Three-Phase Power Calculator

About the Three-Phase Power Calculator

Calculates three-phase power from line voltage, line current and power factor, or solves for the current given the power.

How to use this calculator

1. Choose the quantity you want to find, power, current, or voltage.
2. Enter the known values for line voltage, line current, and power factor.
3. Use the calculator to apply the three-phase power relationship.
4. Read the result and make sure the units match your setup, such as watts, amperes, and volts.

The formula explained

The formula \(P = \sqrt{3}\\,V_L I_L \cos\phi\) computes real three-phase power. Here, \(\cos\phi\) is the power factor, which tells you how much of the electrical power is doing useful work.

• P = real power, in watts \(\\text{W}\)
• \(V_L\) = line voltage, in volts \(\\text{V}\)
• \(I_L\) = line current, in amperes \(\\text{A}\)
• \(\cos\phi\) = power factor, a unitless number between \(0\) and \(1\)
• \(\phi\) = phase angle between voltage and current

Step by step method

1. Write down the known values for \(P\), \(V_L\), \(I_L\), and \(\cos\phi\).
2. If you are finding power, substitute into \(P = \sqrt{3}\\,V_L I_L \cos\phi\). If you are finding current or voltage, rearrange the formula first.
3. Multiply the values carefully, then use \(\sqrt{3} \\approx 1.732\).
4. Check that the result has the right unit, and round to a sensible precision.

Worked example

Problem. A three-phase motor has a line voltage of \(400\\,\\text{V}\), a line current of \(12\\,\\text{A}\), and a power factor of \(0.85\). Find the real power.

1. Start with \(P = \sqrt{3}\\,V_L I_L \cos\phi\).
2. Substitute the values: \(P = \sqrt{3} \\times 400 \\times 12 \\times 0.85\).
3. Calculate: \(P \\approx 1.732 \\times 400 \\times 12 \\times 0.85 = 7064.64\\,\\text{W}\).

Tips and common mistakes

• Make sure you use line values, not phase values, unless the calculator specifically asks for phase values.
• A power factor of \(1\) means all the power is real power, while lower power factors mean more current is needed for the same real power.

Frequently asked questions