Power-Motor Calculator (AC3)

Overview

Calculates electrical parameters for three-phase AC motors (AC3 duty): motor full-load current, apparent power, reactive power, and phase angle. Essential for motor circuit design, cable sizing, and switchgear selection.

AC3 duty covers the normal starting, running, and stopping cycle of squirrel-cage induction motors — the most common motor duty classification for industrial applications.

Input Parameters

Parameter	Type	Default	Range	Description
`power`	Number	90	min 0.01	Motor active power rating
`voltage`	Number	400 V	100 – 1000 V	Phase-to-phase voltage
`powerfactor`	Number	0.85	0.01 – 1.0	Motor power factor (\(\cos\phi\))
`efficiency`	Number	97%	1 – 100%	Motor efficiency (\(\eta\))
`powertype`	Enum: kW / watt / HP	kW	—	Power unit type

Output Results

Output	Unit	Description
`current`	A	Motor full-load current (3 sig figs)
`powerS`	kVA (or VA for watt mode)	Apparent power (3 sig figs)
`powerQ`	kVAR (or VAR for watt mode)	Reactive power (3 sig figs)
`phaseangle`	degrees	Phase angle between voltage and current (3 sig figs)

Formulas

Motor Full-Load Current (kW input)

\[ I = \frac{P \times 1000}{\eta \times \sqrt{3} \times V \times \cos\phi} \]

where \(P\) = active power (kW), \(V\) = phase-to-phase voltage (V), \(\cos\phi\) = power factor, \(\eta\) = efficiency (decimal).

Motor Full-Load Current (watt input)

\[ I = \frac{P}{\eta \times \sqrt{3} \times V \times \cos\phi} \]

where \(P\) = active power (W).

Motor Full-Load Current (HP input)

\[ I = \frac{P \times 1000 \times 0.7457}{\eta \times \sqrt{3} \times V \times \cos\phi} \]

where the HP-to-kW conversion factor is \(\text{HP\_TO\_KW} = 0.745699872\).

Apparent Power

\[ S = \frac{\sqrt{3} \times V \times I}{1000} \quad \text{(kVA)} \]

For watt mode, the result is in VA (the \(\div 1000\) factor is omitted).

Reactive Power

\[ Q = \sqrt{S^2 - P_{\text{elec}}^2} \]

where \(P_{\text{elec}} = P_{\text{shaft}} / \eta\) is the real electrical power drawn from the supply (shaft power divided by efficiency).

Phase Angle

\[ \theta = \arccos(\cos\phi) \times \frac{180}{\pi} \]

Historical bug fixes:

Reactive power was originally \(\sqrt{S^2 - (P + P)}\), then corrected to \(\sqrt{S^2 - P^2}\), and finally corrected to use \(P_{\text{elec}} = P_{\text{shaft}} / \eta\) instead of \(P_{\text{shaft}}\). The power triangle relates S, Q, and the real electrical input power, not the mechanical shaft power.
Phase angle used \(\cos(\text{PF})\) instead of \(\arccos(\text{PF})\) — corrected to use the inverse cosine function.
HP apparent power incorrectly divided by the HP→kW conversion factor — corrected to use the same \(S = \sqrt{3} \times V \times I / 1000\) formula as kW mode. Apparent power is always in kVA regardless of input power unit.

Worked Examples

The 11 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_motor; worked examples for every calculation).

#	Case	Rating	V (V)	pf / η	I (A)	S (kVA)	Q (kVAr)
1	90 kW motor at 400 V	90 kW	400	0.85 / 97 %	157	109	57.5
2	50 hp motor at 415 V	50 hp	415	0.8 / 92 %	70.4	50.6	30.3
3	2.2 kW motor at 400 V	2.2 kW	400	0.9 / 85 %	4.15	2.87	1.25
4	0.75 kW motor, low power factor	0.75 kW	415	0.72 / 77.4 %	1.87	1.34	0.933
5	7.5 kW IE3 motor at 400 V	7.5 kW	400	0.84 / 89.6 %	14.3	9.96	5.4
6	11 kW motor on a 230 V three-phase supply	11 kW	230	0.85 / 91 %	35.6	14.2	7.49
7	20 hp motor at 208 V	20 hp	208	0.84 / 91 %	54.1	19.5	10.5
8	100 hp motor at 460 V	100 hp	460	0.85 / 94.5 %	116	92.8	48.9
9	37 kW motor at 690 V	37 kW	690	0.86 / 94 %	38.2	45.7	23.3
10	315 kW motor at 1000 V	315 kW	1,000	0.88 / 96.5 %	214	370	176
11	15 kW at unity power factor	15 kW	400	1 / 90 %	24	16.6	0

Example 1 — 90 kW motor at 400 V

Baseline nameplate calculation: a 90 kW, 400 V motor at 0.85 pf and 97 % efficiency — current, input kW, kVA, kVAr and phase angle.

Given: Three-phase motor, rated output 90 kW, supply 400 V line-to-line, full-load power factor 0.85, efficiency 97 %.

Input power	P_in = P / η = 90.000 / 0.97 = 92.784 kW
Full-load current	I = P_in × 1000 / (√3 × V × cos φ) = 92,783.5 / (√3 × 400 × 0.85) = 92,783.5 / 588.90 = 157.55 A; engine reports 157 A (rounded down to 3 s.f.)
Apparent power	S = √3 × V × I / 1000 = √3 × 400 × 157.55 / 1000 = 109.157 kVA; engine reports 109 kVA (rounded down to 3 s.f.)
Reactive power	Q = √(S² − P_in²) = √(109.157² − 92.784²) = 57.502 kVAr; engine reports 57.5 kVAr (rounded down to 3 s.f.)
Phase angle	φ = acos(0.85) = 31.79°; engine reports 31.7° (rounded down to 3 s.f.)
Result	full-load current 157 A, 109 kVA at 400 V
Reference	Hand calculation: P_in = 90/0.97 = 92.784 kW; I = 92783.5 / (√3 × 400 × 0.85) = 157.55 A; S = √3 × 400 × 157.55 / 1000 = 109.16 kVA; Q = √(S² − P_in²) = 57.50 kVAr (= P_in·tan φ = 57.50); φ = acos(0.85) = 31.79° (validation case `c0004-90kw-400v-full-load-current`)

Example 2 — 50 hp motor at 415 V

A horsepower-rated motor: the output is converted at 1 hp = 0.7457 kW before the current is calculated.

Given: Three-phase motor, rated output 50 hp, supply 415 V line-to-line, full-load power factor 0.8, efficiency 92 %.

Output power	P = 50 hp × 0.7457 kW/hp = 37.285 kW
Input power	P_in = P / η = 37.285 / 0.92 = 40.527 kW
Full-load current	I = P_in × 1000 / (√3 × V × cos φ) = 40,527.2 / (√3 × 415 × 0.8) = 40,527.2 / 575.04 = 70.48 A; engine reports 70.4 A (rounded down to 3 s.f.)
Apparent power	S = √3 × V × I / 1000 = √3 × 415 × 70.48 / 1000 = 50.659 kVA; engine reports 50.6 kVA (rounded down to 3 s.f.)
Reactive power	Q = √(S² − P_in²) = √(50.659² − 40.527²) = 30.395 kVAr; engine reports 30.3 kVAr (rounded down to 3 s.f.)
Phase angle	φ = acos(0.8) = 36.87°; engine reports 36.8° (rounded down to 3 s.f.)
Result	full-load current 70.4 A, 50.6 kVA at 415 V
Reference	Hand calculation: P = 50 hp × 0.7457 = 37.285 kW; P_in = 37.285/0.92 = 40.527 kW; I = 40527.2 / (√3 × 415 × 0.8) = 70.48 A; S = √3 × 415 × 70.48 / 1000 = 50.66 kVA; Q = √(S² − P_in²) = 30.40 kVAr (= P_in·tan φ = 30.40); φ = acos(0.8) = 36.87° (validation case `c0004-50hp-415v`)

Example 3 — 2.2 kW motor at 400 V

A small motor where the lower efficiency (85 %) noticeably raises the input power above the shaft rating.

Given: Three-phase motor, rated output 2.2 kW, supply 400 V line-to-line, full-load power factor 0.9, efficiency 85 %.

Input power	P_in = P / η = 2.200 / 0.85 = 2.588 kW
Full-load current	I = P_in × 1000 / (√3 × V × cos φ) = 2,588.2 / (√3 × 400 × 0.9) = 2,588.2 / 623.54 = 4.15 A; engine reports 4.15 A (rounded down to 3 s.f.)
Apparent power	S = √3 × V × I / 1000 = √3 × 400 × 4.15 / 1000 = 2.876 kVA; engine reports 2.87 kVA (rounded down to 3 s.f.)
Reactive power	Q = √(S² − P_in²) = √(2.876² − 2.588²) = 1.254 kVAr; engine reports 1.25 kVAr (rounded down to 3 s.f.)
Phase angle	φ = acos(0.9) = 25.84°; engine reports 25.8° (rounded down to 3 s.f.)
Result	full-load current 4.15 A, 2.87 kVA at 400 V
Reference	Hand calculation: P_in = 2.2/0.85 = 2.588 kW; I = 2588.2 / (√3 × 400 × 0.9) = 4.15 A; S = √3 × 400 × 4.15 / 1000 = 2.88 kVA; Q = √(S² − P_in²) = 1.25 kVAr (= P_in·tan φ = 1.25); φ = acos(0.9) = 25.84° (validation case `c0004-2p2kw-400v`)

Example 4 — 0.75 kW motor, low power factor

A fractional-kilowatt motor at 0.72 pf: reactive power exceeds 60 % of the input kW and the phase angle passes 40°.

Given: Three-phase motor, rated output 0.75 kW, supply 415 V line-to-line, full-load power factor 0.72, efficiency 77.4 %.

Input power	P_in = P / η = 0.750 / 0.774 = 0.969 kW
Full-load current	I = P_in × 1000 / (√3 × V × cos φ) = 969.0 / (√3 × 415 × 0.72) = 969.0 / 517.54 = 1.87 A; engine reports 1.87 A (rounded down to 3 s.f.)
Apparent power	S = √3 × V × I / 1000 = √3 × 415 × 1.87 / 1000 = 1.346 kVA; engine reports 1.34 kVA (rounded down to 3 s.f.)
Reactive power	Q = √(S² − P_in²) = √(1.346² − 0.969²) = 0.934 kVAr; engine reports 0.933 kVAr (rounded down to 3 s.f.)
Phase angle	φ = acos(0.72) = 43.95°; engine reports 43.9° (rounded down to 3 s.f.)
Result	full-load current 1.87 A, 1.34 kVA at 415 V
Reference	Hand calculation: P_in = 0.75/0.774 = 0.969 kW; I = 969.0 / (√3 × 415 × 0.72) = 1.87 A; S = √3 × 415 × 1.87 / 1000 = 1.35 kVA; Q = √(S² − P_in²) = 0.93 kVAr (= P_in·tan φ = 0.93); φ = acos(0.72) = 43.95° (validation case `c0004-0p75kw-415v-low-pf`)

Example 5 — 7.5 kW IE3 motor at 400 V

A typical 4-pole IE3 motor using catalogue power factor and efficiency.

Given: Three-phase motor, rated output 7.5 kW, supply 400 V line-to-line, full-load power factor 0.84, efficiency 89.6 %.

Input power	P_in = P / η = 7.500 / 0.896 = 8.371 kW
Full-load current	I = P_in × 1000 / (√3 × V × cos φ) = 8,370.5 / (√3 × 400 × 0.84) = 8,370.5 / 581.97 = 14.38 A; engine reports 14.3 A (rounded down to 3 s.f.)
Apparent power	S = √3 × V × I / 1000 = √3 × 400 × 14.38 / 1000 = 9.965 kVA; engine reports 9.96 kVA (rounded down to 3 s.f.)
Reactive power	Q = √(S² − P_in²) = √(9.965² − 8.371²) = 5.407 kVAr; engine reports 5.4 kVAr (rounded down to 3 s.f.)
Phase angle	φ = acos(0.84) = 32.86°; engine reports 32.8° (rounded down to 3 s.f.)
Result	full-load current 14.3 A, 9.96 kVA at 400 V
Reference	Hand calculation: P_in = 7.5/0.896 = 8.371 kW; I = 8370.5 / (√3 × 400 × 0.84) = 14.38 A; S = √3 × 400 × 14.38 / 1000 = 9.96 kVA; Q = √(S² − P_in²) = 5.41 kVAr (= P_in·tan φ = 5.41); φ = acos(0.84) = 32.86° (validation case `c0004-7p5kw-400v-ie3`)

Example 6 — 11 kW motor on a 230 V three-phase supply

The same output at 230 V line-to-line draws √3 times the 400 V current — the delta-connected low-voltage rating of a 230/400 V motor.

Given: Three-phase motor, rated output 11 kW, supply 230 V line-to-line, full-load power factor 0.85, efficiency 91 %.

Input power	P_in = P / η = 11.000 / 0.91 = 12.088 kW
Full-load current	I = P_in × 1000 / (√3 × V × cos φ) = 12,087.9 / (√3 × 230 × 0.85) = 12,087.9 / 338.62 = 35.70 A; engine reports 35.6 A (rounded down to 3 s.f.)
Apparent power	S = √3 × V × I / 1000 = √3 × 230 × 35.70 / 1000 = 14.221 kVA; engine reports 14.2 kVA (rounded down to 3 s.f.)
Reactive power	Q = √(S² − P_in²) = √(14.221² − 12.088²) = 7.491 kVAr; engine reports 7.49 kVAr (rounded down to 3 s.f.)
Phase angle	φ = acos(0.85) = 31.79°; engine reports 31.7° (rounded down to 3 s.f.)
Result	full-load current 35.6 A, 14.2 kVA at 230 V
Reference	Hand calculation: P_in = 11/0.91 = 12.088 kW; I = 12087.9 / (√3 × 230 × 0.85) = 35.70 A; S = √3 × 230 × 35.70 / 1000 = 14.22 kVA; Q = √(S² − P_in²) = 7.49 kVAr (= P_in·tan φ = 7.49); φ = acos(0.85) = 31.79° (validation case `c0004-11kw-230v-delta`)

Example 7 — 20 hp motor at 208 V

A North American 208 V service: horsepower input and a low voltage give a high current for the rating.

Given: Three-phase motor, rated output 20 hp, supply 208 V line-to-line, full-load power factor 0.84, efficiency 91 %.

Output power	P = 20 hp × 0.7457 kW/hp = 14.914 kW
Input power	P_in = P / η = 14.914 / 0.91 = 16.389 kW
Full-load current	I = P_in × 1000 / (√3 × V × cos φ) = 16,389.0 / (√3 × 208 × 0.84) = 16,389.0 / 302.62 = 54.16 A; engine reports 54.1 A (rounded down to 3 s.f.)
Apparent power	S = √3 × V × I / 1000 = √3 × 208 × 54.16 / 1000 = 19.511 kVA; engine reports 19.5 kVA (rounded down to 3 s.f.)
Reactive power	Q = √(S² − P_in²) = √(19.511² − 16.389²) = 10.586 kVAr; engine reports 10.5 kVAr (rounded down to 3 s.f.)
Phase angle	φ = acos(0.84) = 32.86°; engine reports 32.8° (rounded down to 3 s.f.)
Result	full-load current 54.1 A, 19.5 kVA at 208 V
Reference	Hand calculation: P = 20 hp × 0.7457 = 14.914 kW; P_in = 14.914/0.91 = 16.389 kW; I = 16389.0 / (√3 × 208 × 0.84) = 54.16 A; S = √3 × 208 × 54.16 / 1000 = 19.51 kVA; Q = √(S² − P_in²) = 10.59 kVAr (= P_in·tan φ = 10.59); φ = acos(0.84) = 32.86° (validation case `c0004-20hp-208v`)

Example 8 — 100 hp motor at 460 V

A NEMA 100 hp motor at 460 V calculated from its nameplate pf and efficiency.

Given: Three-phase motor, rated output 100 hp, supply 460 V line-to-line, full-load power factor 0.85, efficiency 94.5 %.

Output power	P = 100 hp × 0.7457 kW/hp = 74.570 kW
Input power	P_in = P / η = 74.570 / 0.945 = 78.910 kW
Full-load current	I = P_in × 1000 / (√3 × V × cos φ) = 78,910.0 / (√3 × 460 × 0.85) = 78,910.0 / 677.23 = 116.52 A; engine reports 116 A (rounded down to 3 s.f.)
Apparent power	S = √3 × V × I / 1000 = √3 × 460 × 116.52 / 1000 = 92.835 kVA; engine reports 92.8 kVA (rounded down to 3 s.f.)
Reactive power	Q = √(S² − P_in²) = √(92.835² − 78.910²) = 48.904 kVAr; engine reports 48.9 kVAr (rounded down to 3 s.f.)
Phase angle	φ = acos(0.85) = 31.79°; engine reports 31.7° (rounded down to 3 s.f.)
Result	full-load current 116 A, 92.8 kVA at 460 V
Reference	Hand calculation: P = 100 hp × 0.7457 = 74.570 kW; P_in = 74.57/0.945 = 78.910 kW; I = 78910.0 / (√3 × 460 × 0.85) = 116.52 A; S = √3 × 460 × 116.52 / 1000 = 92.84 kVA; Q = √(S² − P_in²) = 48.90 kVAr (= P_in·tan φ = 48.90); φ = acos(0.85) = 31.79° (validation case `c0004-100hp-460v`)

Example 9 — 37 kW motor at 690 V

A 690 V mining/industrial supply: the current is 58 % of the 400 V value for the same motor.

Given: Three-phase motor, rated output 37 kW, supply 690 V line-to-line, full-load power factor 0.86, efficiency 94 %.

Input power	P_in = P / η = 37.000 / 0.94 = 39.362 kW
Full-load current	I = P_in × 1000 / (√3 × V × cos φ) = 39,361.7 / (√3 × 690 × 0.86) = 39,361.7 / 1,027.80 = 38.30 A; engine reports 38.2 A (rounded down to 3 s.f.)
Apparent power	S = √3 × V × I / 1000 = √3 × 690 × 38.30 / 1000 = 45.769 kVA; engine reports 45.7 kVA (rounded down to 3 s.f.)
Reactive power	Q = √(S² − P_in²) = √(45.769² − 39.362²) = 23.356 kVAr; engine reports 23.3 kVAr (rounded down to 3 s.f.)
Phase angle	φ = acos(0.86) = 30.68°; engine reports 30.6° (rounded down to 3 s.f.)
Result	full-load current 38.2 A, 45.7 kVA at 690 V
Reference	Hand calculation: P_in = 37/0.94 = 39.362 kW; I = 39361.7 / (√3 × 690 × 0.86) = 38.30 A; S = √3 × 690 × 38.30 / 1000 = 45.77 kVA; Q = √(S² − P_in²) = 23.36 kVAr (= P_in·tan φ = 23.36); φ = acos(0.86) = 30.68° (validation case `c0004-37kw-690v`)

Example 10 — 315 kW motor at 1000 V

A large motor at the 1000 V upper limit of the tool's voltage range.

Given: Three-phase motor, rated output 315 kW, supply 1,000 V line-to-line, full-load power factor 0.88, efficiency 96.5 %.

Input power	P_in = P / η = 315.000 / 0.965 = 326.425 kW
Full-load current	I = P_in × 1000 / (√3 × V × cos φ) = 326,424.9 / (√3 × 1,000 × 0.88) = 326,424.9 / 1,524.20 = 214.16 A; engine reports 214 A (rounded down to 3 s.f.)
Apparent power	S = √3 × V × I / 1000 = √3 × 1,000 × 214.16 / 1000 = 370.937 kVA; engine reports 370 kVA (rounded down to 3 s.f.)
Reactive power	Q = √(S² − P_in²) = √(370.937² − 326.425²) = 176.185 kVAr; engine reports 176 kVAr (rounded down to 3 s.f.)
Phase angle	φ = acos(0.88) = 28.36°; engine reports 28.3° (rounded down to 3 s.f.)
Result	full-load current 214 A, 370 kVA at 1,000 V
Reference	Hand calculation: P_in = 315/0.965 = 326.425 kW; I = 326424.9 / (√3 × 1000 × 0.88) = 214.16 A; S = √3 × 1000 × 214.16 / 1000 = 370.94 kVA; Q = √(S² − P_in²) = 176.19 kVAr (= P_in·tan φ = 176.19); φ = acos(0.88) = 28.36° (validation case `c0004-315kw-1000v`)

Example 11 — 15 kW at unity power factor

Boundary case pf = 1: the apparent power equals the input power, reactive power and phase angle are zero.

Given: Three-phase motor, rated output 15 kW, supply 400 V line-to-line, full-load power factor 1, efficiency 90 %.

Input power	P_in = P / η = 15.000 / 0.9 = 16.667 kW
Full-load current	I = P_in × 1000 / (√3 × V × cos φ) = 16,666.7 / (√3 × 400 × 1) = 16,666.7 / 692.82 = 24.06 A; engine reports 24 A (rounded down to 3 s.f.)
Apparent power	S = √3 × V × I / 1000 = √3 × 400 × 24.06 / 1000 = 16.667 kVA; engine reports 16.6 kVA (rounded down to 3 s.f.)
Reactive power	Q = √(S² − P_in²) = √(16.667² − 16.667²) = 0.000 kVAr; engine reports 0 kVAr (rounded down to 3 s.f.)
Phase angle	φ = acos(1) = 0.00°; engine reports 0° (rounded down to 3 s.f.)
Result	full-load current 24 A, 16.6 kVA at 400 V
Reference	Hand calculation: P_in = 15/0.9 = 16.667 kW; I = 16666.7 / (√3 × 400 × 1) = 24.06 A; S = √3 × 400 × 24.06 / 1000 = 16.67 kVA; Q = √(S² − P_in²) = 0.00 kVAr (= P_in·tan φ = 0.00); φ = acos(1) = 0.00° (validation case `c0004-15kw-unity-pf`)

Standards & References

This calculator implements three-phase motor analysis per IEC/AS/NZS standards. The AC3 duty classification is defined in IEC 60947-4-1 and covers the normal starting, running, and stopping of squirrel-cage induction motors.

For cable sizing or protection coordination based on the calculated full-load current, refer to the applicable national wiring standard (e.g., AS/NZS 3008, IEC 60364, NEC).

Tips & Limitations

Use nameplate values: Where available, use the motor nameplate values for power factor and efficiency rather than generic estimates. Nameplate data reflects the specific motor design and operating point.

Full-load current only: Starting current for squirrel-cage motors is typically 5–8× the full-load current. This calculator provides the steady-state full-load value only — do not use these results for starting duty (e.g., DOL starter coordination).

HP conversion: For HP-rated motors, the conversion factor used is \(1 \; \text{HP} = 0.7457 \; \text{kW}\) (more precisely, 0.745699872).

Efficiency as divisor: Efficiency is applied as a divisor (\(P / \eta\)), so a lower efficiency results in a higher current draw from the supply for the same mechanical output.

Balanced three-phase operation: Results assume balanced three-phase conditions — equal voltages on all three phases and a symmetrical motor load. Voltage unbalance or single-phasing conditions are not covered.

Architecture Developer

File Layout

components/c0004_power_motor/component.py — PowerMotorCalculator class, parameter schema, component registration
components/c0004_power_motor/calculations.py — pure calculation functions for motor current, apparent power, reactive power, and phase angle

The motor current logic in this calculator is shared with c0005 (Cable Voltage Drop), which reuses the same current calculation as a first step in its voltage drop analysis.

Code Paths Developer

Key Functions in `calculations.py`

calculate_motor_current(power, voltage, powerfactor, efficiency, powertype) — returns current in Amps. Handles kW, watt, and HP power types with appropriate unit conversion.
calculate_apparent_power(voltage, current, powertype) — returns \(S\) in kVA (or VA for watt mode).
calculate_reactive_power(apparent_power, active_power) — returns \(Q\) using the power triangle relationship.
calculate_phase_angle(powerfactor) — returns \(\theta\) in degrees via \(\arccos(\text{PF})\).

Testing Developer

Run the component tests with:

pytest calcs/test_c0004_power_motor.py -v

Tests cover all three power type modes (kW, watt, HP), boundary conditions, power triangle identities, and verify the corrected formulas for reactive power, apparent power, and phase angle.