text{scale} \) (e.g. 15 kW → 15,000 W)
Then: \( I = P_{\text{base}} / (\sqrt{3} \times V \times \cos\phi \times \eta) \)
Apparent power input — pure scaling to VA:
\( S_{\text{base}} = S_{\text{input}} \times \text{scale} \) (e.g. 17.6 kVA → 17,600 VA)
Then: \( I = S_{\text{base}} / (\sqrt{3} \times V) \) — no PF or η in the formula
Active power is derived as: \( P = S \times \cos\phi \)
Auto-scaling prefixes are chosen based on the magnitude of the apparent power:
- Up to 1,000 VA → no prefix (W, VA, VAr)
- Above 1,000 VA up to 1,000,000 VA → k (kW, kVA, kVAr)
- Above 1,000,000 VA → M (MW, MVA, MVAr)
Worked Examples
The 15 examples below are the validation cases for this calculation. Each one was produced by the calculation engine from the inputs shown, every figure was checked against the independent reference quoted at the end of the example, and the same cases are run against every engine build (see validation scorecard, tool fe_calc_power; worked examples for every calculation).
| # | Case | Solve for | V (V) | I (A) | P (kW) | S (kVA) | Q (kVAr) | pf | η (%) | Result |
|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 15 kW at 415 V, pf 0.85 — find the current | current | 415.0 | 24.55 | 15.00 | 17.64 | 9.296 | 0.85 | 100 | |
| 2 | 100 A at 400 V, pf 0.8 — find the power | power | 400.0 | 100.0 | 55.42 | 69.28 | 41.56 | 0.8 | 100 | |
| 3 | 50 A motor at 400 V, pf 0.80, η 92 % — find the output power | power | 400.0 | 50.00 | 25.49 | 34.64 | 20.78 | 0.8 | 92 | |
| 4 | 100 kW output at 200 A, pf 0.90, η 95 % — find the voltage | voltage | 337.6 | 200.0 | 100.0 | 116.9 | 50.98 | 0.9 | 95 | |
| 5 | 36 kW resistive heater at 415 V, pf 1.0 | current | 415.0 | 50.08 | 36.00 | 36.00 | 0 | 1 | 100 | |
| 6 | 250 W at 400 V, pf 0.5 — a very small load | current | 400.0 | 0.7216 | 0.2500 | 0.5000 | 0.4330 | 0.5 | 100 | |
| 7 | 75 kVA transformer secondary at 480 V | current | 480.0 | 90.21 | 67.50 | 75.00 | 32.69 | 0.9 | 100 | |
| 8 | 250 kW pump motor at 690 V, pf 0.86, η 95 % | current | 690.0 | 256.0 | 250.0 | 305.9 | 156.1 | 0.86 | 95 | |
| 9 | 1 MW at 200 A, pf 0.87 — find the voltage | voltage | 3,318 | 200.0 | 1,000 | 1,149 | 566.7 | 0.87 | 100 | |
| 10 | 400 A feeder at 6.6 kV, pf 0.88 | power | 6,600 | 400.0 | 4,023 | 4,572 | 2,171 | 0.88 | 100 | |
| 11 | 2.5 MVA transformer primary at 11 kV | current | 11,000 | 131.2 | 2,125 | 2,500 | 1,316 | 0.85 | 100 | |
| 12 | 20 MW at 33 kV, pf 0.95 — a very large load | current | 33,000 | 368.3 | 20,000 | 21,050 | 6,573 | 0.95 | 100 | |
| 13 | 500 kVA at 695.6 A — find the voltage | voltage | 415.0 | 695.6 | 400.0 | 500.0 | 300.0 | 0.8 | 100 | |
| 14 | 55 kW resistive load at 480 V, pf 1.0 | current | 480.0 | 66.15 | 55.00 | 55.00 | 0 | 1 | 100 | |
| 15 | 50 A heater bank at 400 V, pf 1.0, η 92 % | power | 400.0 | 50.00 | 31.86 | 34.64 | 0 | 1 | 92 |
Example 1 — 15 kW at 415 V, pf 0.85 — find the current
The basic three-phase conversion I = P / (√3·V·pf), and the power triangle that follows from it.
Given: Balanced three-phase circuit, solve for current: VLL = 415 V, P = 15 kW (active power), pf = cos φ = 0.85 (lagging), η = 100 %.
| Base units | P = 15 kW = 15,000 W |
|---|---|
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
| Current | I = P / (√3 × V × pf) = 15,000 / (√3 × 415 × 0.85) = 15,000 / 610.98 = 24.551 A → 24.55 A |
| Apparent power | S = √3 × V × I = √3 × 415 × 24.551 = 17,647 VA → 17.64 kVA |
| Active power | P = 15 kW as given → 15.00 kW |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 17,647 × 0.5268 = 9,296.2 VAr → 9.296 kVAr |
| Phase angle | φ = acos 0.85 = 31.788° → 31.78°; triangle check: P / S = 15,000 / 17,647 = 0.8500 = pf |
| Display | S = 17,647 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 15.00 kW, S = 17.64 kVA, Q = 9.296 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 415.0 V, I = 24.55 A, P = 15.00 kW, S = 17.64 kVA, Q = 9.296 kVAr, φ = 31.78° |
| Reference | Hand calculation: I = P / (√3 × V × pf) = 15,000 / (√3 × 415 × 0.85) = 24.55 A; S = √3 × V × I = √3 × 415 × 24.55 = 17.65 kVA; Q = S × sin φ = 17.65 × 0.5268 = 9.296 kVAr; φ = acos 0.85 = 31.79° (validation case c0001-15kw-415v-solve-current) |
Example 2 — 100 A at 400 V, pf 0.8 — find the power
Solving for power from a measured current: S = √3·V·I, then P = S·pf and Q = S·sin φ (the 3-4-5 triangle at pf 0.8).
Given: Balanced three-phase circuit, solve for power: VLL = 400 V, I = 100 A, pf = cos φ = 0.8 (lagging), η = 100 %.
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
|---|---|
| Apparent power | S = √3 × V × I = √3 × 400 × 100 = 69,282 VA → 69.28 kVA |
| Active power | P = S × pf = 69,282 × 0.8 = 55,426 W → 55.42 kW |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 69,282 × 0.6000 = 41,569 VAr → 41.56 kVAr |
| Phase angle | φ = acos 0.8 = 36.870° → 36.86°; triangle check: P / S = 55,426 / 69,282 = 0.8000 = pf |
| Display | S = 69,282 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 55.42 kW, S = 69.28 kVA, Q = 41.56 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 400.0 V, I = 100.0 A, P = 55.42 kW, S = 69.28 kVA, Q = 41.56 kVAr, φ = 36.86° |
| Reference | Hand calculation: S = √3 × 400 × 100 = 69.28 kVA; P = S × pf = 69.28 × 0.8 = 55.43 kW; Q = S × sin φ = 69.28 × 0.6000 = 41.57 kVAr; φ = acos 0.8 = 36.87° (validation case c0001-400v-100a-solve-power) |
Example 3 — 50 A motor at 400 V, pf 0.80, η 92 % — find the output power
With an efficiency the solved active power is the equipment's output: P = √3·V·I·pf·η.
Given: Balanced three-phase circuit, solve for power: VLL = 400 V, I = 50 A, pf = cos φ = 0.8 (lagging), η = 92 %.
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
|---|---|
| Apparent power | S = √3 × V × I = √3 × 400 × 50 = 34,641 VA → 34.64 kVA |
| Active power | P = S × pf × η = 34,641 × 0.8 × 0.92 = 25,496 W → 25.49 kW (output power) |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 34,641 × 0.6000 = 20,785 VAr → 20.78 kVAr (S is the input, so Q = Pin × tan φ with Pin = P / η = 27,713 W; not √(S² − Pout²)) |
| Phase angle | φ = acos 0.8 = 36.870° → 36.86° |
| Display | S = 34,641 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 25.49 kW, S = 34.64 kVA, Q = 20.78 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 400.0 V, I = 50.00 A, P = 25.49 kW, S = 34.64 kVA, Q = 20.78 kVAr, φ = 36.86° |
| Reference | Hand calculation: S = √3 × 400 × 50 = 34.64 kVA; P = S × pf × η = 34.64 × 0.8 × 92 % = 25.5 kW; Q = S × sin φ = 34.64 × 0.6000 = 20.78 kVAr; check: P_in = P / η = 27.71 kW, Q = P_in × tan φ = 27.71 × 0.7500 = 20.78 kVAr; φ = acos 0.8 = 36.87° (validation case c0001-400v-50a-eta92-solve-power) |
Example 4 — 100 kW output at 200 A, pf 0.90, η 95 % — find the voltage
Solving for line-to-line voltage from an output-power rating: V = P / (√3·I·pf·η).
Given: Balanced three-phase circuit, solve for voltage: I = 200 A, P = 100 kW (active power, equipment output), pf = cos φ = 0.9 (lagging), η = 95 %.
| Base units | P = 100 kW = 100,000 W |
|---|---|
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
| Voltage | V = P / (√3 × I × pf × η) = 100,000 / (√3 × 200 × 0.9 × 0.95) = 100,000 / 296.18 = 337.63 V → 337.6 V |
| Apparent power | S = √3 × V × I = √3 × 337.63 × 200 = 116,959 VA → 116.9 kVA |
| Active power | P = 100 kW as given → 100.0 kW (output power); input power = S × pf = 116,959 × 0.9 = 105,263 W; losses = 5,263.2 W |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 116,959 × 0.4359 = 50,981 VAr → 50.98 kVAr (S is the input, so Q = Pin × tan φ with Pin = P / η = 105,263 W; not √(S² − Pout²)) |
| Phase angle | φ = acos 0.9 = 25.842° → 25.84° |
| Display | S = 116,959 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 100.0 kW, S = 116.9 kVA, Q = 50.98 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 337.6 V, I = 200.0 A, P = 100.0 kW, S = 116.9 kVA, Q = 50.98 kVAr, φ = 25.84° |
| Reference | Hand calculation: V = P / (√3 × I × pf × η) = 100,000 / (√3 × 200 × 0.9 × 95 %) = 337.6 V; S = √3 × V × I = √3 × 337.6 × 200 = 117 kVA; Q = S × sin φ = 117 × 0.4359 = 50.98 kVAr; check: P_in = P / η = 105.3 kW, Q = P_in × tan φ = 105.3 × 0.4843 = 50.98 kVAr; φ = acos 0.9 = 25.84° (validation case c0001-100kw-200a-eta95-solve-voltage) |
Example 5 — 36 kW resistive heater at 415 V, pf 1.0
At unity power factor S = P, Q = 0 and φ = 0°: the smallest current for a given active power.
Given: Balanced three-phase circuit, solve for current: VLL = 415 V, P = 36 kW (active power), pf = cos φ = 1 (lagging), η = 100 %.
| Base units | P = 36 kW = 36,000 W |
|---|---|
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
| Current | I = P / (√3 × V × pf) = 36,000 / (√3 × 415 × 1) = 36,000 / 718.80 = 50.083 A → 50.08 A |
| Apparent power | S = √3 × V × I = √3 × 415 × 50.083 = 36,000 VA → 36.00 kVA |
| Active power | P = 36 kW as given → 36.00 kW |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 36,000 × 0.0000 = 0 VAr → 0 kVAr |
| Phase angle | φ = acos 1 = 0.000° → 0.00°; triangle check: P / S = 36,000 / 36,000 = 1.0000 = pf |
| Display | S = 36,000 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 36.00 kW, S = 36.00 kVA, Q = 0 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 415.0 V, I = 50.08 A, P = 36.00 kW, S = 36.00 kVA, Q = 0 kVAr, φ = 0.00° |
| Reference | Hand calculation: I = P / (√3 × V × pf) = 36,000 / (√3 × 415 × 1) = 50.08 A; S = √3 × V × I = √3 × 415 × 50.08 = 36 kVA; Q = S × sin φ = 36 × 0.0000 = 0 kVAr; φ = acos 1 = 0.00° (validation case c0001-36kw-heater-415v-pf1) |
Example 6 — 250 W at 400 V, pf 0.5 — a very small load
Below 1 kVA the component scales outputs in W/VA/VAr; the tool still reports kW/kVA/kVAr. Poor pf doubles the current.
Given: Balanced three-phase circuit, solve for current: VLL = 400 V, P = 250 W (active power), pf = cos φ = 0.5 (lagging), η = 100 %.
| Base units | P = 250 W = 250 W |
|---|---|
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
| Current | I = P / (√3 × V × pf) = 250 / (√3 × 400 × 0.5) = 250 / 346.41 = 0.72169 A → 0.7216 A |
| Apparent power | S = √3 × V × I = √3 × 400 × 0.72169 = 500.00 VA → 0.5000 kVA |
| Active power | P = 250 W as given → 0.2500 kW |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 500.00 × 0.8660 = 433.01 VAr → 0.4330 kVAr |
| Phase angle | φ = acos 0.5 = 60.000° → 60.00°; triangle check: P / S = 250 / 500.00 = 0.5000 = pf |
| Display | S = 500.00 VA is up to 1,000 VA, so the calculator shows base units: P = 250.0 W, S = 500.0 VA, Q = 433.0 VAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 400.0 V, I = 0.7216 A, P = 0.2500 kW, S = 0.5000 kVA, Q = 0.4330 kVAr, φ = 60.00° |
| Reference | Hand calculation: I = P / (√3 × V × pf) = 250 / (√3 × 400 × 0.5) = 0.7217 A; S = √3 × V × I = √3 × 400 × 0.7217 = 0.5 kVA; Q = S × sin φ = 0.5 × 0.8660 = 0.433 kVAr; φ = acos 0.5 = 60.00° (validation case c0001-250w-400v-pf05-small-load) |
Example 7 — 75 kVA transformer secondary at 480 V
An apparent-power rating gives I = S / (√3·V) with no pf or η; the pf only splits S into P = S·pf and Q.
Given: Balanced three-phase circuit, solve for current: VLL = 480 V, S = 75 kVA (apparent power), pf = cos φ = 0.9 (lagging), η = 100 %.
| Base units | S = 75 kVA = 75,000 VA |
|---|---|
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
| Current | I = S / (√3 × V) = 75,000 / (√3 × 480) = 75,000 / 831.38 = 90.211 A → 90.21 A (an apparent-power rating: pf and η do not enter) |
| Apparent power | S = √3 × V × I = √3 × 480 × 90.211 = 75,000 VA → 75.00 kVA |
| Active power | P = S × pf = 75,000 × 0.9 = 67,500 W → 67.50 kW |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 75,000 × 0.4359 = 32,692 VAr → 32.69 kVAr |
| Phase angle | φ = acos 0.9 = 25.842° → 25.84°; triangle check: P / S = 67,500 / 75,000 = 0.9000 = pf |
| Display | S = 75,000 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 67.50 kW, S = 75.00 kVA, Q = 32.69 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 480.0 V, I = 90.21 A, P = 67.50 kW, S = 75.00 kVA, Q = 32.69 kVAr, φ = 25.84° |
| Reference | Hand calculation: I = S / (√3 × V) = 75,000 / (√3 × 480) = 90.21 A; S = √3 × V × I = √3 × 480 × 90.21 = 75 kVA; P = S × pf = 75 × 0.9 = 67.5 kW; Q = S × sin φ = 75 × 0.4359 = 32.69 kVAr; φ = acos 0.9 = 25.84° (validation case c0001-75kva-480v-solve-current) |
Example 8 — 250 kW pump motor at 690 V, pf 0.86, η 95 %
An LV motor rated by shaft output: the line current includes the losses, I = P / (√3·V·pf·η).
Given: Balanced three-phase circuit, solve for current: VLL = 690 V, P = 250 kW (active power, equipment output), pf = cos φ = 0.86 (lagging), η = 95 %.
| Base units | P = 250 kW = 250,000 W |
|---|---|
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
| Current | I = P / (√3 × V × pf × η) = 250,000 / (√3 × 690 × 0.86 × 0.95) = 250,000 / 976.41 = 256.04 A → 256.0 A |
| Apparent power | S = √3 × V × I = √3 × 690 × 256.04 = 305,998 VA → 305.9 kVA |
| Active power | P = 250 kW as given → 250.0 kW (output power); input power = S × pf = 305,998 × 0.86 = 263,158 W; losses = 13,158 W |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 305,998 × 0.5103 = 156,149 VAr → 156.1 kVAr (S is the input, so Q = Pin × tan φ with Pin = P / η = 263,158 W; not √(S² − Pout²)) |
| Phase angle | φ = acos 0.86 = 30.683° → 30.68° |
| Display | S = 305,998 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 250.0 kW, S = 305.9 kVA, Q = 156.1 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 690.0 V, I = 256.0 A, P = 250.0 kW, S = 305.9 kVA, Q = 156.1 kVAr, φ = 30.68° |
| Reference | Hand calculation: I = P / (√3 × V × pf × η) = 250,000 / (√3 × 690 × 0.86 × 95 %) = 256 A; S = √3 × V × I = √3 × 690 × 256 = 306 kVA; Q = S × sin φ = 306 × 0.5103 = 156.1 kVAr; check: P_in = P / η = 263.2 kW, Q = P_in × tan φ = 263.2 × 0.5934 = 156.1 kVAr; φ = acos 0.86 = 30.68° (validation case c0001-250kw-690v-eta95-solve-current) |
Example 9 — 1 MW at 200 A, pf 0.87 — find the voltage
Solving for voltage with a power given in MW lands on a 3.3 kV MV system.
Given: Balanced three-phase circuit, solve for voltage: I = 200 A, P = 1 MW (active power), pf = cos φ = 0.87 (lagging), η = 100 %.
| Base units | P = 1 MW = 1,000,000 W |
|---|---|
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
| Voltage | V = P / (√3 × I × pf) = 1,000,000 / (√3 × 200 × 0.87) = 1,000,000 / 301.38 = 3,318.1 V → 3,318 V |
| Apparent power | S = √3 × V × I = √3 × 3,318.1 × 200 = 1,149,425 VA → 1,149 kVA |
| Active power | P = 1 MW as given → 1,000 kW |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 1,149,425 × 0.4931 = 566,726 VAr → 566.7 kVAr |
| Phase angle | φ = acos 0.87 = 29.541° → 29.54°; triangle check: P / S = 1,000,000 / 1,149,425 = 0.8700 = pf |
| Display | S = 1,149,425 VA is above 1,000,000 VA, so the calculator shows M units: P = 1.000 MW, S = 1.149 MVA, Q = 0.5667 MVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 3,318 V, I = 200.0 A, P = 1,000 kW, S = 1,149 kVA, Q = 566.7 kVAr, φ = 29.54° |
| Reference | Hand calculation: V = P / (√3 × I × pf) = 1,000,000 / (√3 × 200 × 0.87) = 3,318 V; S = √3 × V × I = √3 × 3,318 × 200 = 1,149 kVA; Q = S × sin φ = 1,149 × 0.4931 = 566.7 kVAr; φ = acos 0.87 = 29.54° (validation case c0001-1mw-200a-3300v-solve-voltage) |
Example 10 — 400 A feeder at 6.6 kV, pf 0.88
An MV feeder above 1 MVA: the component switches to M-scaling, the tool still reports kW/kVA/kVAr.
Given: Balanced three-phase circuit, solve for power: VLL = 6,600 V, I = 400 A, pf = cos φ = 0.88 (lagging), η = 100 %.
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
|---|---|
| Apparent power | S = √3 × V × I = √3 × 6,600 × 400 = 4,572,614 VA → 4,572 kVA |
| Active power | P = S × pf = 4,572,614 × 0.88 = 4,023,900 W → 4,023 kW |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 4,572,614 × 0.4750 = 2,171,871 VAr → 2,171 kVAr |
| Phase angle | φ = acos 0.88 = 28.358° → 28.35°; triangle check: P / S = 4,023,900 / 4,572,614 = 0.8800 = pf |
| Display | S = 4,572,614 VA is above 1,000,000 VA, so the calculator shows M units: P = 4.023 MW, S = 4.572 MVA, Q = 2.171 MVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 6,600 V, I = 400.0 A, P = 4,023 kW, S = 4,572 kVA, Q = 2,171 kVAr, φ = 28.35° |
| Reference | Hand calculation: S = √3 × 6,600 × 400 = 4,573 kVA; P = S × pf = 4,573 × 0.88 = 4,024 kW; Q = S × sin φ = 4,573 × 0.4750 = 2,172 kVAr; φ = acos 0.88 = 28.36° (validation case c0001-6600v-400a-solve-power) |
Example 11 — 2.5 MVA transformer primary at 11 kV
Transformer full-load current from its MVA rating at 11 kV, I = S / (√3·V).
Given: Balanced three-phase circuit, solve for current: VLL = 11,000 V, S = 2.5 MVA (apparent power), pf = cos φ = 0.85 (lagging), η = 100 %.
| Base units | S = 2.5 MVA = 2,500,000 VA |
|---|---|
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
| Current | I = S / (√3 × V) = 2,500,000 / (√3 × 11,000) = 2,500,000 / 19,053 = 131.22 A → 131.2 A (an apparent-power rating: pf and η do not enter) |
| Apparent power | S = √3 × V × I = √3 × 11,000 × 131.22 = 2,500,000 VA → 2,500 kVA |
| Active power | P = S × pf = 2,500,000 × 0.85 = 2,125,000 W → 2,125 kW |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 2,500,000 × 0.5268 = 1,316,957 VAr → 1,316 kVAr |
| Phase angle | φ = acos 0.85 = 31.788° → 31.78°; triangle check: P / S = 2,125,000 / 2,500,000 = 0.8500 = pf |
| Display | S = 2,500,000 VA is above 1,000,000 VA, so the calculator shows M units: P = 2.125 MW, S = 2.500 MVA, Q = 1.316 MVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 11,000 V, I = 131.2 A, P = 2,125 kW, S = 2,500 kVA, Q = 1,316 kVAr, φ = 31.78° |
| Reference | Hand calculation: I = S / (√3 × V) = 2,500,000 / (√3 × 11,000) = 131.2 A; S = √3 × V × I = √3 × 11,000 × 131.2 = 2,500 kVA; P = S × pf = 2,500 × 0.85 = 2,125 kW; Q = S × sin φ = 2,500 × 0.5268 = 1,317 kVAr; φ = acos 0.85 = 31.79° (validation case c0001-2500kva-11kv-solve-current) |
Example 12 — 20 MW at 33 kV, pf 0.95 — a very large load
A sub-transmission load: 20 MW at 33 kV draws only a few hundred amps.
Given: Balanced three-phase circuit, solve for current: VLL = 33,000 V, P = 20 MW (active power), pf = cos φ = 0.95 (lagging), η = 100 %.
| Base units | P = 20 MW = 20,000,000 W |
|---|---|
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
| Current | I = P / (√3 × V × pf) = 20,000,000 / (√3 × 33,000 × 0.95) = 20,000,000 / 54,300 = 368.33 A → 368.3 A |
| Apparent power | S = √3 × V × I = √3 × 33,000 × 368.33 = 21,052,632 VA → 21,050 kVA |
| Active power | P = 20 MW as given → 20,000 kW |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 21,052,632 × 0.3122 = 6,573,682 VAr → 6,573 kVAr |
| Phase angle | φ = acos 0.95 = 18.195° → 18.19°; triangle check: P / S = 20,000,000 / 21,052,632 = 0.9500 = pf |
| Display | S = 21,052,632 VA is above 1,000,000 VA, so the calculator shows M units: P = 20.00 MW, S = 21.05 MVA, Q = 6.573 MVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 33,000 V, I = 368.3 A, P = 20,000 kW, S = 21,050 kVA, Q = 6,573 kVAr, φ = 18.19° |
| Reference | Hand calculation: I = P / (√3 × V × pf) = 20,000,000 / (√3 × 33,000 × 0.95) = 368.3 A; S = √3 × V × I = √3 × 33,000 × 368.3 = 21,053 kVA; Q = S × sin φ = 21,053 × 0.3122 = 6,574 kVAr; φ = acos 0.95 = 18.19° (validation case c0001-20mw-33kv-pf095-solve-current) |
Example 13 — 500 kVA at 695.6 A — find the voltage
Voltage from an apparent-power rating and current, V = S / (√3·I): a 500 kVA, 415 V transformer.
Given: Balanced three-phase circuit, solve for voltage: I = 695.6 A, S = 500 kVA (apparent power), pf = cos φ = 0.8 (lagging), η = 100 %.
| Base units | S = 500 kVA = 500,000 VA |
|---|---|
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
| Voltage | V = S / (√3 × I) = 500,000 / (√3 × 695.6) = 500,000 / 1,204.8 = 415.00 V → 415.0 V (an apparent-power rating: pf and η do not enter) |
| Apparent power | S = √3 × V × I = √3 × 415.00 × 695.60 = 500,000 VA → 500.0 kVA |
| Active power | P = S × pf = 500,000 × 0.8 = 400,000 W → 400.0 kW |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 500,000 × 0.6000 = 300,000 VAr → 300.0 kVAr |
| Phase angle | φ = acos 0.8 = 36.870° → 36.86°; triangle check: P / S = 400,000 / 500,000 = 0.8000 = pf |
| Display | S = 500,000 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 400.0 kW, S = 500.0 kVA, Q = 300.0 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 415.0 V, I = 695.6 A, P = 400.0 kW, S = 500.0 kVA, Q = 300.0 kVAr, φ = 36.86° |
| Reference | Hand calculation: V = S / (√3 × I) = 500,000 / (√3 × 695.6) = 415 V; S = √3 × V × I = √3 × 415 × 695.6 = 500 kVA; P = S × pf = 500 × 0.8 = 400 kW; Q = S × sin φ = 500 × 0.6000 = 300 kVAr; φ = acos 0.8 = 36.87° (validation case c0001-500kva-695a-solve-voltage) |
Example 14 — 55 kW resistive load at 480 V, pf 1.0
Unity power factor where S and P agree only to floating-point precision: Q = S·sin 0° = 0 exactly.
Given: Balanced three-phase circuit, solve for current: VLL = 480 V, P = 55 kW (active power), pf = cos φ = 1 (lagging), η = 100 %.
| Base units | P = 55 kW = 55,000 W |
|---|---|
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
| Current | I = P / (√3 × V × pf) = 55,000 / (√3 × 480 × 1) = 55,000 / 831.38 = 66.155 A → 66.15 A |
| Apparent power | S = √3 × V × I = √3 × 480 × 66.155 = 55,000 VA → 55.00 kVA |
| Active power | P = 55 kW as given → 55.00 kW |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 55,000 × 0.0000 = 0 VAr → 0 kVAr |
| Phase angle | φ = acos 1 = 0.000° → 0.00°; triangle check: P / S = 55,000 / 55,000 = 1.0000 = pf |
| Display | S = 55,000 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 55.00 kW, S = 55.00 kVA, Q = 0 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 480.0 V, I = 66.15 A, P = 55.00 kW, S = 55.00 kVA, Q = 0 kVAr, φ = 0.00° |
| Reference | Hand calculation: I = P / (√3 × V × pf) = 55,000 / (√3 × 480 × 1) = 66.15 A; S = √3 × V × I = √3 × 480 × 66.15 = 55 kVA; Q = S × sin φ = 55 × 0.0000 = 0 kVAr; φ = acos 1 = 0.00° (validation case c0001-55kw-480v-pf1-solve-current) |
Example 15 — 50 A heater bank at 400 V, pf 1.0, η 92 %
Unity power factor with an efficiency: P_out = √3·V·I·η is below S, yet Q is still S·sin 0° = 0.
Given: Balanced three-phase circuit, solve for power: VLL = 400 V, I = 50 A, pf = cos φ = 1 (lagging), η = 92 %.
| Three-phase factor | √3 = 1.7321; balanced three-phase S = √3·VLL·IL, P = √3·VLL·IL·cos φ·η |
|---|---|
| Apparent power | S = √3 × V × I = √3 × 400 × 50 = 34,641 VA → 34.64 kVA |
| Active power | P = S × pf × η = 34,641 × 1 × 0.92 = 31,870 W → 31.86 kW (output power) |
| Reactive power | Q = S × sin φ = S × √(1 − pf²) = 34,641 × 0.0000 = 0 VAr → 0 kVAr (S is the input, so Q = Pin × tan φ with Pin = P / η = 34,641 W; not √(S² − Pout²)) |
| Phase angle | φ = acos 1 = 0.000° → 0.00° |
| Display | S = 34,641 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 31.86 kW, S = 34.64 kVA, Q = 0 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr |
| Result | V = 400.0 V, I = 50.00 A, P = 31.86 kW, S = 34.64 kVA, Q = 0 kVAr, φ = 0.00° |
| Reference | Hand calculation: S = √3 × 400 × 50 = 34.64 kVA; P = S × pf × η = 34.64 × 1 × 92 % = 31.87 kW; Q = S × sin φ = 34.64 × 0.0000 = 0 kVAr; check: P_in = P / η = 34.64 kW, Q = P_in × tan φ = 34.64 × 0.0000 = 0 kVAr; φ = acos 1 = 0.00° (validation case c0001-400v-50a-eta92-pf1-solve-power) |
Diagrams
The calculator generates two SVG diagrams alongside the numerical results:
- Phasor Diagram — Shows three voltage phasors (VA, VB, VC) separated by 120°, drawn as solid coloured arrows (red, yellow, blue). Current phasors (IA, IB, IC) are drawn as dashed arrows, each lagging its corresponding voltage phasor by the phase angle \(\phi\). An arc at the origin marks the angle between VA and IA.
- Power Triangle — A right triangle with active power \(P\) on the horizontal axis (green), reactive power \(Q\) on the vertical axis (cyan), and apparent power \(S\) as the hypotenuse (purple). The angle \(\theta\) between \(P\) and \(S\) equals the phase angle.
Standards & References
This calculator implements general three-phase power theory based on standard balanced three-phase circuit analysis. No specific standard is mandated — the equations are fundamental electrical engineering relationships derived from Kirchhoff's laws applied to symmetrical three-phase systems.
The formulas are consistent with the treatment found in:
- IEEE Std 141 (Red Book) — Electric Power Distribution for Industrial Plants
- IEC 60038 — Standard Voltages
- Any university-level power systems textbook (e.g., Glover, Sarma & Overbye)
Tips & Limitations
Tips
- Efficiency < 100% means the load draws more current than the ideal case — useful for motor circuits where shaft output is less than electrical input.
- VA-family units (VA, kVA, MVA) represent apparent power at the electrical terminals. When using these units, the calculator uses \(I = S / (\sqrt{3} \times V)\) directly — efficiency and power factor do not affect the solved current or voltage.
- Auto-scaling switches at the 1,000 and 1,000,000 boundaries, so results are always displayed in convenient units.
Limitations
- Balanced system only — assumes equal voltages and equal loading on all three phases. For unbalanced loads, use per-phase analysis instead.
- Power factor must be between 0 and 1. Lagging power factor is assumed (inductive load).
- Efficiency must be between 1% and 100%. A value of 100% means no losses in the load.
- The calculator returns zeroes when inputs are missing or produce mathematically invalid results (e.g., division by zero).
Developer Guide
Architecture
File layout:
components/c0001_vip_3_phase/component.py— Component registration, parameter schema, SVG generators (generate_phasor_diagram_svg,generate_power_triangle_svg)components/c0001_vip_3_phase/calculations.py— Pure math functions (no Django dependencies)components/c0001_vip_3_phase/equations.py— MathJax equation templates and unified article builder for PDF/HTML rendering
The component plugs into the calculator registry via
register_component(), which is called during
auto-discovery. The calculator class is
ThreePhaseCalculator, inheriting from
ElectricalCalculator.
Code Paths
Request flow:
URL
→ calculate_api view
→ registry.create_calculator(calc_type, params)
→ ThreePhaseCalculator.calculate()
→ solve_three_phase_ac(data) # calculations.py
→ results dictKey functions in calculations.py:
| Function | Purpose |
|---|---|
normalize_power(power, units) |
Pure unit scaling to base units (W or VA). No PF multiplication. |
is_apparent_power(units) |
Returns True for VA/KVA/MVA, False
for W/kW/MW. Used to select formula path. |
solve_three_phase_ac(data) |
Main calculation entry point. Branches on
solve_for and whether input is apparent or active
power to compute the unknown variable, then derives S, Q,
and phase angle. |
Auto-scaling is handled inline within
solve_three_phase_ac: if apparent power exceeds
1,000,000 the scale factor is M; above 1,000 it is
k; otherwise no prefix.
SVG generators in component.py:
| Function | Purpose |
|---|---|
generate_phasor_diagram_svg(parameters, results) |
Three-phase voltage/current phasor diagram |
generate_power_triangle_svg(parameters, results) |
P-Q-S power triangle |
generate_qr_code_svg(parameters, results, url) |
QR code linking to the calculation online |
Testing
Running tests:
pytest calcs/test_c0001_vip_3_phase.py -v
pytest calcs/ -k "vip_3_phase"How to extend:
- Add a new option to the
solve_forenum inPARAMETER_SCHEMAinsidecomponent.py. - Add a corresponding branch in
solve_three_phase_ac()incalculations.py. - Update
validate_parameters()inThreePhaseCalculatorto enforce the required inputs for the new mode. - Add equation templates in
equations.pywith a matching unified article entry. - Write tests covering the new solve mode, including edge cases (zero inputs, boundary values).