Utility Power Factor Penalty Calculation Simplified

The effects of low power factor impact electric utilities in many different ways. I’ll show you how through a utility power factor penalty calculation.

You’ll see why all large utility customers do everything they can to maintain a good power factor.

To generalize, residential customers pay for real power. While commercial and industrial customers pay for apparent power. I’ll explain this more in the depth in the following sections.

Before we do any math, let’s better understand the types of electrical power.

Power factor and types of electrical power

What is power factor? Power factor is just the name given to the ratio of real power to apparent power. In other words, the ratio of real power used in a circuit to the power that is apparently drawn from the source.

I also like to think of power factor as the degree a load matches a resistive load. A purely resistive load has a power factor of 1.

While some loads have a power factor of less than 1. These loads require reactive power to operate. Such as transformers and induction motors.

Let’s dive into the three types of electrical power to make more sense of everything.

Real power (P)

The energy a load consumes. In other words, the electrical energy converted into another energy form. The electrical energy is thus “consumed,” and isn’t returning.

For example, the converted energy can be heat, light, or mechanical power.

You measure real power in watts (W).

Reactive power (Q)

Represents the electrical energy stored and then released. The storage is in the form of a magnetic field or electrostatic field, for an inductor and capacitor respectively.

An example is an induction motor. Induction motors require reactive current to function. The current’s reactive component creates the rotating magnetic field in the motor. While the real current component converts the real electrical power to mechanical power.

More specifically, the motor’s magnetic field stores energy. This energy transfers to and from the power source. This happens every time the magnetic field polarity reverses. Similarly, the energy can transfer to and from power factor compensation capacitors. This transfer of energy is reactive power.

You measure reactive power in volt-amperes reactive (VARs).

Apparent power (S)

It’s what the power source inputs into a circuit. It’s the total power in an AC circuit, both dissipated and absorbed.

In other words, the combination of reactive power and true power.

S = \surd(P^{2} + Q^{2})

You measure apparent power in volt-amperes (VAs).

Now, take a look at the below power factor triangle. It’ll help you better visualize these power relationships.

Notice how apparent power is the hypotenuse. Also, you can apply the Pythagorean Theorem.

power triangle

Effects of low power factor on utilities

Let’s switch gears and see why utilities want their customers to maintain a good power factor. So much so, they’ll penalize you if your power factor gets too low.


A generator is normally rated in kVA.

Also, you’ll notice most generators have a 0.8 power factor rating. This is to support some lagging power factor load. For example, a generator that feeds a bunch of induction motors in a water treatment facility.

The 0.8 generator power factor means the full load kW capacity is 80% of the load kVA. Of course, as long as you maintain or keep the power factor higher.

Imagine you have a 100 kVA generator with a power factor of 0.8.

P.F. = \dfrac{P}{S}

\Rightarrow 0.8 = \dfrac{P}{100 \: kVA}

P = 80 kW

\Rightarrow \dfrac{80 \: kW}{100 \: kVA} \times 100 = 80%

Now, what happens when the power factor drops below 0.8? The increased total kVA reduces the generator’s capacity. This is because of the increased lagging reactive kVA component. The current flow increases and causes a voltage drop.

This commonly happens when a facility is blindly retrofitted. For example, if you connect more induction motors than designed to the generator. The generator would then need to support more magnetic fields versus “doing real work.”

Important Note: reactive power isn’t useless. Yes, reactive power doesn’t do real work.

But, AC machines wouldn’t operate without reactive power. So how do you do real work without operating machines? You don’t!

Think of electric motors. They rotate and do work through the interaction of magnetizing fields created in the motor windings. These magnetic fields need a source of magnetism.

The magnetization for an induction motor comes from the AC line it’s connected to. More specifically, this magnetism comes from the power source. 

Without the supplied magnetizing kVAR, the induction motor wouldn’t operate. The catch is, the magnetizing kVAR reduces the power factor. 

cutaway view of induction motor
Cutaway view of induction motor (Photo Credit: Stunteltje)


Transformers need a magnetic field for power conversion. Converting power from a primary to secondary voltage.

The power factor of a load directly affects the real power capacity of a transformer though. Refer back to the generator we just discussed.

Also similarly, the voltage drop experienced by a transformer through an increase in load is greater with a lagging power factor.

As a result, the voltage regulation of a transformer suffers from a poor power factor.

Distribution line losses

Utility’s size wires in electric circuits to carry the required current.

But what happens when the connected customer’s power factor drops?

At a low power factor, the circuit needs to carry a reactive kVA component on top of the kW. More current will now run through the wires.

The utility will now need to make the wires larger. For a given kW, a low power factor will draw more current than a high power factor.

Also, keep in mind, the resistance losses (I^{2}R) of the conductor. These losses are proportional to the current flowing through them.

So, the voltage drop at the load end of lines can be a huge problem. Voltage drop without a doubt is a common theme when it comes to a poor power factor.

To drill a little deeper, the same problem happens with current limiting reactors. These are just large coils of wire made to produce a magnetic field.

You use current limiting reactors to reduce short circuit current. But they’ll also reduce your power system’s power factor. I go over how to size a current limiting reactor for a substation. This will help you better understand inductive loads. Also, the working principles at play.

Power cost

A utility kilowatt-hour (kWh) meter tracks energy usage. This is how utilities determine how much to bill you per your energy consumption.

Now, on top of the kWh charge, utilities have a penalty for a low power factor in their billing rates. At the same time, they have incentives for high power factors.

As a scenario, the line losses vary per the square of the line current (I^{2}R). The line current is higher at a lower power factor. The utility certainly isn’t going to eat the cost on behalf of the customer.

Important Note: there are benefits to purchasing power through a power factor clause. The customer benefits when they have a high power factor.

Electric utilities reducing power costs

Electric utilities need to provide power to all their customers. This also includes the reactive power needs of their customers.

To accommodate everyone, electric utilities need to upgrade their infrastructure a lot. They need to spend money on the following higher rated equipment:

  • Generators
  • Transformers
  • Power lines
  • Other electrical equipment

Everything comes at a cost!

To pay for these upgrades, utilities have a power factor penalty. And for some customers, the penalty cost of reactive power is a big portion of their bill.

But if a customer installs capacitors, they can have huge cost savings. They’ll eliminate the utility power factor penalty.

Important Note: inductive loads require reactive power (kVAR). Again, a motor requires inductive power for magnetizing. In return, the apparent power increases from the lagging reactive power. 

S = \surd(P^{2} + Q^{2})

Now, power factor correction capacitors improve power factor. They provide leading reactive power that cancels out the lagging reactive power. So, they’re a great way to improve a power system’s power factor. 

In short, capacitive loads improve the power factor. While inductive loads reduce the power factor.

Utility power factor penalty calculation

Finally, let’s calculate the benefits of installing capacitors.

I’m going to use the following typical utility rates you’d find in most U.S. states:

Demand charge:

Rates (Demand Charge)Per Meter, Per Month
First 1,000 kW of billing demand$10,500.00
Over 1,000 kW of billing demand, per kW$12.00

Energy charge:

Rates (Energy Charge)Per Meter, Per Month
First 100 kWh per kW of billing demand per kWh$0.22
Next 300 kWh per kW of billing demand per kWh$0.28
All excess, per kWh$0.35

Important Note: let’s define the power factor charge and fuel cost adjustment. 

Power Factor Charge: we’ll use the above-listed rates in our calculation. The total charge for any month decreases or increases by 0.1% for each 1% that the average power factor of the customer is greater or less than 85%. 

We’ll compute the average power factor using the ratio of lagging kilovolt ampere-hours to kilowatt-hours. The calculation will be over a single month. 

Fuel Cost Adjustment: this is the total billed kilowatt-hours multiplied by $0.006 per kWh. This figure isn’t impacted by adjustments for voltage or power factor either. 

This $0.006 adjustment captures the amount of required revenue to cover the cost to produce power. The utility compares the actual cost of fuel and purchased power to the costs they had projected. Because energy costs fluctuate. 

EXAMPLE: imagine Factory Z has 5,000 kW of load at 0.65 power factor. Their monthly electricity consumption is 3,750,000 kWh/month. They purchase their power from Utility X.

Demand charge

First 1,000 kW = $10,500.00

Over 1,000 kW (bill at $12.00 per kW) =  (5,000 kW – 1,000 kW) x $12.00/kW = $48,000.00

Demand charge subtotal: $10,500.00 + $48,000.00 = $58,500.00

Energy charge

First 100 kWh/kW demand (100 x 5,000 x $0.22) = $110,000.00

Next 200 kWh/kW demand (300 x 5,000 x $0.28) = $420,000.00

All excess kWh (3,750,000 – [100 x 5,000] + [300 x 5,000]) x $0.35 = 1,750,000 x $0.35 = $612,500.00

Energy charge subtotal: $110,000.00 + $420,000.00 + $612,500.00 = $1,142,500.00

Now, we add the demand subtotal with the energy subtotal.

$58,500.00 + $1,142,500.00 = $1,201,000.00

Power factor penalty 

(85-65) x 0.001 x $1,201,000.00

20 x 0.001 x $1,201,000.00 = $24,020.00

Fuel cost adjustment

3,750,000 kWh/month x $0.006/kWh = $22,500.00

Total monthly bill

Now, we add all the following charges together:

  • Demand charge
  • Energy charge
  • Power factor penalty
  • Fuel cost adjustment

$1,201,000.00 + $24,020.00 + $22,500.00 = $1,247,520.00

Utility power factor calculation with capacitors installed

Here, we’ll do a second round of calculations with capacitors. Our goal is to improve the power factor to 0.95.

Also, we’ll calculate the power cost reduction and payoff in years for the capacitors.

We’ll use the same 5,000 kW load, and 3,750,000 kWh per month electricity consumption.

Calculating capacitor size to improve power factor to 0.95

At 0.65 P.F._{1}, cos^{-1}(0.65) \Rightarrow \theta = 49.5^\circ, where \theta = P.F. angle

\Rightarrow sin(\theta) = sin(49.5^\circ) = 0.76

At 0.95 P.F._{2}, cos^{-1}(0.95) \Rightarrow \theta = 18.2^\circ

\Rightarrow sin(\theta) = sin(18.2^\circ) = 0.31

Let’s calculate the apparent power using the power factor of 0.65. We already know the real power is 5,000 kW.

P.F._{1} = \dfrac{P}{S} \Rightarrow S = \dfrac{P}{P.F._{1}}

= \dfrac{5,000}{0.65} = 7,692.30 kVA

sin(\theta) = \dfrac{Q}{S}

Q = S \times sin(\theta) = 7,692.30 kVA x 0.76 = 5,846.15 kVAR

We’ll now do the same calculation using the power factor of 0.95.

P.F._{2} = \dfrac{P}{S_{2}} \Rightarrow S_{2} = \dfrac{P}{P.F._{2}}

= \dfrac{5,000}{0.95} = 5,263.16 kVA

sin(\theta) = \dfrac{Q}{S}

Q_{2} = S \times sin(\theta) = 5,263.16 kVA x 0.31 = 1,631.58 kVAR

Now, take a look at the below power triangle. This will piece together everything we just did in our calculation.

You can see how the added kVAR improves the power factor. Also, how the kVA drastically drops.

power factor correction using a capacitor

Capacitor size required

Q_{1} - Q_{2} = 5,263.16 kVAR – 1,631.58 kVAR = 3,631.58 kVAR

Capacitor cost 

The installation cost of a capacitor per kVAR let’s estimate to be $40.00.

$40.00 x 3,631.57 kVAR = $145,263.20

Total cost reduction in the monthly bill

Power factor penalty savings (from P.F. 0.65 to 0.85) = $24,020.00

Power factor credit (from P.F. 0.85 to 0.95) = $12,010.00

Now, we calculate the loss reduction using the following math:

(Monthly energy usage) x (Estimated 2% loss) x (Loss varies per I^{2}R) x (kWh + Fuel cost adjustment) = 3,750,000 x 0.02 x 0.38 x ($0.35 + $0.006) = $10,146.00

Now, let’s add up the total savings per month.

Total savings per month: $24,020.00 + $12,010.00 + $10,146.00 = $46,176.00

Payoff for capacitor installation

\dfrac{Capacitor \: cost}{Monthly \: saving} = $145,263.20 / $46,176.00 = 3.15 months.

So it’s a no brainer to install a capacitor in this scenario. As the utility power factor penalty is extremely costly.

The payback from installing a capacitor will improve profit margins right away for Factory Z.

At the same time, the utility power factor penalty cost shows the impact on Utility X.


A utility power factor penalty can greatly affect the bottom line of a company. These days we heavily rely on AC machines. As we learned, AC machines need magnetism from their power source to operate.

Now by eliminating power factor surcharges through capacitors, you can reduce your electricity bill. All the while, maintain the operations in your facility.

This is even more important for businesses in states like California. As the cost of doing business is increasing year after year.

What’s more, you increase the load-carrying capability of your existing circuits. Also, you improve the supplied voltage at your equipment. A win-win!

What are your thoughts on the utility power factor penalty? What’s your view on the future of energy billing by utilities?

Featured Image Photo Credit: Stunteltje (image cropped)


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