Electrical Power Calculator delivers accurate AC and DC load results, including real, apparent, reactive power, current, impedance, PF correction, and daily energy with precision.
How It’s Calculated
This is not an NEC ampacity or voltage-drop calculator — there’s no wire sizing or conductor table involved. It’s built on Ohm’s Law extended through the AC power triangle, plus the standard three-phase power formula. For DC circuits, power is simply voltage times current:
$$P_{DC} = V \times I$$
For single-phase AC, apparent power (S) is voltage times current, and real power (P) is apparent power scaled by power factor (PF):
$$S = V \times I \qquad P = S \times PF \qquad Q = S\sqrt{1-PF^2} \qquad \theta = \cos^{-1}(PF)$$
where Q is reactive power and θ is the phase angle between voltage and current. For three-phase AC, the same relationships apply, but apparent power picks up a $\sqrt{3}$ multiplier that accounts for the phase relationship across three conductors:
$$S = \sqrt{3} \times V_{LL} \times I_L \qquad P = \sqrt{3} \times V_{LL} \times I_L \times PF$$
$V_{LL}$ is line-to-line voltage and $I_L$ is line current — both are what you read directly off a meter or nameplate, not the internal phase voltage of a wye-connected source.
From there, the calculator derives an equivalent load impedance using the same active and reactive power values. For single-phase and DC circuits this is a direct Ohm’s Law impedance; for three-phase it’s a wye-equivalent impedance per phase, because a single Ω value can’t otherwise represent three conductors carrying current simultaneously:
$$Z = \frac{S}{I^2} \quad R = \frac{P}{I^2} \quad X = \frac{Q}{I^2} \quad \text{(single-phase / DC)}$$
$$Z = \frac{S}{3I_L^2} \quad R = \frac{P}{3I_L^2} \quad X = \frac{Q}{3I_L^2} \quad \text{(three-phase, per-phase wye equivalent)}$$
Finally, for AC circuits below a 0.95 power factor, the calculator estimates the capacitive correction needed to bring a lagging (inductive) load up to that target, along with the resulting drop in line current:
$$Q_c = P\left(\tan(\cos^{-1}(PF)) – \tan(\cos^{-1}(0.95))\right) \qquad I_{corrected} = \frac{P}{V \times k \times 0.95}$$
where $k = \sqrt{3}$ for three-phase and $k = 1$ for single-phase. Voltage and current can each be entered in base units or in kilo-units (kV, kA); switching that just multiplies the entered value by 1,000 before it reaches any of the formulas above.
Worked Example: 480V Three-Phase Load at 50A, 0.85 PF
Take a 480V three-phase circuit drawing 50A at a 0.85 lagging power factor — a realistic reading for a mid-size induction motor.
Apparent power first: $S = \sqrt{3} \times 480 \times 50 = 41{,}569\ \text{VA} \approx 41.57\ \text{kVA}$.
Real power scales that by the power factor: $P = 41.57 \times 0.85 \approx 35.33\ \text{kW}$. That’s the number that shows up on the utility bill.
Reactive power fills the rest of the triangle: $Q = 41.57 \times \sqrt{1-0.85^2} \approx 21.90\ \text{kVAR}$, at a phase angle of $\theta = \cos^{-1}(0.85) \approx 31.79°$.
The wye-equivalent per-phase impedance comes from dividing by $3I^2 = 3 \times 50^2 = 7{,}500$: $Z \approx 5.54\ \Omega$, $R \approx 4.71\ \Omega$, $X \approx 2.92\ \Omega$.
Since 0.85 is below the calculator’s 0.95 correction target, it estimates the capacitor bank needed to close that gap: about 10.28 kVAR, which would drop the line current from 50A to roughly 44.74A — a 5.26A reduction for the same 35.33 kW of real work.
| Quantity | Formula | Result |
|---|---|---|
| Apparent Power (S) | √3 × V × I | 41.57 kVA |
| Active Power (P) | S × PF | 35.33 kW |
| Reactive Power (Q) | S × sin(θ) | 21.90 kVAR |
| Phase Angle (θ) | cos⁻¹(PF) | 31.79° |
| Wye-Equivalent Impedance (Z) | S ÷ 3I² | 5.54 Ω |
| Resistance (R) | P ÷ 3I² | 4.71 Ω |
| Reactance (X) | Q ÷ 3I² | 2.92 Ω |
| Correction to 0.95 PF | P × (tanθ₁ − tanθ₂) | 10.28 kVAR |
| Corrected Line Current | P ÷ (V × √3 × 0.95) | 44.74 A |
| Continuous Energy (24 hr) | P × 24 | 848.01 kWh |
What the Result Means
Active power (kW) is the only figure that reflects work actually done — heat, torque, light. Apparent power (kVA) is what your service, transformer, or generator has to be sized for, because reactive current still flows through the conductors even though it does no net work. Apparent power is always equal to or greater than active power, never less.
The power factor correction section runs against a fixed internal target of 0.95, not an NEC or utility-specific citation — it’s simply the threshold the calculator uses to decide what to show. Below 0.95 PF, it reports the capacitive kVAR needed to reach that target and the resulting line current drop. At 0.95 PF or above, it instead reports how much current headroom exists before the load would fall back to that threshold at the same active power.
Power factor itself must fall between 0.01 and 1.00 — it’s a ratio of real to apparent power, so it can’t physically exceed 1.0. In DC mode, power factor is fixed at 1.00 and locked, since there’s no phase relationship between DC voltage and current and therefore no reactive component at all.
The continuous-energy figures (24-hour, 30-day, 365-day) only mean something if the load actually runs at that exact power level the whole time. Most real loads cycle, so treat the 30- and 365-day numbers as an upper bound, not a bill forecast.
What Changes the Result
- Phase selection. Three-phase applies a √3 multiplier that single-phase and DC don’t. For identical voltage and current entries, three-phase apparent power comes out about 73% higher than single-phase — this is the single biggest swing factor in the tool.
- Power factor. At fixed voltage and current, PF doesn’t change apparent power at all — S = V × I × (phase factor) is independent of PF. What PF changes is how that fixed apparent power splits between active and reactive power, and whether the correction section shows a required kVAR or a margin.
- Voltage/current unit selection. Switching an input from base units to kilo-units (V→kV, A→kA) multiplies the entered number by 1,000 before it hits any formula. Leaving a unit toggle on “k” by mistake is a thousand-fold error, not a rounding one.
- Wye-equivalent impedance in three-phase mode. The reported Ω, resistance, and reactance values describe a per-phase equivalent computed from line voltage and line current — not a physical winding measurement — and they only hold for a balanced three-phase system.
- Invalid or out-of-range entries. Voltage and current must be finite positive numbers; power factor must be 0.01–1.00 for AC modes. Anything outside those bounds halts the calculation rather than returning a distorted number.
Frequently Asked Questions
How do you calculate three-phase power from voltage and current?
Multiply line-to-line voltage by line current by √3, then by power factor for real power: $P = \sqrt{3} \times V_{LL} \times I_L \times PF$. Drop the √3 for single-phase circuits, where it’s just $P = V \times I \times PF$.
What’s the difference between kW and kVA?
kW is real power — the portion actually converted into work or heat. kVA is apparent power, the raw product of voltage and current before power factor is applied. The two are only equal when power factor is exactly 1.0; otherwise kVA is always the larger number.
Why is my apparent power higher than my real power?
Inductive loads like motors and transformers draw current that shifts out of phase with voltage. That out-of-phase share shows up as reactive power (kVAR) and still contributes to the voltage × current product, even though it does no net work — which is exactly why apparent power (kVA) comes out higher than real power (kW).
What power factor should I use if I don’t know mine?
A nameplate rating or a clamp-meter reading is always more accurate than an assumption. If neither is available, 0.85 is a common starting point for general inductive equipment, but changing this number only reallocates a fixed apparent power between real and reactive — it won’t change your kVA figure.
How much capacitor correction do I need to reach 0.95 power factor?
$Q_c = P(\tan(\cos^{-1}(PF)) – \tan(\cos^{-1}(0.95)))$. For a 35.33 kW load at 0.85 PF, that works out to about 10.28 kVAR of capacitive correction, which also drops line current by roughly 5.26A at the same real-power output.
Does this calculator work for DC circuits?
Yes. Selecting DC mode locks power factor at 1.00 and reactive power at zero, since DC has no phase relationship between voltage and current. It then reports power as $P = V \times I$, along with the load’s equivalent resistance, conductance, and source current required per kilowatt.
What voltage do I enter for three-phase power — line-to-line or phase voltage?
Enter line-to-line voltage — the value you’d read directly off a meter between two phase conductors. The three-phase formula ($\sqrt{3} \times V_{LL} \times I_L$) is built around that line value, and the calculator derives the per-phase wye-equivalent impedance internally rather than asking for it.
Why won’t the calculator accept a power factor above 1.00?
Power factor is a ratio of real power to apparent power, and real power can never exceed apparent power — so a PF above 1.0 isn’t physically possible. Entries above 1.00, below 0.01, or non-numeric, zero, or negative values for voltage, current, or power factor are all rejected and stop the calculation until corrected.
These results are for planning and estimation only. Any wiring, breaker sizing, or installation work should be verified against your local electrical code and performed or inspected by a licensed electrician.