Capacitor size calculator for motor

Capacitor size calculator for motor

Motor Capacitor Size Calculator

Motor Capacitor Size Calculator

Causes of Low Power Factor and Accurate Capacitor Size Calculation for Power Factor Correction:

Low power factor in electrical systems is primarily caused by inductive loads, where the current lags behind the voltage by 90° in a purely inductive circuit. This significant phase angle difference between current and voltage leads to a power factor of zero.

The following factors contribute to low power factor:

  1. Single-phase and three-phase induction motors: Induction motors often operate at poor power factors, ranging from 0.2 to 0.9, depending on the load condition.
  2. Varying load in power systems: Lightly loaded power systems experience reduced power factors due to a decrease in the ratio of real power to reactive power.
  3. Industrial heating furnaces.
  4. Electrical discharge lamps and arc lamps: High-intensity discharge lighting operates at low power factors.
  5. Transformers.
  6. Harmonic currents.

To achieve better efficiency and optimize power factor, capacitor banks must be sized accurately. Incorrectly sized capacitor banks can lead to cable overheating with an oversized bank or provide little benefit with an undersized bank, resulting in high electricity bills.

The power factor correction calculator parameters include:

  1. Power (in kW).
  2. Connection type: Single-phase or three-phase.

For three-phase calculations, additional parameters include:

  1. Voltage (line-to-line or line-to-neutral in volts).
  2. Load type (Y or delta).
  3. Old power factor (in units or percentage).
  4. Required power factor (in units or percentage).
  5. Frequency (in Hz).

For single-phase calculations, the following parameters are required:

  1. Voltage (in volts).
  2. Old power factor (in units or percentage).
  3. Required power factor (in units or percentage).
  4. Frequency (in Hz).

Steps for power factor correction calculation:

  1. Convert given power factors into angles using the formula: CosØ = power factor, Ø = Cos⁻¹(power factor).
  2. Calculate the angle of the old and new power factors required.
  3. Determine the required capacitance reactive power using the formula: Qc = P * (tan⁡Ø₁ – tan⁡Ø₂).
  4. Calculate the capacitance (C) using the formula: C = Qc / (V² * 2 * π * f).


  • Power factor is expressed in units ranging from 0 to 1 (e.g., 0.8, 0.9). If it’s given in percentage, convert it to units by dividing by 100 before using it in the formulas.
  • For Y-connected loads and line-to-line voltage, convert it to line-to-neutral voltage (phase voltage) by dividing by 1.73 (square root of 3).

Example for single-phase system:

Given: Voltage (V) = 230 V Power (P) = 1.5 kW Old power factor (p.f₁) = 0.7 (unit) New power factor (p.f₂) = 0.9 (unit) Frequency (f) = 50 Hz


  1. Calculate angles: Ø₁ = Cos⁻¹(0.7) ≈ 45.6°, Ø₂ = Cos⁻¹(0.9) ≈ 25.85°.
  2. Calculate required reactive power (Qc): Qc = 1.5 * 1000 * (tan⁡(45.6°) – tan⁡(25.85°)) ≈ 804.193 VARs or 0.804 KVARs.
  3. Calculate capacitance (C): C = 804.193 / (230² * 2 * 3.14 * 50) ≈ 48.41 µF.

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