3 Phase Power Calculator

Compute three-phase real, apparent, and reactive power from line voltage, line current, and power factor, with steps and examples instantly.

RMS VL-L in volts. Typical: 208, 400, 415, 480. Must be > 0.
RMS IL in amperes per line/conductor. Must be > 0.
Unitless cosine of phase angle. Typical: 0.8–0.95. Must be between 0 and 1.

Equation Preview

Helping Notes

  • Balanced three-phase: use line-to-line voltage and line current. Apparent power \(S=\sqrt{3}\,VI\).
  • Real power \(P=\sqrt{3}\,VI\cdot \text{PF}\); reactive \(Q=S\sin\varphi=S\sqrt{1-\text{PF}^2}\), with \(\varphi=\cos^{-1}(\text{PF})\).
  • RMS values assumed; results shown in W/kW, VA/kVA, and var/kvar.
  • For delta or wye, formulas hold with line quantities. (No extra inputs needed.)

Results

Apparent Power (S)

Real / Active Power (P)

Reactive Power (Q)

Power Triangle & Angle

Error

Steps


              

What is 3‑Phase Power Calculator?

A 3‑Phase Power Calculator evaluates real power (kW), apparent power (kVA), and reactive power (kVAR) in balanced or practical three‑phase circuits. It supports both wye (Y) and delta (Δ) connections and accepts either line quantities (line‑to‑line voltage \(V_L\), line current \(I_L\)) or per‑phase values (\(V_P\), \(I_P\)). For balanced loads the core relationships are:

About the 3‑Phase Power Calculator

The tool computes kW, kVA, and kVAR from any sensible set of inputs: voltage, current, and power factor; or any two of \(P\), \(Q\), \(S\). It lets you choose connection (Y/Δ), units, and frequency (for nameplate context) and shows every step—conversion, substitution, and simplification—in responsive formula blocks. Advanced modes include per‑phase entry for unbalanced diagnostics and the classic two‑wattmeter measurement summary for three‑wire systems.

How to Use this 3‑Phase Power Calculator

  1. Select the system type (Wye or Delta) and input mode (line or per‑phase).
  2. Enter \(V\), \(I\), and power factor \(\cos\varphi\) (lag/lead) or provide any two among \(P\), \(Q\), \(S\).
  3. Click calculate to obtain kW, kVA, kVAR, and (optionally) phase values using the wye/delta relationships.
  4. Review the step‑by‑step formula substitutions and power‑triangle relations \(S^{2}=P^{2}+Q^{2}\).
  5. Copy the results and steps for design checks, sizing conductors, or verifying nameplate data.

Examples

Example 1: From voltage, current, and PF (balanced)

\(V_L=400\,\text{V},\; I_L=32\,\text{A},\; \cos\varphi=0.85\;\text{(lag)}\).

Example 2: Find current from kW

\(P=15\,\text{kW},\; V_L=208\,\text{V},\; \cos\varphi=0.90\;\Rightarrow\; I_L=\dfrac{P}{\sqrt{3}V_L\cos\varphi}\approx46.26\,\text{A}\).

Example 3: Delta load

\(V_L=480\,\text{V},\; I_L=20\,\text{A},\; \cos\varphi=0.95\).

Example 4: Two‑wattmeter readings

\(W_1=8\,\text{kW},\; W_2=5\,\text{kW}\Rightarrow P=13\,\text{kW},\; Q=\sqrt{3}(8-5)\approx5.196\,\text{kVAr},\; S=\sqrt{P^2+Q^2}=14\,\text{kVA}\Rightarrow \cos\varphi=13/14\approx0.929\).

FAQs

What is the difference between kW, kVA, and kVAr?

kW is real power, kVAr is reactive power, and kVA is apparent power with \(S^{2}=P^{2}+Q^{2}\).

Do I use line or phase values?

Use line quantities with \(\sqrt{3}\) formulas; per‑phase values need the Y/Δ relationships shown above.

What is power factor?

\(\cos\varphi=P/S\). Lagging implies inductive loads (positive \(Q\)); leading implies capacitive compensation (negative \(Q\)).

How do I get current from kW?

\(I_L=P/(\sqrt{3}V_L\cos\varphi)\) for balanced systems.

What’s the difference between wye and delta?

In Y: \(V_L=\sqrt{3}V_P\) and \(I_L=I_P\). In Δ: \(V_L=V_P\) and \(I_L=\sqrt{3}I_P\).

Does frequency (50/60 Hz) change the formulas?

No; it affects equipment ratings and speed, not the algebra for \(P,Q,S\).

Can the calculator handle unbalanced loads?

Enter per‑phase data and sum results; strict \(\sqrt{3}\) shortcuts apply only to balanced systems.

How do I estimate power factor correction?

Use \(Q_c=P(\tan\varphi_1-\tan\varphi_2)\) to size capacitors for a target PF.

How are two‑wattmeter readings interpreted?

Total real power is \(W_1+W_2\); reactive power is \(\sqrt{3}(W_1-W_2)\).

Can I mix units?

Keep voltage in volts, current in amperes; outputs appear in kW, kVA, and kVAr with clear conversions.

Is single‑phase power just \(P=VI\cos\varphi\)?

Yes; three‑phase balanced adds the \(\sqrt{3}\) multiplier with line quantities.

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