How Capacitor Banks Work: kvar, Power Factor and Why They Are Installed

What a shunt capacitor bank does

Shunt capacitor banks are used for power factor correction and voltage support. A widely cited protection paper puts it from the system operator's side: a bank provides voltage support, power factor correction, and/or harmonic filtering. The bank is connected in parallel (shunt) with the load, so it supplies the reactive current that inductive loads such as motors and transformers draw, and that current no longer has to flow all the way from the generator or substation.

USDA Rural Utilities Service Bulletin 1724D-112, The Application of Capacitors on Rural Electric Systems, names four consequences of low power factor: voltage regulation problems, increased system losses, power factor penalties in wholesale power contracts, and reduced system capacity. Every one of those is a cost of moving reactive current through conductors and transformers. A capacitor bank placed near the load removes much of it.

Power factor and the power triangle

The RUS bulletin defines power factor as the ratio of real power (kW) to total power (kVA), and notes it is also equal to the cosine of the phase angle between the voltage and current vectors. Apparent power is the vector sum, not the arithmetic sum, of real power and reactive power, so kvar = sqrt(kVA^2 - kW^2). Real power does the work; reactive power sets up magnetic fields in motors and transformers and flows back and forth every cycle.

The bulletin also gives a rule of thumb: most uncorrected loads sit at a power factor angle of around 30 degrees, a power factor of about 87%, and at that angle the reactive power is roughly half the apparent power. The table below repeats the bulletin's angle to power factor values.

How much kvar: Q = P (tan φ1 - tan φ2)

One manufacturer's plant engineer guide gives the correction formula in motor form: kvar required = (hp x 0.746 / efficiency) x [sqrt(1 - PFa^2)/PFa - sqrt(1 - PFt^2)/PFt], where PFa is the actual power factor and PFt the target. The first term is just real power in kW, and each square-root term is the tangent of the power factor angle, so the formula reduces to Q = P (tan φ1 - tan φ2).

Worked example. A plant draws 400 kW at 0.87 power factor and wants 0.97. tan(arccos 0.87) = 0.5667 and tan(arccos 0.97) = 0.2506. The difference is 0.316, so Q = 400 x 0.316 = 126 kvar. The same guide prints 0.316 in its multiplier table for that step and rounds the result up to a 140 kvar standard size. A second step from its table: 0.75 to 0.95 uses a multiplier of 0.553 (0.8819 - 0.3287), so a 500 kW load would need 500 x 0.553 = 277 kvar.

The RUS bulletin works a feeder-sized case. A 1,200 kVA load at 87% power factor is 1,044 kW and 592 kvar. Adding 600 kvar of capacitors leaves -8 kvar: power factor is virtually 1.00 but slightly leading, apparent power falls to about 1,044 kVA, and current on the 7,200 V system drops from 56 A to 48 A. The bulletin also warns that line losses at 80% leading power factor are just as detrimental as losses at 80% lagging, so overcorrection is not free.

Capacitor unit ratings: kvar, voltage, current and capacitance

A bank is assembled from single-phase capacitor units. IEEE 18-2025, the shunt power capacitor standard, covers units rated 216 V or higher and 2.5 kvar or more. A protection paper gives the commercial range as 216 V to 24,940 V and 2.5 kvar to around 800 kvar, and the RUS bulletin notes that the smallest standard unit for primary use is now 50 kvar, with some manufacturers not making anything below 100 kvar.

The RUS bulletin states the unit current formula as capacitor current = kvar per phase / kV line to neutral, and gives the example that each 100 kvar capacitor draws 13.9 A on a 7.2 kV system. Capacitance follows from kvar = 2 pi f C V^2 / 1000. Keep the units straight: with V in kV, C in microfarads = kvar x 1000 / (2 pi f kV^2). For the bulletin's 50 kvar, 7,200 V example: 50 x 1000 / (377 x 7.2^2) = 2.56 uF.

Because kvar depends on V^2, a unit run off its rated voltage delivers different kvar. One manufacturer's formula sheet gives actual kvar = rated kvar x (actual V / rated V)^2. One maker's owner guide works an example: a 900 kvar bank rated 13.2 kV on a 12.47 kV system gives 900 x (12.47 / 13.2)^2, about 802 kvar, an effective current of 37.2 A, and equipment sized for at least 37.2 x 1.35 = 50.2 A. The RUS bulletin adds a warning in the other direction: putting line-to-line voltage (1.732 times) on a unit rated line-to-neutral raises its effective kvar to three times its rating.

How units are arranged into a bank

A protection paper describes the building blocks: units are connected in parallel (parallel groups) to meet the var specification, and parallel groups are connected in series to meet the nameplate voltage rating. Most distribution and transmission banks are wye connected, grounded or ungrounded; distribution banks may also be delta.

The connection sets the unit voltage. One maker's installation instructions state it directly: capacitors connected in delta must be rated at the line-to-line system voltage, and capacitors connected in wye at the line-to-neutral system voltage. On a 12.47 kV grounded-wye feeder, 12.47 / 1.732 = 7.20 kV, which is why 7,200 V units are common there. One utility's pole bank standard uses pre-assembled banks of 300, 450, 600, 900, 1,200 and 1,350 kvar on a 12,470 V grounded-wye system with 7.2 kV units, for example 600 kvar as one 200 kvar unit per phase or two 100 kvar units per phase.

The RUS bulletin adds a practical rule: install equal kvar per phase, and take the whole bank out of service when one phase is out.

What the bank buys you: voltage rise, losses, capacity and billing

These are the reasons a bank gets installed, and each has a simple formula that can be checked against measured data.

• Voltage rise: rise = capacitor current x system inductive reactance. The RUS example: 300 kvar three phase on 12.47/7.2 kV with 13.3 ohms of reactance gives 13.9 A x 13.3 = 184.7 V, which is 2.57%, or 3.08 V on a 120 V base. The bulletin says a 2 to 3 V rise with 300 kvar on 7.2 kV systems is common, and about half that on 14.4 kV.
• Losses: one manufacturer's formula sheet gives percent loss reduction = 100 - 100 x (original PF / improved PF)^2. Going from 0.70 to 0.90 gives 100 - 100 x (0.778)^2 = 39.5% lower losses in the conductors upstream of the bank.
• Capacity: a plant engineer guide states that raising power factor from 70% to 90% releases 0.32 kVA per kW (1/0.70 - 1/0.90), so a 400 kW load frees 128 kVA of transformer and cable capacity.
• Billing: the RUS bulletin shows a power factor penalty clause, billing demand = metered demand x (target PF / measured PF). At 0.95 target and 0.83 measured, 0.95 / 0.83 = 1.145, a 14.5% increase in billed demand. It notes utilities typically want 90 to 95% power factor.

Fixed and switched banks

The RUS bulletin's rule is that the kvar needed at minimum load should be permanently connected (fixed, unswitched), and heavier loads are covered by switched banks. A switched bank needs a switching device, control equipment and a control power source. The bulletin says switched-bank voltage change is commonly limited to 3 V on a 120 V base, up to 6 V on long, lightly loaded feeders, and that fixed banks on uniformly loaded feeders are usually placed 1/2 to 2/3 of the way down the line.

Controls can respond to voltage, current, temperature, kvar, power factor or time, or a combination. Switching matters to fuse selection because every energization drives a high-frequency inrush through the fuses, covered in the inrush and outrush article.

Where the fuses come in

In an externally fused bank each unit has its own fuse, and that fuse starts from the unit current. Take the 200 kvar, 7,200 V unit above: 200 / 7.2 = 27.8 A. That current is not the whole story. IEEE's 2005 interpretation of IEEE 18-2002 lists continuous contingency limits of 110% of rated rms voltage, 120% of rated peak voltage including harmonics, 135% of nominal rms current and 135% of rated kvar. Harmonics add current because capacitive reactance, Xc = 1/(2 pi f C), falls as frequency rises.

Published manufacturer guidance therefore applies a margin. One manufacturer's application method uses fuse current = capacitor current x 1.35 at a 25 °C ambient, which here gives 27.8 x 1.35 = 37.5 A. One manufacturer's table maps 7,200 V units to the 8.3 kV fuse class. DENCO's catalog lists DNXC8 full-range current-limiting capacitor fuses at 10, 18, 25, 30, 40, 45 and 65 A, so the arithmetic points toward 40 A before any temperature derating, outrush I2t or case rupture check. That is a starting point for a conversation with DENCO and the capacitor maker, not a selection.

The sizing article walks through the full method, including enclosure temperature and harmonic current.

Common mistakes

• Adding kW and kvar arithmetically. kVA is the vector sum, sqrt(kW^2 + kvar^2).
• Overcorrecting. A leading power factor costs as much in line losses as the same lagging value, and it can raise voltage at light load.
• Using line-to-line voltage for a wye-connected unit. Unit current is kvar per phase divided by kV line to neutral; a unit rated line to neutral and fed line to line sees three times its kvar.
• Ignoring off-nominal voltage. kvar scales with voltage squared, so a 13.2 kV bank on a 12.47 kV system delivers about 11% less than nameplate.
• Sizing the fuse at nameplate current. Capacitors are expected to run continuously up to the contingency limits above, and harmonics raise current further.
• Leaving one phase of a bank in service with another phase out. The RUS bulletin says to take the whole bank out.

Frequently asked questions