Principle: Power factor (PF) correction is the practice of reducing the reactive power in an AC system to improve the power factor toward unity (1.0). Loads with inductance—such as motors, transformers, and fluorescent lighting ballasts—draw current that lags the voltage, resulting in a PF less than 1. While the real power (kW) does useful work, reactive power (kVAR) causes additional current flow, increasing losses and loading generators and transformers. Correcting PF typically means adding capacitors (which supply leading reactive power), using synchronous condensers, or employing active filters so that the net reactive power drawn from the source is minimized. This not only reduces total current but also improves voltage stability.
To raise PF from an initial value cosφ₁
to a desired cosφ₂
, you must supply the difference in reactive power. The fundamental relation is:
Qc = P (tanφ₁ − tanφ₂)
where Qc
is the reactive power (in kVAR) of the capacitors needed, and P
is the real power (in kW) of the load. (Recall that tanφ = Q/P
.)
For example, consider a 100 kW load at 0.75 PF (lagging). Here, cosφ₁ = 0.75
, so φ₁ = arccos(0.75) ≈ 41.4°
. To correct to 0.95 PF (φ₂ ≈ 18.2°
):
tanφ₁ = tan(41.4°) ≈ 0.88
tanφ₂ = tan(18.2°) ≈ 0.33
Qc = 100 × (0.88 − 0.33) = 55 kVAR
.
Approximately 55 kVAR of capacitors would be installed across that load or on the distribution bus to ideally raise the PF to about 0.95.
A factory has a measured demand of 500 kW at 0.8 PF, and the utility imposes a penalty for PF below 0.9. How much capacitance is needed to achieve 0.9 PF?
P = 500 kW
and cosφ₁ = 0.8
so that φ₁ = 36.87°
and tanφ₁ ≈ 0.75
.
cosφ₂ = 0.90
, φ₂ ≈ 25.84°
and tanφ₂ ≈ 0.484
.
Qc = 500 × (0.75 − 0.484) = 500 × 0.266 = 133 kVAR
.
Approximately 133 kVAR of capacitors are needed. If standard capacitor units are, say, 50 kVAR each, then installing 3 × 50 = 150 kVAR is common. This slight overshoot may raise the PF a bit above 0.9, which is acceptable (typically up to ~0.95–0.98 is beneficial, while one avoids exactly 1.0 to prevent a leading PF when loads drop).
Check: Originally, the reactive power is Q = 500 × tanφ₁ ≈ 500 × 0.75 = 375 kVAR
. With 150 kVAR added, the new reactive power is 375 − 150 = 225 kVAR
. The new PF becomes:
PF = 500 / √(500² + 225²) ≈ 0.91
(which is acceptable).
Qc
demand.
Poor power factor results in higher currents for the same amount of useful power, which causes:
Companies invest in capacitor banks to mitigate these issues, and the return on investment can be calculated by comparing the cost of the capacitors with the annual savings on utility bills.
Qc = P (tanφ₁ − tanφ₂)
(taking care with degree/radian conversion).
In summary, power factor correction calculations revolve around balancing reactive power. By supplying the required Qc
locally through capacitors (or other devices), the system reduces the reactive demand on the source. This not only improves efficiency and voltage stability but also helps avoid unnecessary capacity burdens on the infrastructure.