A low power factor means a network carrying a large current to deliver a small amount of useful power: higher losses, lower voltage, and equipment occupied without producing output. Capacitors are the classic remedy — here you'll understand the concept and learn how to calculate the dose.
The Problem: Reactive Power
Inductive loads — motors, air conditioners, transformers — need, in addition to their real power (kW), reactive power (kVAr) to build their magnetic fields. This power oscillates between the source and the load without performing any work, but it occupies the network with real current. The power factor (PF) is the ratio of real to apparent power: the lower it is, the more current is needed to perform the same useful work.
The Solution: Compensation with Capacitors
A capacitor draws a leading current that opposes the lagging inductive current — so by installing it in parallel near the loads, reactive power is exchanged locally between the capacitor and the load instead of being drawn from the substation across the entire network. The results:
- Reduced total current in lines and transformers → lower I²R losses.
- Improved voltage at the ends of feeders.
- Freed-up capacity in transformers and cables for new loads.
- Avoidance of power-factor penalties in utility tariffs.
How Do You Calculate the Required Capacitor Rating?
In practice, ready-made tables are used that give the multiplication factor directly from the current and target power factors.
A load of 400 kW with a power factor of 0.8 is to be improved to 0.95. The multiplication factor from the tables ≈ 0.421:
So we select a capacitor bank with a nearby standard rating (such as 175 kVAr, in steps).
Try the power factor improvement calculator in the encyclopedia for instant calculations with any values.
Engineering Notes
- Don't over-compensate: a leading (capacitive) power factor raises the voltage and may cause resonance with harmonics.
- In networks rich in harmonics, detuned-reactor banks are used to avoid resonance.
- Staged compensation with an automatic power factor controller (APFC) keeps pace with load variation throughout the day.
Sample answer: Using the multiplication factor from the tables (≈0.421 for improvement from 0.8 to 0.95): Q = 400 × 0.421 ≈ 168 kVAr. The benefit for the transformer: the apparent power drops from 400/0.8 = 500 kVA to 400/0.95 ≈ 421 kVA, freeing up about 79 kVA of transformer capacity for additional loads, and the current drops by the same proportion, reducing copper losses and heating and improving the voltage.
Installing the entire compensation as a fixed block on a variable load. At night, when loads decrease, the compensation becomes excessive, the power factor becomes leading, and the voltage rises — the solution is staged banks with an automatic controller that switches stages in and out as needed.
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