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## Power factor definition | Calculating Power Factor

### Power Factor Values

In a purely resistive circuit, the power factor is 1 (perfect), because the reactive power equals 0. Here, the power triangle would look like a horizontal line because the opposite (reactive power) side would have 0 length.

In a purely inductive circuit, the power factor is 0, because true power equals 0. Here, the power triangle would look like a vertical line, because the adjacent (true power) side would have 0 length.

The same could be said for a purely capacitive circuit. If there are no dissipative (resistive) components in the circuit, then the true power must be equal to 0, making any power in the circuit purely reactive.

The power triangle for a purely capacitive circuit would again be a vertical line (pointing down instead of up as it was for the purely inductive circuit).

### Importance of Power Factor

Power factor can be an important aspect to consider in an AC circuit because any power factor less than 1 means that the circuit’s wiring has to carry more current than what would be necessary with 0 reactance in the circuit to deliver the same amount of (true) power to the resistive load.

If our last example circuit had been purely resistive, we would have been able to deliver a full 269.236 watts to the load with the same 1.410 amps of current, rather than the mere 190.947 watts that it is presently dissipating with that same current quantity.

The poor power factor (pf) makes for an inefficient power delivery system.

### Power Factor

In AC circuits, the power factor is the ratio of the real power that is used to do work and the apparent power that is supplied to the circuit Current flow.

When all the power is reactive power with no real power P (usually inductive load) – the power factor is 0. The power factor can get values in the range from 0 to 1.

When all the power is real power P with no reactive power (resistive load) – the power factor is 1.

### Power factor definition

A power factor is equal to the real or true Real power P in watts (W) divided by the apparent power S in volt-ampere (VA):

PF = P(W) / S(VA)

PF – power factor.

P – Real power in watts (W).

S – apparent power – the magnitude of the complex power in volt-amps (VA).

### Power factor calculations

For sinusoidal current flow, the power factor PF is equal to the absolute value of the cosine of the apparent power VA phase angle φ (which is also is impedance phase angle):

PF = |cos φ|

PF – is the power factor.

φ – is the apprent power phase angle.

The real power P in watts (W) is equal to the apparent power |S| in volt-ampere (VA) times the power factor PF:

P(W) = S(VA) × PF = S(VA) × cos φ

When the circuit has a resistive impedance load, the real power P is equal to the apparent power |S| and the power factor PF is equal to 1:

PF(resistive load) = P / S = 1

The reactive power Q in volt-amps reactive (VAR) is equal to the apparent power |S| in volt-ampere (VA) times the sine of the phase angle φ:

Q(VAR) = S(VA) × sin φ

Single phase circuit calculation from real power meter reading P in kilowatts (kW), voltage V in volts (V), and current I in amps (A):

PF = cos φ = 1000 × P(kW) / (V(V) × I(A))

Three-phase circuit calculation from real power meter reading P in kilowatts (kW), line to line voltage VL-L in volts (V) and current I in amps (A):

PF = cos φ = 1000 × P(kW) / (3 × VL-L(V) × I(A))

Three-phase circuit calculation from real power meter reading P in kilowatts (kW), line to line neutral VL-N in volts (V) and current I in amps (A):

PF = cos φ = 1000 × P(kW) / (3 × VL-N(V) × I(A))

### Power factor correction calculation

The apparent power |S| in volt-amps (VA) is equal to the Voltage V in volts (V) times the current I in amps (A):

|S(VA)| = V(V) × I(A)

The reactive power Q in volt-amps reactive (VAR) is equal to the square root of the square of the apparent power |S| in volt-ampere (VA) minus the square of the real power P in watts (W) (Pythagorean theorem):

Q(VAR) = √(|S(VA)|2 P(W)2)

Qc (kVAR) = Q(kVAR)Qcorrected (kVAR)

The reactive power Q in volt-amps reactive (VAR) is equal to the square of voltage V in volts (V) divided by the reactance Xc:

Qc (VAR) = V(V)2 / Xc = V(V)2 / (1 / (2πf(Hz)×C(F))) = 2πf(Hz)×C(F)× V(V)2

So the power factor correction capacitor in Farad (F) that should be added to the circuit in parallel is equal to the reactive power Q in volt-amps reactive (VAR) divided by 2π times the frequency f in Hertz (Hz) times the squared voltage V in volts (V):

C(F) = Qc (VAR) / (2πf(Hz)× V(V)2)