Power Calculator

This professional-grade power calculator helps engineers, electricians, students, and makers compute electrical power accurately for DC and AC circuits (single- and three-phase). Enter any two known values (V&I, V&R, or I&R) and, when applicable, a power factor to instantly get real power in watts and kilowatts and apparent power in volt‑amperes.

Interactive Calculator

Select circuit type

For three-phase, enter line-to-line voltage and line current.

Choose known pair
V
Enter RMS line-to-line voltage for three-phase AC systems.
A
Enter RMS line current.
0–1
Power factor is the cosine of the phase angle between voltage and current. Use PF = 1 for purely resistive loads; typical motor loads range 0.7–0.95.
Ready

Results

Real Power (P) 0.00 W
Real Power 0.000 kW
Apparent Power (S) 0.00 VA
Derived quantity —
Equation used —

Data Source and Methodology

Authoritative reference: IEEE Std 1459-2010 — Definitions for the Measurement of Electric Power Quantities under Sinusoidal, Non-sinusoidal, Balanced, or Unbalanced Conditions (2010). IEEE Standards Association. View the standard.

This calculator implements the standard RMS power relationships for DC and AC circuits (single- and three‑phase) and Ohm’s law identities for resistive loads.

Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.

The Formula Explained

DC: $$P = V \cdot I$$
Resistive identity: $$P = I^2 R = \frac{V^2}{R}$$
Single-phase AC (real power): $$P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi$$
Three-phase AC (real power, 3-wire balanced): $$P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi$$
Single-phase apparent power: $$S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}$$
Three-phase apparent power: $$S = \sqrt{3}\; V_{LL} \cdot I_{L}$$

Glossary of Variables

    - V (V): RMS voltage in volts. For three-phase use line-to-line voltage V_LL.
    - I (A): RMS current in amperes. For three-phase use line current I_L.
    - R (Ω): Resistance for purely resistive loads.
    - P (W): Real (active) power in watts.
    - S (VA): Apparent power in volt-amperes.
    - PF (dimensionless): Power factor, PF = cos(φ), 0 ≤ PF ≤ 1.
    - φ (rad): Phase angle between voltage and current phasors.

Worked Example

How It Works: A Step-by-Step Example

Suppose you have a three-phase motor supplied at V_LL = 400 V drawing I_L = 12 A with PF = 0.86.

  1. Choose “AC 3-Phase” and “Voltage & Current”.
  2. Enter V = 400, I = 12, PF = 0.86.
  3. Apply the formula: P = √3 × V_LL × I_L × PF.
  4. Compute: √3 ≈ 1.732; P ≈ 1.732 × 400 × 12 × 0.86 ≈ 7,144 W = 7.144 kW.
  5. Apparent power: S = √3 × V_LL × I_L ≈ 1.732 × 400 × 12 ≈ 8,317 VA.

Frequently Asked Questions (FAQ)

Do I enter RMS or peak values?

Always use RMS values for voltage and current. The formulas are expressed in RMS terms for AC circuits.

What PF should I use if I don’t know it?

If unknown, you may start with PF = 0.8–0.9 for many motor loads. For heaters or incandescent lamps, PF is typically 1.

Is the three-phase formula valid for any configuration?

The listed formula assumes a balanced three-wire system using line-to-line voltage and line current. For unbalanced or four-wire systems, more detailed analysis is required.

Can I compute power with V and R on AC?

Only for purely resistive loads (PF ≈ 1). If reactance is present, you must use V and I with the correct power factor.

What’s the difference between watts and volt-amperes?

Watts measure real power consumed by the load. Volt-amperes measure apparent power, which combines real and reactive components.

Why is the result zero or NaN?

Ensure all required fields are filled with non‑negative numbers. PF must be between 0 and 1. The tool validates inputs and will highlight any issues.

Audit: Complete
Formula (LaTeX) + variables + units
This section shows the formulas used by the calculator engine, plus variable definitions and units.
Formula (extracted LaTeX)
\[','\]
','
Formula (extracted LaTeX)
\[P = V \cdot I\]
P = V \cdot I
Formula (extracted LaTeX)
\[P = I^2 R = \frac{V^2}{R}\]
P = I^2 R = \frac{V^2}{R}
Formula (extracted LaTeX)
\[P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi\]
P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi
Formula (extracted LaTeX)
\[P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi\]
P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi
Formula (extracted LaTeX)
\[S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}\]
S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}
Formula (extracted text)
DC: $P = V \cdot I$ Resistive identity: $P = I^2 R = \frac{V^2}{R}$ Single-phase AC (real power): $P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi$ Three-phase AC (real power, 3-wire balanced): $P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi$ Single-phase apparent power: $S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}$ Three-phase apparent power: $S = \sqrt{3}\; V_{LL} \cdot I_{L}$
Variables and units
  • No variables provided in audit spec.
Sources (authoritative):
Changelog
Version: 0.1.0-draft
Last code update: 2026-01-19
0.1.0-draft · 2026-01-19
  • Initial audit spec draft generated from HTML extraction (review required).
  • Verify formulas match the calculator engine and convert any text-only formulas to LaTeX.
  • Confirm sources are authoritative and relevant to the calculator methodology.
Verified by Ugo Candido on 2026-01-19
Profile · LinkedIn
, ', svg: { fontCache: 'global' } };

Power Calculator

This professional-grade power calculator helps engineers, electricians, students, and makers compute electrical power accurately for DC and AC circuits (single- and three-phase). Enter any two known values (V&I, V&R, or I&R) and, when applicable, a power factor to instantly get real power in watts and kilowatts and apparent power in volt‑amperes.

Interactive Calculator

Select circuit type

For three-phase, enter line-to-line voltage and line current.

Choose known pair
V
Enter RMS line-to-line voltage for three-phase AC systems.
A
Enter RMS line current.
0–1
Power factor is the cosine of the phase angle between voltage and current. Use PF = 1 for purely resistive loads; typical motor loads range 0.7–0.95.
Ready

Results

Real Power (P) 0.00 W
Real Power 0.000 kW
Apparent Power (S) 0.00 VA
Derived quantity —
Equation used —

Data Source and Methodology

Authoritative reference: IEEE Std 1459-2010 — Definitions for the Measurement of Electric Power Quantities under Sinusoidal, Non-sinusoidal, Balanced, or Unbalanced Conditions (2010). IEEE Standards Association. View the standard.

This calculator implements the standard RMS power relationships for DC and AC circuits (single- and three‑phase) and Ohm’s law identities for resistive loads.

Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.

The Formula Explained

DC: $$P = V \cdot I$$
Resistive identity: $$P = I^2 R = \frac{V^2}{R}$$
Single-phase AC (real power): $$P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi$$
Three-phase AC (real power, 3-wire balanced): $$P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi$$
Single-phase apparent power: $$S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}$$
Three-phase apparent power: $$S = \sqrt{3}\; V_{LL} \cdot I_{L}$$

Glossary of Variables

    - V (V): RMS voltage in volts. For three-phase use line-to-line voltage V_LL.
    - I (A): RMS current in amperes. For three-phase use line current I_L.
    - R (Ω): Resistance for purely resistive loads.
    - P (W): Real (active) power in watts.
    - S (VA): Apparent power in volt-amperes.
    - PF (dimensionless): Power factor, PF = cos(φ), 0 ≤ PF ≤ 1.
    - φ (rad): Phase angle between voltage and current phasors.

Worked Example

How It Works: A Step-by-Step Example

Suppose you have a three-phase motor supplied at V_LL = 400 V drawing I_L = 12 A with PF = 0.86.

  1. Choose “AC 3-Phase” and “Voltage & Current”.
  2. Enter V = 400, I = 12, PF = 0.86.
  3. Apply the formula: P = √3 × V_LL × I_L × PF.
  4. Compute: √3 ≈ 1.732; P ≈ 1.732 × 400 × 12 × 0.86 ≈ 7,144 W = 7.144 kW.
  5. Apparent power: S = √3 × V_LL × I_L ≈ 1.732 × 400 × 12 ≈ 8,317 VA.

Frequently Asked Questions (FAQ)

Do I enter RMS or peak values?

Always use RMS values for voltage and current. The formulas are expressed in RMS terms for AC circuits.

What PF should I use if I don’t know it?

If unknown, you may start with PF = 0.8–0.9 for many motor loads. For heaters or incandescent lamps, PF is typically 1.

Is the three-phase formula valid for any configuration?

The listed formula assumes a balanced three-wire system using line-to-line voltage and line current. For unbalanced or four-wire systems, more detailed analysis is required.

Can I compute power with V and R on AC?

Only for purely resistive loads (PF ≈ 1). If reactance is present, you must use V and I with the correct power factor.

What’s the difference between watts and volt-amperes?

Watts measure real power consumed by the load. Volt-amperes measure apparent power, which combines real and reactive components.

Why is the result zero or NaN?

Ensure all required fields are filled with non‑negative numbers. PF must be between 0 and 1. The tool validates inputs and will highlight any issues.

Audit: Complete
Formula (LaTeX) + variables + units
This section shows the formulas used by the calculator engine, plus variable definitions and units.
Formula (extracted LaTeX)
\[','\]
','
Formula (extracted LaTeX)
\[P = V \cdot I\]
P = V \cdot I
Formula (extracted LaTeX)
\[P = I^2 R = \frac{V^2}{R}\]
P = I^2 R = \frac{V^2}{R}
Formula (extracted LaTeX)
\[P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi\]
P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi
Formula (extracted LaTeX)
\[P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi\]
P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi
Formula (extracted LaTeX)
\[S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}\]
S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}
Formula (extracted text)
DC: $P = V \cdot I$ Resistive identity: $P = I^2 R = \frac{V^2}{R}$ Single-phase AC (real power): $P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi$ Three-phase AC (real power, 3-wire balanced): $P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi$ Single-phase apparent power: $S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}$ Three-phase apparent power: $S = \sqrt{3}\; V_{LL} \cdot I_{L}$
Variables and units
  • No variables provided in audit spec.
Sources (authoritative):
Changelog
Version: 0.1.0-draft
Last code update: 2026-01-19
0.1.0-draft · 2026-01-19
  • Initial audit spec draft generated from HTML extraction (review required).
  • Verify formulas match the calculator engine and convert any text-only formulas to LaTeX.
  • Confirm sources are authoritative and relevant to the calculator methodology.
Verified by Ugo Candido on 2026-01-19
Profile · LinkedIn
]], displayMath: [['\\[','\\]']] }, svg: { fontCache: 'global' } };, svg: { fontCache: 'global' } };

Power Calculator

This professional-grade power calculator helps engineers, electricians, students, and makers compute electrical power accurately for DC and AC circuits (single- and three-phase). Enter any two known values (V&I, V&R, or I&R) and, when applicable, a power factor to instantly get real power in watts and kilowatts and apparent power in volt‑amperes.

Interactive Calculator

Select circuit type

For three-phase, enter line-to-line voltage and line current.

Choose known pair
V
Enter RMS line-to-line voltage for three-phase AC systems.
A
Enter RMS line current.
0–1
Power factor is the cosine of the phase angle between voltage and current. Use PF = 1 for purely resistive loads; typical motor loads range 0.7–0.95.
Ready

Results

Real Power (P) 0.00 W
Real Power 0.000 kW
Apparent Power (S) 0.00 VA
Derived quantity —
Equation used —

Data Source and Methodology

Authoritative reference: IEEE Std 1459-2010 — Definitions for the Measurement of Electric Power Quantities under Sinusoidal, Non-sinusoidal, Balanced, or Unbalanced Conditions (2010). IEEE Standards Association. View the standard.

This calculator implements the standard RMS power relationships for DC and AC circuits (single- and three‑phase) and Ohm’s law identities for resistive loads.

Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.

The Formula Explained

DC: $$P = V \cdot I$$
Resistive identity: $$P = I^2 R = \frac{V^2}{R}$$
Single-phase AC (real power): $$P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi$$
Three-phase AC (real power, 3-wire balanced): $$P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi$$
Single-phase apparent power: $$S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}$$
Three-phase apparent power: $$S = \sqrt{3}\; V_{LL} \cdot I_{L}$$

Glossary of Variables

    - V (V): RMS voltage in volts. For three-phase use line-to-line voltage V_LL.
    - I (A): RMS current in amperes. For three-phase use line current I_L.
    - R (Ω): Resistance for purely resistive loads.
    - P (W): Real (active) power in watts.
    - S (VA): Apparent power in volt-amperes.
    - PF (dimensionless): Power factor, PF = cos(φ), 0 ≤ PF ≤ 1.
    - φ (rad): Phase angle between voltage and current phasors.

Worked Example

How It Works: A Step-by-Step Example

Suppose you have a three-phase motor supplied at V_LL = 400 V drawing I_L = 12 A with PF = 0.86.

  1. Choose “AC 3-Phase” and “Voltage & Current”.
  2. Enter V = 400, I = 12, PF = 0.86.
  3. Apply the formula: P = √3 × V_LL × I_L × PF.
  4. Compute: √3 ≈ 1.732; P ≈ 1.732 × 400 × 12 × 0.86 ≈ 7,144 W = 7.144 kW.
  5. Apparent power: S = √3 × V_LL × I_L ≈ 1.732 × 400 × 12 ≈ 8,317 VA.

Frequently Asked Questions (FAQ)

Do I enter RMS or peak values?

Always use RMS values for voltage and current. The formulas are expressed in RMS terms for AC circuits.

What PF should I use if I don’t know it?

If unknown, you may start with PF = 0.8–0.9 for many motor loads. For heaters or incandescent lamps, PF is typically 1.

Is the three-phase formula valid for any configuration?

The listed formula assumes a balanced three-wire system using line-to-line voltage and line current. For unbalanced or four-wire systems, more detailed analysis is required.

Can I compute power with V and R on AC?

Only for purely resistive loads (PF ≈ 1). If reactance is present, you must use V and I with the correct power factor.

What’s the difference between watts and volt-amperes?

Watts measure real power consumed by the load. Volt-amperes measure apparent power, which combines real and reactive components.

Why is the result zero or NaN?

Ensure all required fields are filled with non‑negative numbers. PF must be between 0 and 1. The tool validates inputs and will highlight any issues.

Audit: Complete
Formula (LaTeX) + variables + units
This section shows the formulas used by the calculator engine, plus variable definitions and units.
Formula (extracted LaTeX)
\[','\]
','
Formula (extracted LaTeX)
\[P = V \cdot I\]
P = V \cdot I
Formula (extracted LaTeX)
\[P = I^2 R = \frac{V^2}{R}\]
P = I^2 R = \frac{V^2}{R}
Formula (extracted LaTeX)
\[P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi\]
P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi
Formula (extracted LaTeX)
\[P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi\]
P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi
Formula (extracted LaTeX)
\[S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}\]
S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}
Formula (extracted text)
DC: $P = V \cdot I$ Resistive identity: $P = I^2 R = \frac{V^2}{R}$ Single-phase AC (real power): $P = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cdot \cos\varphi$ Three-phase AC (real power, 3-wire balanced): $P = \sqrt{3}\; V_{LL} \cdot I_{L} \cdot \cos\varphi$ Single-phase apparent power: $S = V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}$ Three-phase apparent power: $S = \sqrt{3}\; V_{LL} \cdot I_{L}$
Variables and units
  • No variables provided in audit spec.
Sources (authoritative):
Changelog
Version: 0.1.0-draft
Last code update: 2026-01-19
0.1.0-draft · 2026-01-19
  • Initial audit spec draft generated from HTML extraction (review required).
  • Verify formulas match the calculator engine and convert any text-only formulas to LaTeX.
  • Confirm sources are authoritative and relevant to the calculator methodology.
Verified by Ugo Candido on 2026-01-19
Profile · LinkedIn