NIST Technical Note 1654 

Regarding Electric Energy Savings, 

Power Factors, and Carbon 

Footprints: A Primer 

Martin Misakian 

Thomas L. Nelson 

William E. Feero 
NIST Technical Note 1654 

Regarding Electric Energy Savings, 

Power Factors, and Carbon 

Footprints: A Primer 

Martin Misakian 

Thomas L. Nelson 

 

Electronics and Electrical Engineering Laboratory 

National Institute of Standards and Technology 

Gaithersburg, MD 20899 

William E. Feero 

 

P.O Box 489 

Reedsville, PA 17084 

October 2009 

U.S. Department of Commerce 

Gary Locke, Secretary 

National Institute of Standards and Technology 

Patrick D. Gallagher, Deputy Director 
National Institute of Standards and Technology Technical Note 1654 

Natl. Inst. Stand. Technol. Technical Note 1654, 9 pages (October 2009) 

CODEN: NSPUE2 
Regarding Electric Energy Savings, Power Factors, 

and Carbon Footprints: A Primer 

Martin Misakian, Thomas L. Nelson, and William E. Feero 

Abstract – A short primer is presented which describes the underlying physical theory of 

certain devices that reduce the current drawn from power distribution lines by improving 

the power factor of residential electric circuits. A brief discussion is provided of the 

associated energy savings, change in power factor, and reduction of “carbon footprint.” 

Introduction 

“Power factor correction devices improve power quality but do not generally improve 

energy efficiency (meaning they won't reduce your energy bill)”. This remark is taken 

from the Energy Star website [1] as part of a discussion on saving the cost of electrical 

energy for residences. We present a short primer on the physical theory underlying such 

devices and provide additional information related to the above statement. The devices 

that we consider introduce electrical capacitance in parallel with activated electrical loads 

that typically contain both resistance and inductance, e.g., an electric motor or 

transformer. The question of whether providers and users experience savings for the cost 

of electric energy or just reduce their carbon footprint is briefly discussed. 

The Model 

For our model we consider a simplified version of an electric motor as our electrical load. 

The motor is represented as a resistor R in parallel with an inductor L as shown in Figure 

1(a). The wires connecting the two elements are assumed to have no resistance. Energy 

is dissipated in the resistor, but not in the inductor which provides a magnetic field for 

operation of the motor. 

Further details of the electric circuit are shown in Figure 1(b) with the addition of a 

sinusoidal alternating voltage source, i.e., the power distribution line with voltage 

Vosin(ωt) and the watt-hour meter which utilities use for billing purposes. Our model 

considers watt-hour meters that monitor the product of voltage, current, power factor 

(PF), and time to determine the cost of the electric energy. Vo is the voltage amplitude, t

is the time, and ω is the angular frequency 2πf, where f is the power frequency 60 Hz. 

For our purposes, the PF can be found by calculating the cosine of the phase difference 

between the current and voltage waveforms. The “energy savings” device is shown as 

the capacitor C, but initially is not connected into the circuit. Introduction of the 

capacitor increases (“corrects”) the PF and, as shown below, less current is drawn from 

the power line. We assume the capacitor is disengaged automatically when the load is 

turned off. Otherwise the capacitor would continue to draw current from the power line 

and a potential would exist for adverse effects on the distribution system during light load 

periods. Ground wires are not indicated as we also assume no leakage currents. 
Calculations 

Mesh analysis for determining currents in circuits is described in many texts on circuit 

analyses and is not reviewed here [2, 3]. We use the approach to set up simultaneous 

equations that are solved to determine circulating currents I1, I2, and I3 in the separate 

loops shown in Figures 1(b) and 1(c). The simultaneous equations that are developed and 

their solutions are presented below. We choose specific values of the circuit elements (R, 

L, C) to demonstrate how the current from the power distribution line is affected with and 

without the inclusion of the capacitance. Since the voltage remains essentially constant, 

changes in the current mainly influence the amount of electric energy provided by the 

power line. Because the number of equations will be few, they can be solved by the 

substitution method or by using matrices and Cramer’s Rule. 

Mesh analysis for the circuit in Figure 1(b) yields the equations 

,0)(

sin(ω and )

2 3

2 3

=++−

− =

L LR

L oL

ZZIZI

tVZIZI

where the impedances ZL = ωLi and ZR = R; i = (-1)0.5. 

Assuming the appliance is a central air conditioner with values for L, R, and Vo of 0.065

henries, 10.4 ohms, and 325 peak amplitude volts (230 volts rms) respectively, the 

solutions for the currents in amperes (A) are 

sin(25.31 ω .A )

sin(25.31 ω sin(263.13A) ω andA i)

3

21

I t

II t t

=

== −

The phase difference between the current and voltage waveforms from the power line is 

determined by examining the angle associated with the tangent of I1 as represented in the 

complex plane, i.e., -23 degrees. Therefore the PF is cos(-23o

) or 0.92. The amplitude of 

the current from the power line is [(31.25)2 + (13.26)2]

0.5 or ~34 A, and the current 

through the inductor, IL, is just I2 – I3 or -13.263sin (ωt)i A. 

Mesh analysis for the circuit that includes the capacitor C [Fig. 1(c)] yields the equations 

,0)(

and ,0)(

sin(ω ),

2 3

1 2 3

1 2

=++−

=−++−

− =

L LR

C CL L

C oC

ZZIZI

ZIZZIZI

tVZIZI

where the impedance ZC = -i/ωC. 

We assume a value for capacitance that will demonstrate the maximum effect for 

reducing the current from the power line. That is we choose a value of C that will satisfy 

the resonance condition with the inductor, i.e., C = 1/ω2

L = ~108.2 μF. 
Solutions for the currents in Figure 1(c) are 

sin(25.31 ω A. )

sin(25.31 ω sin(257.13A) ω and A, )i

sin(25.31 ω sin(1041.5A) ω sin(25.31A i) ω A, )

3

2

3

1

I t

I t t

I t t t

=

= −

= ⋅− ≈ −

The current provided by the power line, I1, is reduced, i.e., the amplitude is reduced to 

~31.3 A from 34 A. Yet the current I2 necessary to operate the motor is unchanged. The 

source of what appears to be “extra” current in I2, -13.257sin(ωt)i A, can be understood 

by first examining the other currents in the circuit [Fig. 1(c)]. The current through the 

inductor, IL, equals -13.257sin(ωt)i A and is unchanged. However, there is now a current 

through the capacitor, IC, equal to I1 - I2 or +13.257sin(ωt)i A, which is 180 degrees out of 

phase with respect to the current in the inductor. 

The underlying physics provides an understanding of what is happening with the 

introduction of the capacitor. When voltage is applied to the circuit in Figure 1(c), the 

capacitor is instantly charged and this charge oscillates between the capacitor and 

inductor [4] creating the current through C and L indicated by the mesh analysis. The 

oscillating charge is the origin of the “extra current” in I2 when the current from the 

power line is reduced. Appendix A shows this oscillatory relationship between the 

charge on the capacitor and current through the inductor as a function of time. 

For the ideal (lossless) conditions we have assumed, the power line does not need to 

provide additional current to charge the capacitor-inductor network. This last point can 

be verified by performing a mesh analysis of the circuit in Fig. 1(c) with the resistor 

removed. The currents through C and L will be the same (±13.257sin(ωt)i A), but the 

current from the power line will be ~zero. The zero current from the power line is also 

expected since the impedance of the L-C circuit is ~infinite. In a more realistic treatment

of the problem there will be some resistive losses in the L-C circuit which will be 

compensated for by additional charging by the power line (see below). 

 

Also very noteworthy is that under the above conditions the PF has increased to nearly 

unity from 0.92 because there is essentially phase congruence between the current and 

voltage waveforms, i.e., the phase of I1 is the same as V0 sin(ωt). 

While the above calculations are for very ideal conditions, including the use of a 

matching capacitor, reductions in power line current will occur even when the capacitor 

is poorly matched. For example, if the capacitor is ±50 μF different from the matched 

value (108.2 μF), the amplitude of I1 will be 31.8 A (versus ~34 A) and the power factor 

will be 0.98 (versus 0.92). Similarly, introducing small resistances of the connecting 

wires will not change the results significantly. 
Discussion 

Residential loads that decrease the power factor include refrigerators, air conditioners, 

and washing machines. Typically these appliances are operated only intermittently, so 

that average residential power factors may not be very low. If an energy saving device of 

the type we have considered is used to increase the power factor, the utility will not have 

to supply as much current when certain electrical appliances are operated. However, the 

homeowner’s electric bill will not decrease because, as noted above, the cost is 

determined in part by the product of current, voltage, and power factor. That is, as the 

current from the power line is reduced by the introduction of capacitance, the power 

factor is increased and the product, I x V x PF, remains essentially the same. The utility 

will save a small amount of money (see discussion of possible savings below) because it 

will not have to supply as much current to the residence, will have smaller resistive losses 

along the power line, will not emit as much greenhouse gas, etc. Will the homeowners 

reduce their “carbon footprint”, are they only helping the utilities reduce their carbon 

footprint, or can both take some credit? In addition, should the homeowner consider the 

CO2 generated during fabrication of the capacitive device and the associated “payback 

time”? We leave the answer to these questions up to the reader. 

It should be noted that some utilities may charge a penalty for power factors that are 

considered too low, for example below 0.95 [5]. In such cases there could be some 

savings by installing devices of the type that has been considered. For example, 

industrial users of electric energy with large inductive loads might be penalized if steps 

are not taken to increase their power factor. 

Possible utility savings. For our central air conditioner example given above, let us 

estimate approximately how much energy is saved because of lower resistive losses in the 

distribution system when the customer installs the capacitor. The utility’s total resistance 

as seen looking back into the distribution system from the watt-hour meter is assumed for 

our rough calculation to be less than 0.05 ohms. The power consumed in the utility 

circuits for operating the air conditioner without the capacitor is about 28.8 watts (Irms2

R), 

where Irms is the root mean square (rms) value of I1 in Figure 1(b), 24 A. With the 

capacitor installed, the rms value of I1 [Fig. 1(c)] becomes 22.1 A and the power 

consumed by the utility is reduced to about 4.4 watts. Therefore, installation of the 

capacitor reduces the energy from the utility by 4.4 watts multiplied by the running time 

of the air conditioner. If the air conditioning unit runs for 12 hours each day, the energy 

savings will be about 52.8 watt-hours per day. At 20 cents per kilowatt-hour, the money 

saved by the utility would be approximately 1 cent per day. Since in most parts of the 

United States air conditioners only operate for less than six months of the year, the 

utility’s annual savings would be about $1.80 for a single residence. 
Appendix A 

The relationship between the charge on the capacitor and the current through the inductor 

as a function of time can be determined from the results that have been calculated above. 

We find the charge in coulombs (Q) on the capacitor, QC, as a function of time by 

integrating the current through the capacitor, IC, i.e., 

dtIQ dtt Qt C C cos(ω )

ω

257.13 == (sin257.13 ω ) −= ∫∫ .

The calculated current through the inductor, IL, is 

I sin(257.13 ωt A. ) L = −

By plotting both the charge and inductor current as a function of time on the same 

coordinates (Figure A1), one can see that the charge on the capacitor goes to zero as the 

current reaches a maximum through the inductor, and reaches a maximum charge as the 

current through the inductor goes to zero. 

Acknowledgement 

We thank Gerald Fitzpatrick for useful conversations which contributed to the 

preparation of this Technical Note. 

References 

[1] http://energystar.custhelp.com/cgibin/energystar.cfg/php/enduser/std_adp.php?p_faqid=4941&p_created=1204908170 

[2] S.L. Oppenheimer, F.R. Hess Jr., and J.P. Borchers, Direct and Alternating Currents, 

McGraw Hill Co., New York (1973). 

[3] R.M. Kerchner and G.F. Corcoran, Alternating Current Circuits, John Wiley & Sons, 

New York (1960). 

[4] D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics, John Wiley & 

Sons, Inc, New York (2005). 

[5] DOE Fact Sheet, http://www.eere.energy.gov/; search for “Reducing Power Factor 

Cost.” 
Figure 1. Model circuits for (a) simplified motor, (b) circuit without 

capacitive device, and (c) circuit with device. 

Figure A1. Relation between charge on the capacitor and inductor current. 

The vertical axis has units of charge and current. Insets are after Ref. [4].