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How to Calculate Power — DC, AC & 3-Phase Formulas

A working reference for how do we calculate power across DC, single-phase, and three-phase circuits, with the right formula for each (P = V × I for DC; V × I × cos φ for 1-φ AC; √3 × V_LL × I × cos φ for 3-φ). Includes calculate power factor, how to calculate power from current and voltage, calculate power in a circuit, and calculate power torque — all in one place with a calculator that handles every combination. Reviewed by a licensed PE.

Power calculator — DC, 1-φ AC, 3-φ AC

The embedded calculate power online tool covers every common combination — solve for P given V and I, for I given P and V, or for V given P and I. Switch between DC, single-phase AC (with PF), and three-phase AC (line-to-line + PF). Outputs both watts and horsepower, plus apparent power (VA) for transformer sizing.

CALC.007 Universal Power · V · I · P · R · 3 modes · 6 solve combos
V
A
Ω

Pure DC: P = V · I. Resistance shown is V/I (Ohm's law equivalent).

Voltage V
— V
Current I
— A
Power P
— W
Resistance R
— Ω
Apparent power S
— kVA
Reactive power Q
— kVAR
Power factor used
—
Mechanical equivalent
— HP
Heat output
— BTU/hr
Show your work
P = V · I = ...
FORMULA · P = V · I (DC) SOURCE · OHM 1827 · IEEE STD 100

Power formulas — by circuit type

Eq. 01 — DC power (Joule's law, three forms) SI
P=V⋅I=I2⋅R=V2RP = V \cdot I = I^2 \cdot R = \frac{V^2}{R}
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V in volts, I in amperes, R in ohms; P in watts.
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No power factor — DC current and voltage are always in phase.
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Three forms are algebraically identical; pick the one whose inputs you have.
Eq. 02 — Single-phase AC real power SI
P=V⋅I⋅cos⁡φP = V \cdot I \cdot \cos \varphi
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V = RMS voltage, I = RMS current; cos φ is the displacement power factor.
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For sinusoidal voltage and current; non-sinusoidal loads need true power factor (includes harmonics).
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For a 120 V × 10 A motor at PF 0.85: P = 120 × 10 × 0.85 = 1 020 W.
Eq. 03 — Three-phase AC real power (balanced) SI
P=3⋅VLL⋅IL⋅cos⁡φP = \sqrt{3} \cdot V_{LL} \cdot I_L \cdot \cos \varphi
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V_LL = line-to-line voltage, I_L = line current.
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Equivalent form using line-to-neutral: P = 3 × V_LN × I_L × cos φ.
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For 480 V × 100 A × PF 0.9: P = 1.732 × 480 × 100 × 0.9 = 74 822 W ≈ 74.8 kW.
Eq. 04 — Mechanical power (torque × speed) SI
P(W)=τ(N⋅m)⋅ω(rad/s);HP=τ(lb⋅ft)⋅N(rpm)5 252P_{(W)} = \tau_{(N \cdot m)} \cdot \omega_{(rad/s)}; \quad HP = \frac{\tau_{(lb \cdot ft)} \cdot N_{(rpm)}}{5\,252}
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For motor sizing: 1 HP = 745.7 W.
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The 5 252 constant ≈ 33 000 / (2π) — links lb·ft·rpm to HP via 33 000 ft·lbf/min.
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Used to size variable-frequency drives by mechanical load HP, not just nameplate.
Eq. 05 — Power triangle (S, P, Q) SI
S2=P2+Q2;PF=PS=cos⁡φS^2 = P^2 + Q^2; \quad PF = \frac{P}{S} = \cos \varphi
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S = apparent power (VA); P = real power (W); Q = reactive power (VAR).
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PF correction adds capacitive Q to cancel inductive Q from motors / transformers.
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See /power-factor/ for the full PF correction calculator.

Standards and authoritative references

StandardScopeRegion
NIST SP 811Definition of the watt and SI power unitsUSA / international
IEC 60050-131International Electrotechnical Vocabulary — power, energy, PF definitionsWorldwide
IEEE Std 1459-2010Definitions for the measurement of electrical power quantities under sinusoidal, non-sinusoidal, balanced, or unbalanced conditionsUSA
NEC / NFPA 70 — Article 220Branch-circuit, feeder, and service load calculationsUSA
NEC Article 430Motor power calculations and protective device sizingUSA
IEEE 519-2022Harmonic limits for non-linear loads — defines "true PF" vs displacement PFUSA

Typical power factors by load type

LoadTypical PFComments
Resistive heater, incandescent lamp1.00No reactive component
LED driver (PF-corrected)0.90–0.99Active PFC required > 25 W per IEC 61000-3-2
LED driver (uncorrected, low-cost)0.50–0.65Common in residential lamps
Switching power supply (server, PC)0.95+Active PFC mandatory above 75 W in EU/US
Induction motor at 100 % load0.85–0.92NEMA Premium motors at the upper end
Induction motor at 50 % load0.70–0.80PF degrades when oversized — common cause for PF correction
Induction motor at no-load0.05–0.30Almost pure magnetizing current
VFD with diode rectifier (no PFC)0.70–0.85 (true PF, low displacement PF ≈ 0.95)True PF reflects harmonic distortion
Welder, arc furnace0.30–0.70Highly variable, large PF correction needed
Capacitor bank (no load)0.0 leadingPure reactive current
  1. Identify the circuit type — DC, single-phase AC, or 3-phase AC. DC circuits: P = V × I (no power factor). Single-phase AC: P = V × I × PF (PF = cos φ, the displacement angle between voltage and current). Three-phase AC: P = √3 × V_LL × I × PF (line-to-line voltage). Always confirm whether the voltage given is line-to-line (V_LL) or line-to-neutral (V_LN) — using V_LN in the 3-phase formula misses a √3 factor.
  2. Measure or look up V, I, and PF (or P, depending on the unknown). Voltage and current come from a multimeter, clamp meter, or nameplate. Power factor for resistive loads (heaters, incandescent lamps) is 1.0. For motors at full load: ~ 0.85–0.95. For lightly-loaded motors: as low as 0.5. For LED drivers and switching power supplies: 0.95+ (for PF-corrected) or as low as 0.5 (uncorrected). Use a power-quality meter for an accurate field reading.
  3. Choose units consistently — W, kW, VA, kVA, or HP. Real power: W or kW. Apparent power: VA or kVA (= V × I without the PF). Mechanical power: HP (1 HP = 745.7 W). For three-phase 480 V motors: nameplate often gives both kW and HP — match the unit your downstream calculation needs (e.g., kVA for transformer sizing, kW for energy billing).
  4. Apply derating for service factor and continuous load. NEC 220 / 430: continuous loads (≥ 3 hours) are computed at 125 % of nameplate. Motors with 1.15 service factor can deliver 15 % more power for short periods (insulation system rated accordingly). Always verify the conductor and protective device are sized to the derated current, not the bare nameplate.
  5. Cross-check against the energy bill (kWh) for big-ticket loads. kWh on the utility bill = average power × hours of operation. A 5 kW continuous load running 24 × 30 = 720 h/month uses 3 600 kWh — at $ 0.15/kWh that's $ 540/month. If the bill doesn't match the calculation, suspect: (a) wrong PF in the equation, (b) intermittent load you didn't count, (c) miscounted phase voltage.

Worked example — 3-phase 480 V motor

A 50 HP induction motor on a 480 V 3-phase service, full-load amps from nameplate = 65 A, full-load PF = 0.88, efficiency 92.5 %. Calculate power 3 phase:

Step 1 — input electrical power: P_in = √3 × V_LL × I × PF = 1.732 × 480 × 65 × 0.88 = 47 597 W ≈ 47.6 kW.

Step 2 — output mechanical power: P_out = P_in × η = 47 597 × 0.925 = 44 027 W = 44.0 kW = 59.0 HP mechanical. Matches the 50 HP nameplate at typical full-load efficiency (motors are usually rated by output HP).

Step 3 — apparent power for transformer sizing: S = √3 × V × I = 1.732 × 480 × 65 = 54 086 VA ≈ 54.1 kVA. The upstream transformer must be sized in kVA, not kW.

Step 4 — reactive power: Q = √(S² − P²) = √(54.1² − 47.6²) = 25.7 kVAR. PF correction at the motor terminals would inject 25.7 kVAR capacitive to bring PF to 1.0.

Step 5 — energy at full load: 47.6 kW × 8 hours/day × 22 days/month = 8 380 kWh/month. At $ 0.12/kWh = $ 1 006/month operating cost.

Real vs apparent vs reactive power — what each one sizes

QuantitySymbol / unitWhat it sizesBill / nameplate?
Real powerP / W (or kW)The energy bill, motor output, heater wattskWh meter, motor nameplate kW
Apparent powerS / VA (or kVA)Transformer kVA, generator kVA, conductor ampacityTransformer / genset nameplate
Reactive powerQ / VAR (or kVAR)PF correction capacitor banks, voltage regulationCapacitor bank rating

The fundamental relationship — sizing wires and transformers from S (kVA) but billing energy as P (kWh) — is why utilities charge a "PF penalty" to large industrial customers: at low PF, the customer draws extra current (sized to S) without using extra energy (P).

Variants and edge cases

Calculate power for 3 phase — line-to-line vs line-to-neutral

The two equivalent formulas: P = √3 × V_LL × I × PF (line-to-line) or P = 3 × V_LN × I × PF (line-to-neutral). For a 480 V Y system: V_LL = 480 V, V_LN = 277 V. Both give the same answer. The most common error is using V_LL in the second formula: 3 × 480 × 65 × 0.88 = 82 kW (incorrectly inflated by √3).

Calculate power torque — for motors and gearboxes

Calculate power torque: P (kW) = τ (N·m) × n (rpm) / 9 549. Or in HP units: HP = τ (lb·ft) × n (rpm) / 5 252. Useful for sizing: a 100 lb·ft load at 1 750 rpm needs at least 33.3 HP at the shaft — pick a 40 HP motor for service-factor margin. For VFD applications, mechanical torque often varies with speed (constant-torque vs variable-torque load curves).

Calculate power cable size — sizing the conductor for a power load

For a known motor power, the conductor sizes from current, not from kW directly. Compute current first: I = P / (V × PF) for 1-φ or I = P / (√3 × V × PF) for 3-φ. Then look up AWG or mm² in NEC 310.16 / IEC 60228. For 50 kW @ 480 V 3-φ PF 0.9: I = 50 000 / (1.732 × 480 × 0.9) = 66.8 A → #4 AWG copper THWN-2 (85 A NEC) — see the wire size calculator.

Frequently asked questions

How do we calculate power?
Power is the rate at which energy is delivered. For DC circuits: P = V × I (volts × amps = watts). For single-phase AC: P = V × I × cos φ where cos φ is the power factor. For three-phase AC: P = √3 × V_LL × I × cos φ where V_LL is the line-to-line voltage. Always use real measurements (or nameplate values) and verify the voltage type (line-to-line vs line-to-neutral) before plugging in.
How to calculate power factor?
Power factor (PF) is the ratio of real power (W) to apparent power (VA): PF = P / S = cos φ, where φ is the phase angle between voltage and current. There are three practical methods. (1) Direct measurement: a power-quality meter or VFD display reads PF directly. (2) From P and S: PF = P (kW) / S (kVA). A 480 V × 100 A 3-phase load = 83.1 kVA apparent; if the wattmeter reads 70 kW then PF = 70 / 83.1 = 0.84. (3) From the impedance angle: cos(arctan(X / R)) where X is reactance and R is resistance of the equivalent circuit. Resistive loads PF = 1.0; induction motors at full load 0.85–0.92; lightly loaded motors as low as 0.5; capacitor banks lead with negative φ.
How to calculate power from current and voltage?
For DC or purely resistive AC: P (W) = V (volts) × I (amps). A 12 V × 5 A DC load = 60 W. For single-phase AC with reactive load: include power factor → P = V × I × cos φ. A 120 V × 8 A motor at PF 0.85 → 120 × 8 × 0.85 = 816 W. For three-phase: P = √3 × V_LL × I × cos φ; a 480 V × 50 A motor at PF 0.9 → 1.732 × 480 × 50 × 0.9 = 37 411 W ≈ 37.4 kW.
How to calculate power in electrical?
Three independent power quantities matter in AC electrical: real power P (W) = V × I × cos φ (does work); apparent power S (VA) = V × I (sizes the conductor and transformer); reactive power Q (VAR) = V × I × sin φ (oscillates between source and load, no net work). Power triangle: S² = P² + Q². For DC: P = V × I = I² × R = V² / R (Joule's law).
How to calculate power in a circuit?
For any single resistive element use Joule's law: P = V × I = I² × R = V² / R. For series circuits, the total power equals the sum of individual element powers (or apply Joule's law to total V and total I). For parallel, the total current is the sum of branch currents at the fixed source voltage, so P_branch = V × I_branch (DC) or V × I_branch × cos φ_branch (AC). For mixed RLC AC circuits compute the complex impedance Z = R + jX, then I = V / |Z|, then P = |V| × |I| × cos(arg Z). The embedded calculator above handles every common case including 3-phase.
What is the formula for calculate power torque?
Mechanical power from torque and rotational speed: P (W) = τ (N·m) × ω (rad/s). In imperial / motor terms: HP = τ (lb·ft) × N (rpm) / 5 252 — the constant 5 252 accounts for the 33 000 ft·lbf/min in 1 HP and the 2π conversion from rpm to rad/s. For a 100 lb·ft motor at 1 800 rpm: HP = 100 × 1 800 / 5 252 = 34.3 HP ≈ 25.6 kW. See the RPM calculator for the full torque-power-speed engine.

The watt — defined by James Watt's steam-engine work

The watt, symbol W, is the unit of power equal to one joule per second. The unit was so named in 1882 by the British Association for the Advancement of Science, in honour of James Watt (1736–1819) and his contribution to the development of the steam engine. The watt is a derived SI unit — equivalent to kg·m²·s⁻³, or one volt-ampere of in-phase electrical power. The horsepower (1 HP = 745.7 W) was defined by Watt himself as the mechanical power one draft horse could sustain over a working day.

11th Conférence Générale des Poids et Mesures (CGPM) → Resolution 12, 1960 — definition of the watt as the SI unit of power

Related calculators and references

Sources and further reading

  1. NIST. SP 811 — Guide for the Use of the International System of Units (SI). Definition of the watt and derived units.
  2. IEEE. IEEE Std 1459-2010 — Definitions for the Measurement of Electric Power Quantities under Sinusoidal, Nonsinusoidal, Balanced, or Unbalanced Conditions.
  3. NFPA. NFPA 70 — National Electrical Code (2023), Article 220 (load calculations) and Article 430 (motor power).
  4. IEC. IEC 60050-131 — International Electrotechnical Vocabulary, Part 131: Circuit theory.
  5. Stevenson, W.D. Elements of Power System Analysis, 4th ed., McGraw-Hill, 1982. Foundational text on real / reactive / apparent power.
  6. Bergen, A.R. & Vittal, V. Power Systems Analysis, 2nd ed., Prentice Hall, 2000. Modern derivation of P, Q, S in three-phase systems.