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Circuits, electronics and noise

Shunt capacitor for single-phase power-factor correction

Capacitance connected across a single-phase load to raise its power factor from PF₁ to PF₂, both lagging or unity, for real power P and RMS voltage V.

Ref
ELN-41
Rev
v1
Field
Circuits, electronics and noise
Result
C [F]

A load with real power P and lagging power factor PF₁ draws a reactive power Q₁ = P tan(arccos PF₁). To raise the power factor to PF₂ a shunt capacitor must supply the difference, Q_C = P[tan(arccos PF₁) − tan(arccos PF₂)]. A capacitor at voltage V and frequency f supplies Q_C = 2πfCV², which gives the capacitance. Real power is not changed; only the current falls.

Capacitance connected across a single-phase load to raise its power factor from PF₁ to PF₂, both lagging or unity, for real power P and RMS voltage V. It is zero when the two power factors are equal.

Calculator

Solve for
F
W
1001,000,000
0.30.99
0.851
Hz
16400
V
241,000
  • Reactive power to remove, Q_C750 var
  • Line current before correction5.435 A
  • Line current after correction4.348 A

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Correction capacitance (F) current value 4.513e-5 F
PReal power of the loadKey driver
4.51e-60.0451
VRMS supply voltage
2.39e-60.00414
PF1Initial lagging power factor
8.57e-61.91e-4
fSupply frequency
5.64e-61.41e-4
PF2Target power factor
7.84e-6
  • Raising a 1,000 W load from PF 0.8 to 1 at 230 V needs 45.13 µF, cutting the line current from 5.435 A to 4.348 A.
  • Correcting all the way to 1.0 is not always the best choice: going past unity makes the load leading and can cause resonance with the supply.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (F)Local stepChange in result
P Real power of the load100 to 1,000,000 W4.513e-60.045130.04512+10%+10 %
V RMS supply voltage24 to 1,000 V2.387e-60.0041450.004142+10%−17.4 %
PF1 Initial lagging power factor0.3 to 0.99 8.574e-61.913e-41.828e-4+10%−28 %
f Supply frequency16 to 400 Hz5.641e-61.41e-41.354e-4+10%−9.09 %
PF2 Target power factor0.85 to 1 7.838e-64.513e-53.729e-5+10%—

Sweep of Target power factor

Target power factorCorrection capacitance (F)
0.857.838e-6
0.86881.083e-5
0.88751.389e-5
0.90631.706e-5
0.9252.041e-5
0.94372.405e-5
0.96252.817e-5
0.98123.331e-5
14.513e-5

Sweep one variable

Correction capacitance (F)Hover, touch, or use the arrow keys on the chart
1e−52e−53e−54e−5 0.860.90.940.98 Target power factor now 4.513e-5

Variables and how each one contributes

C

Correction capacitance

Result · F

What the equation gives you. Everything else feeds it.

P

Real power of the load

W · typical 100 to 1,000,000

The real power of the load, in W. More power means more reactive power to cancel, so the capacitance rises in direct proportion. Raising it by 10 % (to 1,100) moves the result by +10 %, from 4.513×10−5 to 4.964×10−5 F.

PF1

Initial lagging power factor

· typical 0.3 to 0.99

The existing lagging power factor, between 0 and 1. A lower starting power factor needs more correction, so the capacitance rises. Raising it by 10 % (to 0.88) moves the result by −28 %, from 4.513×10−5 to 3.248×10−5 F.

PF2

Target power factor

· typical 0.85 to 1

The target power factor, up to 1. A higher target needs more correction, so the capacitance rises. Raising it by 10 % (to 1.1) gives no valid result at the other default values.

f

Supply frequency

Hz · typical 16 to 400

The supply frequency, in Hz. A higher frequency makes each farad worth more vars, so the capacitance falls in inverse proportion. Raising it by 10 % (to 55) moves the result by −9.09 %, from 4.513×10−5 to 4.103×10−5 F.

V

RMS supply voltage

V · typical 24 to 1,000

The RMS supply voltage across the capacitor, in V. A higher voltage makes each farad worth much more, so the capacitance falls with the square. Raising it by 10 % (to 253) moves the result by −17.4 %, from 4.513×10−5 to 3.73×10−5 F.

Worked example with the values above

  1. Angles: arccos 0.8 = 36.87°, tan = 0.75; arccos 1 gives tan = 0.
  2. Reactive power to remove: 1,000 × (0.75 − 0) = 750 var.
  3. C = 750 / (2π × 50 × 230²) = 4.5129×10−5 F.

Assumptions and limits

  • Sinusoidal steady state with RMS values, and a balanced three-phase load where stated.
  • Accurately connected meters in the standard two-wattmeter arrangement.

Common mistakes

  • Using the apparent power in place of the real power.
  • Using the peak voltage in place of the RMS voltage.
  • Correcting beyond unity, to a leading power factor.

Statistics and uncertainty

Meter errors, supply voltage fluctuation and unbalance all move these results. Wattmeter readings carry the meter’s accuracy class, and for the balanced-load formulas any unbalance gives an error, so check the readings against the total power.

Where it is used

  • Power-factor correction in plants and workshops.
  • Measuring power in three-phase motors and checking delta loads and cables.

Maths you need

  • RMS values: the equivalent steady value of an AC quantity
  • Real, reactive and apparent power: P, Q and S
  • Three-phase systems: line and phase quantities differ by √3

Related formulas

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