Single-Phase V-I-P Calculator — help

Overview

The Single-Phase V-I-P Calculator solves for Voltage, Current, or Power in single-phase AC circuits. It mirrors the Three-Phase calculator (c0001) but without the \(\sqrt{3}\) factor. Use it for residential loads, single-phase equipment, or 240 V circuits.

Given any two of the three primary quantities (plus power factor and efficiency), the calculator derives the missing value and computes all related power quantities: active, reactive, and apparent power, along with the phase angle and a phasor diagram.

Input Parameters

Parameter	Type	Default	Range	Description
`solve_for`	Enum: Current / Power / Voltage	Current	Required	Which quantity to calculate from the other two
`voltage`	Number	240 V	min 0	Phase-to-neutral voltage (V). Hidden when solving for Voltage.
`current`	Number	73.53 A	min 0	Line current (A). Hidden when solving for Current.
`power`	Number	15	min 0	Active power in the selected units
`power_units`	Enum: Watts / Kilowatts / Megawatts / VA / KVA / MVA	Kilowatts	—	Unit for the power input. VA-based units are multiplied by PF internally.
`power_factor`	Number	0.85	0 – 1	Power factor (\(\cos\phi\)). Hidden when solving for Power.
`efficiency`	Number	100%	1 – 100%	Motor/equipment efficiency as a percentage

Output Results

Output	Unit	Description
`voltage`	V	Single-phase system voltage
`current`	A	Load current
`active_power`	W / kW / MW (auto-scaled)	Real power consumed by the load
`apparent_power`	VA / kVA / MVA (auto-scaled)	Total power (vector sum of P and Q)
`reactive_power`	VAr / kVAr / MVAr (auto-scaled)	Reactive (inductive/capacitive) power
`power_factor`	dimensionless	Ratio of active to apparent power
`phase_angle`	degrees	Angle between voltage and current phasors (\(\theta = \cos^{-1}(\text{PF})\))

Auto-scaling: Power outputs are automatically scaled based on the magnitude of the apparent power. Values above 1,000 display as kW/kVA/kVAr; above 1,000,000 as MW/MVA/MVAr.

Formulas

Core Single-Phase Power Equation

\[ P = V \times I \times \cos\phi \times \eta \]

where \(P\) = active power (W), \(V\) = voltage (V), \(I\) = current (A), \(\cos\phi\) = power factor, \(\eta\) = efficiency (decimal).

Solving for Current

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

Solving for Voltage

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

Apparent Power

\[ S = V \times I \]

Reactive Power

\[ Q = S \sin\phi \]

\(S\) 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: \(S^2 = P_{in}^2 + Q^2\)).

Key difference from Three-Phase (c0001): Single-phase formulas use a direct \(V \times I\) relationship. There is no \(\sqrt{3}\) multiplier. In a three-phase system the equations become \(P = \sqrt{3} \times V_{LL} \times I \times \cos\phi \times \eta\).

Power Normalization

When the input power_units is set to a VA-based unit (VA, KVA, MVA), the entered value represents apparent power. The calculator multiplies by the power factor to convert to active power before solving:

\[ P_{\text{active}} = S_{\text{input}} \times \cos\phi \]

Watt-based units (Watts, Kilowatts, Megawatts) are taken directly as active power with only a unit multiplier applied (1, 1000, or 1,000,000).

Worked Examples

The 12 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 (%)
1	2 kW at 10 A, pf 0.9 — find the voltage	voltage	222.2	10.00	2.000	2.222	0.9686	0.9	100
2	15 kW at 240 V, pf 0.85 — find the current	current	240.0	73.52	15.00	17.64	9.296	0.85	100
3	30 A at 230 V, pf 0.95, η 90 % — find the output power	power	230.0	30.00	5.899	6.900	2.154	0.95	90
4	5 kW at 25 A, pf 0.85 — find the voltage	voltage	235.2	25.00	5.000	5.882	3.098	0.85	100
5	1500 W heater at 120 V, pf 1.0	current	120.0	12.50	1.500	1.500	0	1	100
6	2 kVA site transformer at 110 V	current	110.0	18.18	1.600	2.000	1.200	0.8	100
7	10 A at 230 V, pf 0.5 — find the power	power	230.0	10.00	1.150	2.300	1.991	0.5	100
8	9 W LED lamp at 240 V, pf 0.6 — a very small load	current	240.0	0.06250	0.009000	0.01500	0.01200	0.6	100
9	2.2 kW single-phase motor at 240 V, pf 0.78, η 82 %	current	240.0	14.33	2.200	3.439	2.152	0.78	82
10	15 A circuit at 120 V, pf 0.9	power	120.0	15.00	1.620	1.800	0.7846	0.9	100
11	10 MW traction load at 25 kV, pf 0.95 — a very large load	current	25,000	421.0	10,000	10,520	3,286	0.95	100
12	1.85 kW resistive load at 110 V, pf 1.0	current	110.0	16.81	1.850	1.850	0	1	100

Example 1 — 2 kW at 10 A, pf 0.9 — find the voltage

Solving for the supply voltage of a single-phase load: V = P / (I·pf).

Given: Single-phase circuit, solve for voltage: I = 10 A, P = 2 kW (active power), pf = cos φ = 0.9 (lagging), η = 100 %.

Base units	P = 2 kW = 2,000 W
Voltage	V = P / (I × pf) = 2,000 / (10 × 0.9) = 2,000 / 9 = 222.22 V → 222.2 V
Apparent power	S = V × I = 222.22 × 10 = 2,222.2 VA → 2.222 kVA
Active power	P = 2 kW as given → 2.000 kW
Reactive power	Q = S × sin φ = S × √(1 − pf²) = 2,222.2 × 0.4359 = 968.64 VAr → 0.9686 kVAr
Phase angle	φ = acos 0.9 = 25.842° → 25.84°; triangle check: P / S = 2,000 / 2,222.2 = 0.9000 = pf
Display	S = 2,222.2 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 2.000 kW, S = 2.222 kVA, Q = 0.9686 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr
Result	V = 222.2 V, I = 10.00 A, P = 2.000 kW, S = 2.222 kVA, Q = 0.9686 kVAr, φ = 25.84°
Reference	Hand calculation: V = P / (I × pf) = 2,000 / (10 × 0.9) = 222.2 V; S = V × I = 222.2 × 10 = 2.222 kVA; Q = S × sin φ = 2.222 × 0.4359 = 0.9686 kVAr; φ = acos 0.9 = 25.84° (validation case `c0002-2kw-10a-solve-voltage`)

Example 2 — 15 kW at 240 V, pf 0.85 — find the current

The single-phase conversion I = P / (V·pf) — √3 times the three-phase current at the same voltage.

Given: Single-phase circuit, solve for current: V = 240 V, P = 15 kW (active power), pf = cos φ = 0.85 (lagging), η = 100 %.

Base units	P = 15 kW = 15,000 W
Current	I = P / (V × pf) = 15,000 / (240 × 0.85) = 15,000 / 204 = 73.529 A → 73.52 A
Apparent power	S = V × I = 240 × 73.529 = 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 = 240.0 V, I = 73.52 A, P = 15.00 kW, S = 17.64 kVA, Q = 9.296 kVAr, φ = 31.78°
Reference	Hand calculation: I = P / (V × pf) = 15,000 / (240 × 0.85) = 73.53 A; S = V × I = 240 × 73.53 = 17.65 kVA; Q = S × sin φ = 17.65 × 0.5268 = 9.296 kVAr; φ = acos 0.85 = 31.79° (validation case `c0002-15kw-240v-solve-current`)

Example 3 — 30 A at 230 V, pf 0.95, η 90 % — find the output power

With an efficiency the solved active power is the output: P = V·I·pf·η.

Given: Single-phase circuit, solve for power: V = 230 V, I = 30 A, pf = cos φ = 0.95 (lagging), η = 90 %.

Apparent power	S = V × I = 230 × 30 = 6,900 VA → 6.900 kVA
Active power	P = S × pf × η = 6,900 × 0.95 × 0.9 = 5,899.5 W → 5.899 kW (output power)
Reactive power	Q = S × sin φ = S × √(1 − pf²) = 6,900 × 0.3122 = 2,154.5 VAr → 2.154 kVAr (S is the input, so Q = P_in × tan φ with P_in = P / η = 6,555 W; not √(S² − P_out²))
Phase angle	φ = acos 0.95 = 18.195° → 18.19°
Display	S = 6,900 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 5.899 kW, S = 6.900 kVA, Q = 2.154 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr
Result	V = 230.0 V, I = 30.00 A, P = 5.899 kW, S = 6.900 kVA, Q = 2.154 kVAr, φ = 18.19°
Reference	Hand calculation: S = 230 × 30 = 6.9 kVA; P = S × pf × η = 6.9 × 0.95 × 90 % = 5.899 kW; Q = S × sin φ = 6.9 × 0.3122 = 2.155 kVAr; check: P_in = P / η = 6.555 kW, Q = P_in × tan φ = 6.555 × 0.3287 = 2.155 kVAr; φ = acos 0.95 = 18.19° (validation case `c0002-230v-30a-eta90-solve-power`)

Example 4 — 5 kW at 25 A, pf 0.85 — find the voltage

Voltage from active power and current, V = P / (I·pf).

Given: Single-phase circuit, solve for voltage: I = 25 A, P = 5 kW (active power), pf = cos φ = 0.85 (lagging), η = 100 %.

Base units	P = 5 kW = 5,000 W
Voltage	V = P / (I × pf) = 5,000 / (25 × 0.85) = 5,000 / 21.250 = 235.29 V → 235.2 V
Apparent power	S = V × I = 235.29 × 25 = 5,882.4 VA → 5.882 kVA
Active power	P = 5 kW as given → 5.000 kW
Reactive power	Q = S × sin φ = S × √(1 − pf²) = 5,882.4 × 0.5268 = 3,098.7 VAr → 3.098 kVAr
Phase angle	φ = acos 0.85 = 31.788° → 31.78°; triangle check: P / S = 5,000 / 5,882.4 = 0.8500 = pf
Display	S = 5,882.4 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 5.000 kW, S = 5.882 kVA, Q = 3.098 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr
Result	V = 235.2 V, I = 25.00 A, P = 5.000 kW, S = 5.882 kVA, Q = 3.098 kVAr, φ = 31.78°
Reference	Hand calculation: V = P / (I × pf) = 5,000 / (25 × 0.85) = 235.3 V; S = V × I = 235.3 × 25 = 5.882 kVA; Q = S × sin φ = 5.882 × 0.5268 = 3.099 kVAr; φ = acos 0.85 = 31.79° (validation case `c0002-5kw-25a-solve-voltage`)

Example 5 — 1500 W heater at 120 V, pf 1.0

A North American resistive appliance: at unity pf I = P / V, S = P and Q = 0.

Given: Single-phase circuit, solve for current: V = 120 V, P = 1,500 W (active power), pf = cos φ = 1 (lagging), η = 100 %.

Base units	P = 1,500 W = 1,500 W
Current	I = P / (V × pf) = 1,500 / (120 × 1) = 1,500 / 120 = 12.500 A → 12.50 A
Apparent power	S = V × I = 120 × 12.500 = 1,500 VA → 1.500 kVA
Active power	P = 1,500 W as given → 1.500 kW
Reactive power	Q = S × sin φ = S × √(1 − pf²) = 1,500 × 0.0000 = 0 VAr → 0 kVAr
Phase angle	φ = acos 1 = 0.000° → 0.00°; triangle check: P / S = 1,500 / 1,500 = 1.0000 = pf
Display	S = 1,500 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 1.500 kW, S = 1.500 kVA, Q = 0 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr
Result	V = 120.0 V, I = 12.50 A, P = 1.500 kW, S = 1.500 kVA, Q = 0 kVAr, φ = 0.00°
Reference	Hand calculation: I = P / (V × pf) = 1,500 / (120 × 1) = 12.5 A; S = V × I = 120 × 12.5 = 1.5 kVA; Q = S × sin φ = 1.5 × 0.0000 = 0 kVAr; φ = acos 1 = 0.00° (validation case `c0002-1500w-120v-heater-pf1`)

Example 6 — 2 kVA site transformer at 110 V

An apparent-power rating gives I = S / V directly; pf 0.8 splits it into P and Q.

Given: Single-phase circuit, solve for current: V = 110 V, S = 2 kVA (apparent power), pf = cos φ = 0.8 (lagging), η = 100 %.

Base units	S = 2 kVA = 2,000 VA
Current	I = S / (V) = 2,000 / (110) = 2,000 / 110 = 18.182 A → 18.18 A (an apparent-power rating: pf and η do not enter)
Apparent power	S = V × I = 110 × 18.182 = 2,000.0 VA → 2.000 kVA
Active power	P = S × pf = 2,000 × 0.8 = 1,600 W → 1.600 kW
Reactive power	Q = S × sin φ = S × √(1 − pf²) = 2,000.0 × 0.6000 = 1,200.0 VAr → 1.200 kVAr
Phase angle	φ = acos 0.8 = 36.870° → 36.86°; triangle check: P / S = 1,600 / 2,000.0 = 0.8000 = pf
Display	S = 2,000.0 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 1.600 kW, S = 2.000 kVA, Q = 1.200 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr
Result	V = 110.0 V, I = 18.18 A, P = 1.600 kW, S = 2.000 kVA, Q = 1.200 kVAr, φ = 36.86°
Reference	Hand calculation: I = S / V = 2,000 / 110 = 18.18 A; S = V × I = 110 × 18.18 = 2 kVA; P = S × pf = 2 × 0.8 = 1.6 kW; Q = S × sin φ = 2 × 0.6000 = 1.2 kVAr; φ = acos 0.8 = 36.87° (validation case `c0002-2kva-110v-site-transformer`)

Example 7 — 10 A at 230 V, pf 0.5 — find the power

A poor power factor: only half the apparent power is active, and Q = S·sin 60°.

Given: Single-phase circuit, solve for power: V = 230 V, I = 10 A, pf = cos φ = 0.5 (lagging), η = 100 %.

Apparent power	S = V × I = 230 × 10 = 2,300 VA → 2.300 kVA
Active power	P = S × pf = 2,300 × 0.5 = 1,150 W → 1.150 kW
Reactive power	Q = S × sin φ = S × √(1 − pf²) = 2,300 × 0.8660 = 1,991.9 VAr → 1.991 kVAr
Phase angle	φ = acos 0.5 = 60.000° → 60.00°; triangle check: P / S = 1,150 / 2,300 = 0.5000 = pf
Display	S = 2,300 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 1.150 kW, S = 2.300 kVA, Q = 1.991 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr
Result	V = 230.0 V, I = 10.00 A, P = 1.150 kW, S = 2.300 kVA, Q = 1.991 kVAr, φ = 60.00°
Reference	Hand calculation: S = 230 × 10 = 2.3 kVA; P = S × pf = 2.3 × 0.5 = 1.15 kW; Q = S × sin φ = 2.3 × 0.8660 = 1.992 kVAr; φ = acos 0.5 = 60.00° (validation case `c0002-230v-10a-pf05-solve-power`)

Example 8 — 9 W LED lamp at 240 V, pf 0.6 — a very small load

A tiny electronic load (S = 15 VA, below the 1 kVA scaling threshold): the current is tens of milliamps.

Given: Single-phase circuit, solve for current: V = 240 V, P = 9 W (active power), pf = cos φ = 0.6 (lagging), η = 100 %.

Base units	P = 9 W = 9 W
Current	I = P / (V × pf) = 9 / (240 × 0.6) = 9 / 144 = 0.062500 A → 0.06250 A
Apparent power	S = V × I = 240 × 0.062500 = 15 VA → 0.01500 kVA
Active power	P = 9 W as given → 0.009000 kW
Reactive power	Q = S × sin φ = S × √(1 − pf²) = 15 × 0.8000 = 12 VAr → 0.01200 kVAr
Phase angle	φ = acos 0.6 = 53.130° → 53.13°; triangle check: P / S = 9 / 15 = 0.6000 = pf
Display	S = 15 VA is up to 1,000 VA, so the calculator shows base units: P = 9.000 W, S = 15.00 VA, Q = 12.00 VAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr
Result	V = 240.0 V, I = 0.06250 A, P = 0.009000 kW, S = 0.01500 kVA, Q = 0.01200 kVAr, φ = 53.13°
Reference	Hand calculation: I = P / (V × pf) = 9 / (240 × 0.6) = 0.0625 A; S = V × I = 240 × 0.0625 = 0.015 kVA; Q = S × sin φ = 0.015 × 0.8000 = 0.012 kVAr; φ = acos 0.6 = 53.13° (validation case `c0002-9w-led-240v-small-load`)

Example 9 — 2.2 kW single-phase motor at 240 V, pf 0.78, η 82 %

A small motor rated by shaft output: losses and pf together raise the current well above P / V.

Given: Single-phase circuit, solve for current: V = 240 V, P = 2.2 kW (active power, equipment output), pf = cos φ = 0.78 (lagging), η = 82 %.

Base units	P = 2.2 kW = 2,200 W
Current	I = P / (V × pf × η) = 2,200 / (240 × 0.78 × 0.82) = 2,200 / 153.50 = 14.332 A → 14.33 A
Apparent power	S = V × I = 240 × 14.332 = 3,439.6 VA → 3.439 kVA
Active power	P = 2.2 kW as given → 2.200 kW (output power); input power = S × pf = 3,439.6 × 0.78 = 2,682.9 W; losses = 482.93 W
Reactive power	Q = S × sin φ = S × √(1 − pf²) = 3,439.6 × 0.6258 = 2,152.5 VAr → 2.152 kVAr (S is the input, so Q = P_in × tan φ with P_in = P / η = 2,682.9 W; not √(S² − P_out²))
Phase angle	φ = acos 0.78 = 38.739° → 38.73°
Display	S = 3,439.6 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 2.200 kW, S = 3.439 kVA, Q = 2.152 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr
Result	V = 240.0 V, I = 14.33 A, P = 2.200 kW, S = 3.439 kVA, Q = 2.152 kVAr, φ = 38.73°
Reference	Hand calculation: I = P / (V × pf × η) = 2,200 / (240 × 0.78 × 82 %) = 14.33 A; S = V × I = 240 × 14.33 = 3.44 kVA; Q = S × sin φ = 3.44 × 0.6258 = 2.152 kVAr; check: P_in = P / η = 2.683 kW, Q = P_in × tan φ = 2.683 × 0.8023 = 2.152 kVAr; φ = acos 0.78 = 38.74° (validation case `c0002-2200w-motor-240v-eta82`)

Example 10 — 15 A circuit at 120 V, pf 0.9

The active power a fully loaded 15 A, 120 V branch circuit delivers.

Given: Single-phase circuit, solve for power: V = 120 V, I = 15 A, pf = cos φ = 0.9 (lagging), η = 100 %.

Apparent power	S = V × I = 120 × 15 = 1,800 VA → 1.800 kVA
Active power	P = S × pf = 1,800 × 0.9 = 1,620 W → 1.620 kW
Reactive power	Q = S × sin φ = S × √(1 − pf²) = 1,800 × 0.4359 = 784.60 VAr → 0.7846 kVAr
Phase angle	φ = acos 0.9 = 25.842° → 25.84°; triangle check: P / S = 1,620 / 1,800 = 0.9000 = pf
Display	S = 1,800 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 1.620 kW, S = 1.800 kVA, Q = 0.7846 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr
Result	V = 120.0 V, I = 15.00 A, P = 1.620 kW, S = 1.800 kVA, Q = 0.7846 kVAr, φ = 25.84°
Reference	Hand calculation: S = 120 × 15 = 1.8 kVA; P = S × pf = 1.8 × 0.9 = 1.62 kW; Q = S × sin φ = 1.8 × 0.4359 = 0.7846 kVAr; φ = acos 0.9 = 25.84° (validation case `c0002-120v-15a-solve-power`)

Example 11 — 10 MW traction load at 25 kV, pf 0.95 — a very large load

A single-phase 25 kV railway feed above 1 MVA: the component switches to M-scaling.

Given: Single-phase circuit, solve for current: V = 25,000 V, P = 10 MW (active power), pf = cos φ = 0.95 (lagging), η = 100 %.

Base units	P = 10 MW = 10,000,000 W
Current	I = P / (V × pf) = 10,000,000 / (25,000 × 0.95) = 10,000,000 / 23,750 = 421.05 A → 421.0 A
Apparent power	S = V × I = 25,000 × 421.05 = 10,526,316 VA → 10,520 kVA
Active power	P = 10 MW as given → 10,000 kW
Reactive power	Q = S × sin φ = S × √(1 − pf²) = 10,526,316 × 0.3122 = 3,286,841 VAr → 3,286 kVAr
Phase angle	φ = acos 0.95 = 18.195° → 18.19°; triangle check: P / S = 10,000,000 / 10,526,316 = 0.9500 = pf
Display	S = 10,526,316 VA is above 1,000,000 VA, so the calculator shows M units: P = 10.00 MW, S = 10.52 MVA, Q = 3.286 MVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr
Result	V = 25,000 V, I = 421.0 A, P = 10,000 kW, S = 10,520 kVA, Q = 3,286 kVAr, φ = 18.19°
Reference	Hand calculation: I = P / (V × pf) = 10,000,000 / (25,000 × 0.95) = 421.1 A; S = V × I = 25,000 × 421.1 = 10,526 kVA; Q = S × sin φ = 10,526 × 0.3122 = 3,287 kVAr; φ = acos 0.95 = 18.19° (validation case `c0002-10mw-25kv-traction`)

Example 12 — 1.85 kW resistive load at 110 V, pf 1.0

Unity power factor on a 110 V site supply: I = P / V, S = P and Q = 0 exactly.

Given: Single-phase circuit, solve for current: V = 110 V, P = 1.85 kW (active power), pf = cos φ = 1 (lagging), η = 100 %.

Base units	P = 1.85 kW = 1,850 W
Current	I = P / (V × pf) = 1,850 / (110 × 1) = 1,850 / 110 = 16.818 A → 16.81 A
Apparent power	S = V × I = 110 × 16.818 = 1,850.0 VA → 1.850 kVA
Active power	P = 1.85 kW as given → 1.850 kW
Reactive power	Q = S × sin φ = S × √(1 − pf²) = 1,850.0 × 0.0000 = 0 VAr → 0 kVAr
Phase angle	φ = acos 1 = 0.000° → 0.00°; triangle check: P / S = 1,850 / 1,850.0 = 1.0000 = pf
Display	S = 1,850.0 VA is above 1,000 VA up to 1,000,000 VA, so the calculator shows k units: P = 1.850 kW, S = 1.850 kVA, Q = 0 kVAr. Every output is truncated (rounded down) to 4 significant figures; the tool reports kW/kVA/kVAr
Result	V = 110.0 V, I = 16.81 A, P = 1.850 kW, S = 1.850 kVA, Q = 0 kVAr, φ = 0.00°
Reference	Hand calculation: I = P / (V × pf) = 1,850 / (110 × 1) = 16.82 A; S = V × I = 110 × 16.82 = 1.85 kVA; Q = S × sin φ = 1.85 × 0.0000 = 0 kVAr; φ = acos 1 = 0.00° (validation case `c0002-1850w-110v-pf1-solve-current`)

Standards & References

This calculator implements general single-phase AC power theory as found in fundamental electrical engineering textbooks. No specific standard is targeted; the relationships \(P = V I \cos\phi\) and the power triangle \(S^2 = P^2 + Q^2\) are universally accepted.

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

Tips & Limitations

Default voltage: The default of 240 V represents a typical phase-to-neutral residential supply (Australia/UK). Adjust to 120 V for North American circuits or 230 V for IEC-standard regions.

Single-phase vs. three-phase: For three-phase loads, use the Three-Phase V-I-P Calculator (c0001) instead. At the same power level, a single-phase circuit draws \(\sqrt{3}\) times more current than a three-phase circuit, leading to larger conductors.

VA-based power units: When entering apparent power (VA, KVA, MVA), the calculator internally multiplies by the power factor to obtain active power. This is the same normalization behaviour as the three-phase calculator (c0001).

Efficiency: An efficiency of 100% means no losses. For motors, typical efficiencies range from 80% to 95%. Reducing efficiency increases the current drawn from the supply for a given mechanical output.

Division by zero: If any denominator term (voltage, current, power factor, or efficiency) is zero, the calculator returns zeroed results rather than raising an error. Ensure all required inputs are non-zero.

Architecture Developer

File Layout

components/c0002_vip_1_phase/component.py — SinglePhaseCalculator class, parameter schema, SVG generators, component registration
components/c0002_vip_1_phase/calculations.py — pure functions: normalize_power(), solve_single_phase_ac()
components/c0002_vip_1_phase/equations.py — MathJax equation templates and unified article builder
components/c0002_vip_1_phase/excel_generator.py — component-specific Excel report with step-by-step working

The structure is nearly identical to c0001 (Three-Phase). The calculator inherits from ElectricalCalculator (defined in calcs.core.base_calculator).

Code Paths Developer

Calculation Flow

Same flow as c0001:

SinglePhaseCalculator.validate_parameters() — checks required inputs based on solve_for
SinglePhaseCalculator.calculate() — builds a data dict, delegates to solve_single_phase_ac()
solve_single_phase_ac() — normalises power, solves the missing variable, computes S, Q, and phase angle

Key Difference from c0001

The formulas use V × I directly instead of √3 × V × I. This is the only computational difference; all supporting logic (power normalization, auto-scaling, error handling) is shared.

Key Functions

normalize_power(power, units, power_factor) — converts input power to Watts, applying VA → W conversion via PF when needed. Identical to c0001.
solve_single_phase_ac(data) — main calculation. No \(\sqrt{3}\) factor.
Auto-scaling — inline in solve_single_phase_ac(), scales power outputs to k/M based on magnitude of apparent power. Same logic as c0001.

SVG Generators

generate_phasor_diagram_svg() — renders a single voltage/current phasor pair (not three). Red solid line for voltage, blue dashed for current.
generate_power_triangle_svg() — renders the P-Q-S power triangle with phase angle arc.
generate_qr_code_svg() — QR code linking to the calculation URL.

Testing Developer

Run the component tests with:

pytest calcs/test_c0002_vip_1_phase.py -v

Tests cover each solve_for mode (Current, Voltage, Power), edge cases (zero inputs, boundary power factors), power normalization logic for all unit types, equation building, table data visibility, and the Excel generator.