#1 India's Top IT Training Institute
×

About Uncodemy

Know more about who we are and what we stand for.

About Us

About Us

Discover our mission to equip individuals with essential IT skills and industry expertise.

Read More →
UnCodeMy Gallery

UnCodeMy Gallery

Explore highlights from our IT training sessions and success stories.

Read More →
Mentors

Mentors

Meet the industry experts who train and guide you through your course.

Read More →
Free Resources

Free Resources

Free tutorials, interview questions, tools and calculators to help you learn.

Explore →
×

Free Resources

Learn, practice and plan your career — bilkul free.

Tutorials

Tutorials

Step-by-step written tutorials to learn any technology at your own pace.

Read More →
Free Counselling

Free Counselling

Book a free video counselling session and pick the right course for you.

Read More →
Online Compiler

Online Compiler

Write, run and test your code online — no setup required.

Read More →
⚡ FREE · STEP-BY-STEP

Three Phase Power Calculator: P, Q, S and Current

Solve balanced three-phase power problems for star and delta loads.

  • 🔺 Star and delta
  • ⚡ kW, kVA, kVAR
  • 📟 Two-wattmeter method

How to use the Three Phase Power Calculator

  1. 1. Choose what to findPower, line current from power, or two-wattmeter readings.
  2. 2. Enter voltage and pfLine voltage like 415 V and the power factor.
  3. 3. See P, Q, SPlus phase voltage, phase current and impedance per phase.

Three-phase power formulas

For a balanced load, P = √3 VL IL cos φ, Q = √3 VL IL sin φ and S = √3 VL IL. In star, Vph = VL/√3 and Iph = IL. In delta, Vph = VL and Iph = IL/√3. The power formula is the same for both.

Worked example: 415 V, 10 A, pf 0.8. P = 1.732 × 415 × 10 × 0.8 = 5,750 W, S = 7,188 VA and Q = 4,313 VAR. For a 7.5 kW motor with 88% efficiency and pf 0.85 on 415 V, the input is 8,523 W and IL = 8,523 / (1.732 × 415 × 0.85) = 13.95 A.

Two-wattmeter method and exam questions

Two wattmeters measure total power in a three-wire system: P = W1 + W2 and tan φ = √3 (W1 − W2)/(W1 + W2). With W1 = 3,000 W and W2 = 1,000 W, P = 4 kW, tan φ = 0.866, φ = 40.9° and pf = 0.756. At pf 0.5 one meter reads zero; below 0.5 it reads negative.

  • Find the line current of a 415 V, 20 kW load at 0.9 pf.
  • Find phase current and impedance for a delta load.
  • Find the power factor from two wattmeter readings.
  • Compare star and delta power for the same impedances (delta is three times more).

Frequently asked questions

Is the formula different for star and delta?

No. P = √3 VL IL cos φ works for both when you use line values. Only the phase voltage and phase current differ.

Why does one wattmeter read negative?

When the power factor is below 0.5, the phase angle makes one meter's reading negative. Reverse its current coil and subtract the reading.

What voltage should I use in India?

Low-voltage three-phase supply is 415 V line (about 240 V phase), at 50 Hz. Use the value given in your question.

Is this tool free?

Yes. You enter your name and WhatsApp number once to see your result, and all other Uncodemy tools then open without the form.

More free tools

See all free tools