Capacitor Banks and VAR Management

9 min · difficulty 5/10

A feeder pulls 1,000 kW at 0.85 lagging power factor, drawing about 620 kVAR. To reach unity you would add capacitors supplying those VARs. A 1,200 kVAR bank is on the truck. Too big, too small, or about right? Decide first.

It is too big for this load. To cancel 620 kVAR you size the bank near 620 kVAR, so a bank of roughly 600 kVAR brings this feeder close to unity. A 1,200 kVAR bank supplies almost double the reactive power the load wants, which over-corrects the feeder into a leading power factor and can push the voltage up. The rule is simple: size the bank to the lagging kVAR you want to cancel, not larger.

Sizing a capacitor bank on the power triangle A power triangle drawn to scale. The horizontal base is real power P = 1000 kW. The vertical leg is reactive power Q = 620 kVAR lagging, drawn at 0.62 of the base length. The hypotenuse is apparent power S = 1176 kVA at a power-factor angle theta = 31.8 degrees, which equals acos 0.85. A downward gold vector along the reactive leg shows a capacitor supplying about 620 kVAR, pulling the resultant toward the P axis and the power factor toward unity. Sizing a capacitor bank on the power triangle drawn to scale: Q is 0.62 of P, since 620 / 1000 = 0.62 P = 1000 kW real power (does the work) Q = 620 kVAR lagging reactive power S = 1176 kVA apparent power = √(P² + Q²) θ = 31.8° acos 0.85 (PF) capacitor supplies ~620 kVAR (leading, cancels Q) Q cancelled → S → P axis, PF → ~unity sizing the bank tanθ = Q/P = 0.62 S = √(1000²+620²)   = 1176 kVA add ~620 kVAR cap
Adding about 620 kVAR of capacitance cancels the 620 kVAR lagging reactive leg of a 1000 kW load, collapsing the 1176 kVA apparent power onto the P axis and pulling the power factor toward unity.

Work the triangle to see why. Real power, the 1,000 kW, is the base. Reactive power, the 620 kVAR the load draws, is the vertical leg. Apparent power is the hypotenuse, and S = kW / PF, so at 0.85 PF this load is 1,000 / 0.85, about 1,176 kVA. The reactive leg is 1,000 multiplied by the tangent of the angle whose cosine is 0.85, which is 620 kVAR. A capacitor supplies leading kVAR that subtracts from that vertical leg. Cancel the whole 620 and the triangle collapses to its base: unity power factor, with the apparent power dropping back toward 1,000 kVA.

That cancellation happens locally. A shunt capacitor bank supplies leading kVAR right at the load, so the source no longer has to push reactive current down the feeder. The feeder then carries less total current, runs cooler, loses less to resistance, and holds voltage better all the way out. Capacitors fix power factor by making the reactive power near the load instead of dragging it from the substation.

So why not just bolt up one big fixed bank and forget it? Because the reactive load is not constant. It is heavy when motors and air conditioning run at midday and light overnight. A fixed bank, sized to the base reactive load that is present most of the time, makes the same kVAR around the clock. A switched bank handles the variable part: it comes online when kVAR demand is high and drops off at light load, so the feeder never over-corrects into a leading power factor and an unwanted voltage rise. Fixed banks cover the steady base; switched banks follow the load.

Every component on this list, conductor, switchgear, cutouts, regulators, carries its own lead time and supplier landscape. DistroForge report editions track capacity and lead times across the distribution equipment categories covered in this track.

Question 1 of 3

A shunt capacitor bank improves a lagging power factor. How does it actually do that?

Educational material only. This is not engineering, safety, or procurement advice. Confirm any value against manufacturer documentation and a licensed professional before specifying equipment.