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Three Phase Power Calculator: kW, kVA, Amps, Star/Delta & Neutral

Calculate balanced three-phase kW, kVA and amps, convert motor hp, and solve unbalanced star or delta loads with neutral-current phasors and animated waveforms.

Wired White

Published 28. October 2026 · version 1

#motor horsepower #neutral current #phasors #power-factor #star-delta #three-phase power #unbalanced loads

Use this three phase power calculator to convert between electrical kW, kVA and line amps, compare star and delta connections, or find the neutral current of an unbalanced load. It is intended for engineers checking a load schedule, students learning phasors, and technicians interpreting equipment data. All entered voltages and currents are RMS values.

Balanced power and motor conversions

Choose the balanced model and select the quantity you know: line current, total electrical kW, total kVA, or motor shaft horsepower. Enter the line-to-line voltage and displacement power factor. For a kW to amps three phase calculation, the tool divides real power by the product of line voltage, power factor and √3. Leading power factor gives negative reactive power; lagging gives positive reactive power.

The motor option treats horsepower as mechanical shaft output and divides by efficiency to obtain electrical input. Efficiency is a fraction: enter 0.90 for 90%. Mechanical hp is about 745.7 W; it is different from metric horsepower, PS. The displayed electrical hp equivalent is a unit conversion, not a shaft rating. Use actual operating efficiency and power factor rather than assuming they stay constant at every load.

Star, delta and imbalance

In star, the voltage across a load branch is line voltage divided by √3, and branch current equals line current. An unbalanced star calculation assumes a solid neutral holding the load star point at source neutral potential. Enter the current, power factor and leading or lagging direction separately for A–N, B–N and C–N.

In delta, each branch sees the line voltage. Enter AB, BC and CA branch currents; the calculator obtains line currents by phasor subtraction. The familiar √3 ratio between line and branch current applies only to balanced delta loads. Changing the connection here preserves the entered currents or power; it does not simulate reconnecting a fixed set of impedances.

For a neutral current unbalanced load check, the tool adds current phasors rather than current magnitudes. It reports kW, signed kvar and kVA for every branch and the totals. Under unequal power factors, the sum of branch kVA can exceed the magnitude of total complex power; both are shown with explicit definitions.

Reading the animation and limits

Press play to rotate the RMS current phasors through a slowed electrical cycle. The moving waveform cursor shows the same instant. Waveform peaks are √2 times RMS values.

The model covers sinusoidal steady state with balanced, ideal supply voltages and ABC phase sequence. It excludes floating star points, voltage drop, startup, faults and harmonic calculations. Triplen harmonics can add in a neutral instead of cancelling. This tool does not select conductor sizes or protective devices. Installation work belongs to a qualified person; check the applicable standard in force.

How it is calculated

For a balanced load, $S=\sqrt{3}V_{LL}I_L$ , $P=S\cos\varphi$ and $Q=S\sin\varphi$ . Star: $V_{ph}=V_{LL}/\sqrt{3}$ and $I_{ph}=I_L$ . Balanced delta: $V_{ph}=V_{LL}$ and $I_L=\sqrt{3}I_{ph}$ . See Kuphaldt, attached PDF pages 323–325 (printed 314–316).

For unbalanced loads, apply $P_k=V_k I_k\mathrm{PF}_k$ , $Q_k=V_k I_k\sin\varphi_k$ and $S_k=V_k I_k$ to each branch, then sum real and reactive powers. The arithmetic apparent-power sum is $S_{sum}=\sum S_k$ ; the total complex-power magnitude is $S_{net}=\sqrt{(\sum P_k)^2+(\sum Q_k)^2}$ . The displayed aggregate ratio is $P/S_{sum}$ , not a claim about every meter's definition of power factor under imbalance. At no load this ratio displays 0.

Use the sine convention $i(t)=\sqrt{2}I\sin(\theta+\alpha_I)$ . With $V_{AN}$ at 0°, branch voltage angles are [0°, −120°, 120°] in star, and [30°, −90°, 150°] for AB, BC, CA in delta. Each current lags its own voltage by $+\arccos(\mathrm{PF})$ , or leads by the negative angle. Resolve currents into real and imaginary components. In star, neutral return (load → source) equals the phasor sum of incoming line currents; a neutral referenced source → load has the opposite angle but the same magnitude. In delta, $I_A=I_{AB}-I_{CA}$ and cyclically; there is no neutral. See VEMU, attached page 55, Example 5.

Motor shaft hp is converted using mechanical horsepower (approximately 745.7 W/hp): $P_{in}=P_{shaft}/\eta$ . Electrical input expressed in hp is $P_{in}/745.7$ with power in watts. Frequency sets the waveform period, not these RMS power relations.

Assumptions

Balanced, stiff sinusoidal supply with ABC phase sequence; RMS quantities throughout.
Loads consume non-negative real power. Power factor is displacement factor in [0, 1]; direction controls reactive-power sign.
Unbalanced star has an ideal connected neutral. Balanced star also represents a balanced three-wire load.
Delta currents are branch currents AB, BC and CA. Phasor subtraction gives line currents even when the branches differ.
Negligible source and conductor impedance, no harmonics, no neutral displacement. Motor efficiency applies only to the shaft-hp input mode.
Changing star/delta does not hold impedance constant. Zero-current angles and the no-load aggregate ratio display 0 by convention.

Where it applies

Sinusoidal steady state only. Do not apply the neutral result to floating-neutral star, broken-neutral, distorted-waveform, startup or fault conditions. Different leading and lagging angles can make neutral current exceed the largest phase current even without harmonics; the equal-power-factor shortcut must not be used for that situation. Triplen harmonic components can reinforce in the neutral; they are not included here. The arithmetic kVA sum and net complex-power magnitude are explicitly defined load summaries, not a conductor-thermal or universal unbalanced apparent-power standard. No conductor, breaker or neutral size is selected. Installation work belongs to a qualified person; check against the standard in force.

References

Tony R. Kuphaldt, Lessons in Electric Circuits, Volume II – AC, chapter 10, especially printed pp. 296–299 and 314–316 (attached PDF pp. 305–308, 323–325). https://www.ibiblio.org/kuphaldt/electricCircuits/AC/AC.pdf
VEMU Institute of Technology, Electrical Circuits II lecture notes, 2019–2020, Example 5, p. 55: unbalanced 230 V star load with 8, 10 and 20 ohm resistances. https://www.vemu.org/uploads/lecture_notes/04_01_2020_509222796.pdf
Paul Evans, The Engineering Mindset, Three Phase Voltage + Calculations, 30 May 2019: phase separation, RMS and sinusoidal waveform context. https://theengineeringmindset.com/three-phase-voltage-calculations/
Fluke, Determining load horsepower: motor efficiency separates shaft hp from electrical input; its worked example uses 472 V, 20 A, PF 0.85 and efficiency 0.90. The calculator uses exact √3 and mechanical hp rather than rounded 1.73 and 746 W. https://www.fluke.com/en-sg/learn/blog/power-quality/determining-load-horsepower-wiring-and-breaker-size-for-safe-and-efficient-installations

Inputs

Input	Symbol	Unit	Default
Load model	`load_case`		Unbalanced: three branch currents
Load connection Unbalanced star requires a solid neutral. Delta inputs refer to AB, BC and CA branches, not line currents.	`connection`		Star (Y), neutral connected if unbalanced
Line-to-line RMS voltage Use the voltage between two lines, e.g. 400 V or 208 V. Do not enter a line-to-neutral value.	`V_LL`	V	400
Supply frequency Sets the waveform time scale; RMS power results are independent of frequency at the entered current and power factor.	`f_supply`	Hz	50
Known balanced quantity	`known`		Line current → power
Line current, RMS	`I_entered`	A	32
Total electrical real power	`P_entered`	kW	18.8
Total apparent power	`S_entered`	kVA	22.2
Motor shaft output Mechanical horsepower: 1 hp ≈ 745.7 W, not metric PS. Efficiency converts shaft output to electrical input.	`hp_shaft`	hp	25
Motor efficiency (0–1)	`eta_motor`		0.9
Power factor magnitude (0–1)	`pf_bal`		0.85
Power factor direction	`q_bal`		Lagging / inductive (+Q)
RMS branch current Star: line-to-neutral current. Delta: current through this load branch, not the current in a line conductor.	`I_a`	A	32
Power factor magnitude (0–1)	`pf_a`		0.9
Power factor direction	`q_a`		Lagging / inductive (+Q)
RMS branch current Star: line-to-neutral current. Delta: current through this load branch, not the current in a line conductor.	`I_b`	A	20
Power factor magnitude (0–1)	`pf_b`		0.8
Power factor direction	`q_b`		Lagging / inductive (+Q)
RMS branch current Star: line-to-neutral current. Delta: current through this load branch, not the current in a line conductor.	`I_c`	A	12
Power factor magnitude (0–1)	`pf_c`		1
Power factor direction	`q_c`		Lagging / inductive (+Q)
Electrical angle ▶ Play a slowed electrical cycle. Phasors use RMS lengths; their vertical projection × √2 is the instantaneous current.	`theta`	deg	30

Results

With the default inputs above, the calc gives:

Result	Value
Total electrical real power	13.117 kW
Signed reactive power (+ inductive)	5.993 kvar
Sum of branch apparent powers	14.78 kVA
Magnitude of total complex power	14.421 kVA
Highest line current, RMS	32 A
Neutral return current, RMS	12.234 A
Voltage across each load branch	230.94 V
Real power / sum of branch kVA	0.8875
Electrical input power equivalent	17.591 hp

Checked against sources

Each case below was calculated by this calc and agrees with the value from its source (within the tolerance where one is shown).

Published balanced delta: three 10 kW loads at 120 V

Source: Kuphaldt, chapter 10, printed p. 316 / attached p. 325. Published currents are rounded to two decimals; input 30 kW is three 10 kW branches.

Inputs: Load model = Balanced: kW / kVA / amps / motor hp; Load connection = Delta (Δ), three wires; Known balanced quantity = Electrical kW → amps; Line-to-line RMS voltage = 120 V; Total electrical real power = 30 kW; Power factor magnitude (0–1) = 1; the others as above.

Result	Source	This calc
Highest line current, RMS	144.3 A	144.3 A
Three RMS load branch currents [1]	83.33 A	83.333 A
Total electrical real power	30 kW	30 kW

Published SPICE star: 120 V per phase and 30 kW

Source: Kuphaldt, printed pp. 297–299 / attached pp. 306–308, SPICE example. Use exact 120√3 V line voltage rather than its nominal rounded 208 V; SPICE prints 8.333E+01 A.

Inputs: Load model = Balanced: kW / kVA / amps / motor hp; Known balanced quantity = Electrical kW → amps; Line-to-line RMS voltage = 207.8 V; Total electrical real power = 30 kW; Power factor magnitude (0–1) = 1; the others as above.

Result	Source	This calc
Voltage across each load branch	120 V	120 V
Highest line current, RMS	83.33 A	83.333 A
Neutral return current, RMS	0 ± 1 × 10⁻⁸ A	1.421 × 10⁻¹⁴ A

Published unbalanced neutral: 8, 10, 20 ohms at 230 V

Source: VEMU Electrical Circuits II, 2019–2020, p. 55, Example 5. Currents follow the stated 230 V and 8/10/20 ohm resistances. Published neutral 15.21 A is rounded; the unrounded value is 15.21307 A.

Inputs: Line-to-line RMS voltage = 398.4 V; RMS branch current = 28.75 A; RMS branch current = 23 A; RMS branch current = 11.5 A; Power factor magnitude (0–1) = 1; Power factor magnitude (0–1) = 1; the others as above.

Result	Source	This calc
Neutral return current, RMS	15.21 A	15.213 A
Voltage across each load branch	230 V	230 V

Published two loaded phases: neutral is not an arithmetic sum

Source: Kuphaldt, printed pp. 296–297 / attached pp. 305–306, two 10 kW loads at 120 V on phases 120° apart. Reversing phase order changes the neutral angle, not the magnitude.

Inputs: Line-to-line RMS voltage = 207.8 V; RMS branch current = 83.333 A; RMS branch current = 83.333 A; RMS branch current = 0 A; Power factor magnitude (0–1) = 1; Power factor magnitude (0–1) = 1; the others as above.

Result	Source	This calc
Neutral return current, RMS	83.33 A	83.333 A
Total electrical real power	20 kW	20 kW

Published reactive delta example, with power typo corrected

Source: VEMU pp. 58–59, Example 8: 220 V delta, Z = 6+j8 ohm, I_phase = 22 A and line current 38.1 A. The printed 2204 W per phase is an arithmetic/typographical error: 22²×6 = 2904 W. A later 2.908 kW result uses a rounded/inaccurate rectangular current; the exact real component is 13.2 A. Expected real and reactive powers are independently recomputed from the published impedance and current, without altering the circuit equations.

Inputs: Load model = Balanced: kW / kVA / amps / motor hp; Load connection = Delta (Δ), three wires; Line-to-line RMS voltage = 220 V; Line current, RMS = 38.105 A; Power factor magnitude (0–1) = 0.6; the others as above.

Result	Source	This calc
Three RMS load branch currents [1]	22 A	22 A
Highest line current, RMS	38.1 A	38.105 A
Real power of each load branch [1]	2.904 kW	2.904 kW
Signed reactive power of each load branch [1]	3.872 kvar	3.872 kvar