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AC power describes energy transfer associated with time-varying electric potential difference and electric current, including the distinction between instantaneous transfer and its average. With the passive sign convention, current enters the terminal marked positive, and instantaneous power absorbed by the element is $p(t)=v(t)i(t)$. A positive value means energy entering the element; a negative value means energy returning through that terminal pair.

For sinusoidal steady operation, choosewhere $V$ and $I$ are root mean square magnitudes. The cosine-product identity givesAveraging over a complete period removes the second term, giving active power $P=VI\cos \phi$. This derivation separates the energy-transfer average from a twice-frequency oscillation.^[ref]

For an original example, $V=120\mathrm{V}$, $I=5\mathrm{A}$ and $\cos \phi =0.8$ give active power480W. The oscillating term has amplitude600W, so instantaneous power ranges from−120 to1080W. Over ten complete cycles at50Hz, the net energy absorbed is $480(10/50)=96\mathrm{J}$. Temporary negative power can accompany positive net transfer when an element stores and returns energy.

The product $VI=600$ in volt-amperes is the apparent power. In the single-frequency sinusoidal convention, reactive power is $Q=VI\sin \phi$, giving360 in reactive volt-amperes for a lagging current with $\sin \phi =0.6$. Complex electrical power packages $P$ and $Q$, while the power factor $P/(VI)=0.8$ distinguishes net transfer from the terminal magnitude product. The primary MIT treatment derives these quantities and makes the peak-versus-RMS convention explicit.^[ref]

A measurement must retain the relative phase of the two channels. Multiplying separately measured RMS magnitudes gives apparent power, not generally active power. For nonsinusoidal signals, the time average of $v(t)i(t)$ remains the definition of active power, but one phase angle need not describe every harmonic contribution.

Active power, also called real power, is the time average of instantaneous power transferred through a terminal pair. For periodic electric potential difference $v(t)$ and electric current $i(t)$ of common period $T$, the passive sign convention givesThe net energy entering during each period is $PT$. Positive active power denotes net absorption under this convention; it can represent useful conversion to other forms of energy as well as heating.

For sinusoids of RMS magnitudes $V$ and $I$ with relative phase $\phi$, $P=VI\cos \phi$. For a pure resistor with constant resistance $R$, the stronger waveform-independent statement follows from $v(t)=Ri(t)$:This is why the root mean square current is the constant-current value giving the same heating in that resistor.^[ref]

An original harmonic example shows why RMS values alone do not determine active power. Letin volts and amperes. Orthogonality over a period gives $I_{\mathrm{rms}}=\sqrt{3^2+4^2}=5\mathrm{A}$, but the third harmonic averages to zero when multiplied by the fundamental potential difference. Consequently $P=100(3)=300\mathrm{W}$, whereas apparent power is $100(5)=500\mathrm{VA}$. The power factor is0.6 even though the fundamental current is exactly in phase with the fundamental potential difference.

More generally, Fourier analysis of real periodic waveforms with RMS harmonic magnitudes giveswhen the expansion and average are justified. Cross terms of different harmonic orders vanish. A single fundamental-frequency complex electrical power therefore need not capture every harmonic contribution to the total average.

For an ideal inductor or capacitor in periodic steady operation, the stored energy returns to its starting value each cycle, so its net active power is zero despite nonzero exchange during the cycle. Real components can also have losses. The primary MIT treatment explicitly separates resistor dissipation from periodic inductive and capacitive energy exchange.^[ref]

Apparent power is the product of RMS terminal electric potential difference and electric current, conventionally expressed in volt-amperes:For sinusoidal signals at one frequency, it equals the magnitude $|S|$ of complex electrical power and satisfies $S_{\mathrm{app}}^2=P^2+Q^2$. It is not generally the average rate of net energy absorption, which is active power.^[ref]

For an original example, a100V terminal pair carrying5A RMS has500VA of apparent power. If active power is400W, the power factor is0.8. In the sinusoidal case the magnitude of reactive power is $\sqrt{500^2-400^2}=300$ reactive volt-amperes; the magnitude calculation alone does not determine whether the current leads or lags.

Magnitudes do not add like complex quantities. Consider two parallel sinusoidal loads at the same supply, with complex powers $300+j400$ and $300-j400$ in volt-amperes. Each has500VA of apparent power, but the combined terminal complex power is600VA with zero imaginary part. The source therefore sees600VA rather than the1000VA sum of individual magnitudes. Their internal branch currents and individual component requirements still exist; reactive cancellation at the source does not erase those branch conditions.

For nonsinusoidal waveforms, the product definition remains useful, but the simple equality with the magnitude of one fundamental-frequency complex power needs qualification. With $v(t)=\sqrt{2}(100)\cos \omega t$ and $i(t)=\sqrt{2}[3\cos \omega t+4\cos 3\omega t]$, RMS current is5A and apparent power is500VA. The fundamental-frequency complex electrical power is only300VA. The extra RMS current is not represented by that one complex number.

Thus record the waveform convention with an apparent power rating or measurement. Separately measuring RMS potential difference and current gives this product, while recovering active power requires the corresponding average product or adequate phase-resolved information. The Cauchy-Schwarz inequality guarantees $|P|\le S_{\mathrm{app}}$ without assuming a sinusoidal waveform.

Automatic power factor correction adjusts a correction device as the measured load changes, rather than leaving one fixed compensating setting connected. In a staged capacitor bank, a controller uses terminal electric potential difference and electric current information to choose which stages are connected. The controlled quantity and target must distinguish active power, signed reactive power, waveform distortion and the chosen lead/lag convention.

For an original single-phase sinusoidal model, available stages each supply $q_s$ of capacitive reactive power at the operating electric potential difference. With $k$ connected stages,At fixed $P>0$, maximizing the magnitude of the power factor chooses the available integer $k$ minimizing $|Q_L-k{q}_s|$. Away from the end limits, nearest-stage rounding leaves $|Q_{\mathrm{net}}|\le q_s/2$. This capacitor-stage quantization error can be positive or negative. A policy forbidding leading operation instead restricts $k{q}_s\le Q_L$, so it can choose a different setting.

Suppose $P=1000\mathrm{W}$, $Q_L=640\mathrm{var}$ and $q_s=250\mathrm{var}$. Three stages leave $-110\mathrm{var}$ and factor about $0.9940$ leading; two stages leave $+140\mathrm{var}$ and factor about $0.9903$ lagging. Both outcomes are consistent with their respective objectives. If the load changes to $P=200\mathrm{W}$, $Q_L=100\mathrm{var}$, leaving all three stages connected produces $-650\mathrm{var}$; zero stages is closer to cancellation than one. A stored stage decision must therefore be updated when the load changes.

Actual feedback must measure the corrected network. The Eaton paper places the current transformer so it senses the relevant load and capacitor contribution, and explains how a wrong phase, polarity or location produces misleading feedback.^[ref] Measuring only the correction branch would not establish the combined source power factor.

A proposed controller test varies a known resistive-inductive load through several operating points, recording stage state, signed reactive power, RMS electric current and synchronized terminal waveforms. Check both increasing and decreasing demand, including small excursions around a stage threshold. A dead band and timing rules can prevent repeated switching in response to small fluctuations; they must be specified along with the target. Stage size, switching transients and the load's harmonic components require additional checks beyond the ideal rounding model.

Capacitor-stage quantization error is the residual reactive power caused by choosing among discrete settings of a capacitor bank instead of a continuously adjustable correction. In a fixed sinusoidal operating model, let the load absorb $Q_L$ and let the target residual be $Q_t$. The desired capacitive correction is $q_{\ast }=Q_L-Q_t$. Selecting available correction $q$ leavesThe error concerns the selected setting, separately from sensing error, changing load and component tolerance.

For equal stages of magnitude $q_s$ and an available range covering the desired correction, nearest-stage selection chooses an integer $k$ nearest $q_{\ast }/q_s$. HenceAt a half-stage tie either neighbor attains the bound. A policy forbidding a negative residual relative to zero instead rounds downward and gives $0\le Q_L-k{q}_s<q_s$, away from the bank's capacity limits. The two policies solve different constrained problems.

For an original zero-target example, let $Q_L=640\mathrm{var}$ and $q_s=250\mathrm{var}$. Three stages give $r=-110\mathrm{var}$; two give $r=+140\mathrm{var}$. With active power $P=1000\mathrm{W}$, nearest-stage selection guaranteesprovided the zero-reactive target is within range and the sinusoidal fixed-load assumptions hold. At only $P=200\mathrm{W}$, the same worst-case bound falls to about $0.8480$. A stage size suitable at high demand need not give the same numerical power factor at low demand.

Unequal stages require the actual attainable set. Stages of 100 and 400 vars permit $\{0,100,400,500\}$ vars. At a desired 250 vars, the nearest setting leaves 150 vars error, despite the smallest stage being only 100 vars. The relevant bound is half the largest gap between neighboring attainable settings in the tested range, not half the smallest component. Outside the bank's range, clipping to an endpoint can leave a larger error.

At fixed frequency, a fixed capacitance gives $q_s=\omega C_s{V}^2$, so stage magnitudes also change with RMS electric potential difference. Eaton's paper describes discrete stage sizes, combinations and automatic control; the error bounds above are original consequences of that ideal setting model.^[ref] A proposed test records the actual stage states and terminal quantities while slowly varying a known load across thresholds. Compare the observed residual with the attainable-setting prediction before assigning every deviation to quantization.

Complex electrical power represents the active and reactive parts of sinusoidal electrical energy transfer in one complex number. Using phasors $V$ and $I$ whose magnitudes are root mean square values, with the passive sign convention, defineThe star denotes a complex conjugate. The real part is active power; the imaginary part is reactive power. The magnitude $|S|=|V||I|=\sqrt{P^2+Q^2}$ is apparent power. A source using peak-amplitude phasors instead has the factor $1/2$: $S=V_{\mathrm{peak}}{I}_{\mathrm{peak}}^{\ast }/2$.^[ref]

To see why the conjugate matters, write $V=|V|e^{j\alpha }$ and $I=|I|e^{j\beta }$. Then $S=|V||I|e^{j(\alpha -\beta )}$ depends on the relative phase. Changing the common time reference multiplies both phasors by $e^{j\theta }$, leaving $V{I}^{\ast }$ unchanged. The product $VI$ would instead acquire $e^{2j\theta }$ and would not represent a time-reference-independent average-power quantity.

For an original example, choose $V=100\mathrm{V}$ and $I=(4-j3)\mathrm{A}$. ThenThus active power is400W, reactive power is300 reactive volt-amperes and power factor is0.8. The corresponding electrical impedance is $V/I=16+j12\Omega$. With RMS current magnitude5A, the resistive term absorbs $16(5^2)=400\mathrm{W}$, agreeing with the real part.

Under the stated sign convention and $e^{j\omega t}$ time convention, an inductive reactance gives positive $Q$ and a capacitive reactance negative $Q$. Negative $Q$ does not mean a negative dissipated-energy average. For example, adding an ideal parallel capacitive branch contributing $-j300\mathrm{VA}$ cancels this example's reactive part, leaving terminal complex power400VA and RMS source current4A at100V, while the original resistive dissipation stays400W.

This one-frequency representation must not be applied indiscriminately to a distorted waveform. Different harmonics need their own phase relationships; the definition of active power remains the time average of the actual instantaneous power.

The current distortion factor is the ratio $D_I=I_1/I_{\mathrm{rms}}$, where $I_1$ is the root mean square magnitude of the fundamental electric current component and $I_{\mathrm{rms}}>0$ is the root mean square magnitude of the complete electric current. Its square is the fraction of the current's mean square supplied by the fundamental component. This ratio is distinct from displacement power factor, which measures the relative phase of the fundamental potential difference and current.

Use a Fourier series at fundamental frequency $\omega /(2\pi )$:Here $I_0$ is the DC component and the $I_n$ are RMS magnitudes of harmonic components. Averaging over a complete period, orthogonality and the Parseval identity giveFor purely sinusoidal electric potential difference $v(t)=\sqrt{2}V\cos \omega t$, only the fundamental electric current contributes to active power:Thus a displacement power factor of one can coexist with a lower total power factor.

For $I_1>0$, use fundamental-normalized current total harmonic distortion, ${\mathrm{THD}}_I=\sqrt{{\sum }_{n\ge 2}{I}_n^2}/I_1$. The relation between current distortion factor and THD with DC current isWith zero DC component, this reduces to $D_I=(1+{\mathrm{THD}}_I^2{)}^{-1/2}$. The definition of RMS here includes the DC component. For example, a 10 ampere fundamental and 3 ampere DC component, with no higher harmonic components, have zero total harmonic distortion but $D_I=10/\sqrt{109}\approx 0.9578$.

For an original zero-mean waveform, take a 230 volt RMS sinusoidal supply, $I_1=10\mathrm{A}$, $\cos {\varphi }_1=0.8$, and RMS third/fifth harmonic components of 6 and 3 amperes. ThenThe active power is 1840 watts, apparent power is about 2769.57 volt-amperes and total power factor is about 0.6644. This is an illustrative waveform model, not a measurement of a particular rectifier.

Terminology requires care. Overview of Functions: Harmonic Measurements by Schneider Electric uses distortion factor for the ratio of power factor to the fundamental cosine. Power Quality Considerations by Schneider Electric uses the label for harmonic content relative to the fundamental, or total harmonic distortion. The defining ratio identifies the intended quantity.

If the supply also contains harmonic components, use the harmonic active power sum:with matching voltage/current phase conventions. The fundamental-only factorization above assumes a purely sinusoidal supply.

The current reduction interval for shunt reactive compensation is the range of ideal capacitive corrections that draws less source RMS electric current than the uncorrected inductive load. It is wider than the range that avoids reactive power overcompensation.

Assume fixed sinusoidal RMS electric potential difference $V>0$, active power $P>0$, and inductive reactive power $Q_L>0$. Let $q=\omega C{V}^2\ge 0$ be the magnitude of the added capacitive reactive power. Comparing squared source electric currents givesConsequentlyor $0<C<2C_{\ast }$ with $C_{\ast }=Q_L/(\omega V^2)$. The model's capacitive reactive power follows the sinusoidal complex electrical power convention.^[ref]

At $q=Q_L$, source electric current is smallest. Between $Q_L$ and $2Q_L$, the source has a leading power factor but still draws less electric current than before correction. At $q=2Q_L$, the residual reactive power is $-Q_L$: its magnitude matches the original value, restoring the original source-current magnitude. Beyond that point the corrected electric current is larger.

For an original example, choose $P=300\mathrm{W}$, $Q_L=400\mathrm{var}$ and $V=100\mathrm{V}$. Uncorrected electric current is $5\mathrm{A}$. Full cancellation $q=400\mathrm{var}$ gives $3\mathrm{A}$. An overcorrection $q=600\mathrm{var}$ leaves $-200\mathrm{var}$ and draws only $\sqrt{13}\approx 3.606\mathrm{A}$. But $q=800\mathrm{var}$ restores $5\mathrm{A}$, and $q=1000\mathrm{var}$ increases it to $\sqrt{45}\approx 6.708\mathrm{A}$.

Two choices $q$ and $2Q_L-q$ are symmetric around the optimum and give the same squared source electric current. This symmetry explains why power factor magnitude alone cannot identify the sign of the residual reactive power. If $Q_L\le 0$, no positive $q$ reduces current under the same fixed-load assumptions.

With the same upstream electrical resistance, reducing source mean-square electric current also reduces that conductor's Joule heating. It does not erase the current carried by the local capacitor branch or assess its ratings. Harmonic interactions, nonideal components and changing load conditions need their own model; this interval is an exact statement about the stated one-frequency circuit.

The current-transformer burden is the load impedance connected to a current transformer's secondary, including the instrument and secondary connection leads. It determines the electric potential difference needed to drive the secondary electric current. Changing it can change both current-transformer ratio error and current-transformer phase displacement. ABB's application guide explicitly connects these errors to the secondary burden and distinguishes standard burdens from actual connected devices.^[ref]

For sinusoidal RMS secondary current $I_s$ and external burden $Z_b=R_b+i{X}_b$, the secondary terminal relations areHere $S_b$ is complex electrical power; its real part is dissipated active power. A volt-ampere burden magnitude alone does not determine its power factor or its complex complex impedance.

For an original resistive example, an instrument contributes $0.10\Omega$ and the complete outgoing-plus-return leads contribute $0.20\Omega$. At $5\mathrm{A}$ RMS, total burden is $0.30\Omega$, terminal magnitude is $1.5\mathrm{V}$, and burden apparent power is $7.5\mathrm{VA}$. The instrument dissipates $2.5\mathrm{W}$ and the leads dissipate $5\mathrm{W}$. Ignoring the leads would understate the required terminal magnitude by a factor of three. A stipulated rating of 5 VA at 5 A corresponds to only $0.20\Omega$ magnitude, so this setup would not satisfy that stated burden limit.

Equal VA values also need not mean equal complex loading. At 5 A, $Z_b=0.20\Omega$ and $Z_b=0.12+0.16i\Omega$ both have 5 VA magnitude. The second instead absorbs $3\mathrm{W}$ and $4\mathrm{var}$, with power factor0.6. The resulting phase and ratio effects depend on the transformer's own response.

The secondary winding's internal impedance is not part of the external burden, but both enter the induced-voltage and excitation calculation. NIST's equivalent circuit separates these terms explicitly.^[ref] A proposed calibration comparison records the complete connected impedance at the relevant frequency, including leads, and repeats ratio and phase measurements at selected burdens. It does not infer unchanged accuracy merely because the nominal turns remain the same.

Current-transformer phase displacement is the phase difference between the secondary-current waveform and the corresponding primary-current waveform after accounting for the chosen terminal polarity. Define $\alpha >0$ when the secondary reproduction leads the primary. NIST uses this sign convention and reports the angle directly; a positive amplitude correction factor does not specify the angle.^[ref]

For sinusoidal signals, choose primary RMS phasor ${\underline{I}}_p=I_p{e}^{-i\phi }$ relative to a real supply phasor $V$. Let the nominally scaled secondary reproducewhere ${ϵ}_I$ is the current-transformer ratio error. A meter using this current and an otherwise exact supply measurement infersFor $\cos \phi \ne 0$,A lag-defined phase error would have the opposite sign in the last term, so the convention must accompany the number.

For an original example, set $V=100\mathrm{V}$, $I_p=5\mathrm{A}$, $\phi =\pi /3$, ${ϵ}_I=0$ and $\alpha =0.01$ radians. True active power is $250\mathrm{W}$, while the current-shifted indication is about $254.32\mathrm{W}$, an error of $+1.727\%$. At unity displacement power factor, the same phase displacement instead gives $\hat{P}/P=\cos 0.01\approx 0.999950$, a much smaller second-order effect. Low power factor makes phase characterization especially important for relative power accuracy.

At zero true active power, a relative-error ratio is undefined. For $\phi =\pi /2$, the shifted indication is $V{I}_p\sin \alpha$, which can be nonzero despite zero true average transfer. Report absolute error there rather than dividing by zero.

A proposed phase check compares the two current waveforms at the stated frequency, current level and current-transformer burden, with terminal direction and channel delay recorded. A time skew $\Delta t$ contributes a phase shift of magnitude $2\pi f|\Delta t|$; at 50 Hz, 10 microseconds is about 3.142 milliradians. The measurement chain must distinguish that delay from the transformer itself. The NIST publication's historical calibration apparatus measured ratio and phase separately; the examples here are derived illustrations, not reported NIST performance.

Daniel J. Carnovale is an electrical engineer associated with investigating and demonstrating the behavior of electric power systems. His coauthored Capacitor Application Issues by Thomas M. Blooming and Daniel J. Carnovale identifies his Eaton affiliation; its author biography describes a demonstration and test facility for reproducing electrical supply problems and evaluating their mitigation.^[ref]

One important distinction in power factor correction is between reduced source electric current and reduced useful-load active power. For an original idealized example, a load takes $P=1000\mathrm{W}$ and $Q=750\mathrm{var}$ at $V=200\mathrm{V}$ RMS. Before correction,Canceling the fundamental reactive power makes source electric current $I_1=P/V=5\mathrm{A}$, but leaves the stipulated load's $1000\mathrm{W}$ unchanged. With a fixed upstream electrical resistance of $0.2\Omega$, conductor Joule heating falls from $7.8125$ to $5\mathrm{W}$. The reduction is $2.8125\mathrm{W}$, rather than 20% of the load's active demand.

A proposed demonstration measures load active power, source RMS electric current and upstream conductor loss separately while connecting the correction branch. It makes the changed and unchanged quantities visible. The same input-output discipline applies when comparing a device's apparent electrical savings with useful service delivered. The numerical example above is an original fixed-boundary calculation, not a measurement attributed to Carnovale.

His paper with Thomas M. Blooming places such calculations within practical capacitor application. Choosing a nominal correction alone does not describe how variable load and the existing network affect operation. The distinction motivates recording terminal waveforms and operating conditions along with a single power factor number.

The displacement power factor is $\cos {\phi }_1$, where ${\phi }_1$ is the phase difference between the fundamental components of terminal electric potential difference and electric current. It describes phase displacement at that frequency. For purely sinusoidal signals it equals total power factor; with distorted waveforms those quantities can differ. The one-frequency phase relation follows from the RMS average-power expression.^[ref]

For a purely sinusoidal potential difference of RMS magnitude $V$ and zero-mean periodic current, orthogonality of harmonics givesThe ratio $I_1/I_{\mathrm{rms}}$ is the current distortion factor under these assumptions. A displacement factor near one does not establish that the whole current waveform closely follows the supply sinusoid.

For an original example, let the supply be100V RMS. Let the fundamental current be4A RMS at a lagging phase with cosine0.8, and add a3A RMS third harmonic. Total RMS current is5A, active power320W and apparent power500VA. The displacement factor is0.8 while total power factor is0.64.

Now cancel only the fundamental quadrature current, whose magnitude is $4\sqrt{1-0.8^2}=2.4\mathrm{A}$. The remaining in-phase fundamental is3.2A and the third harmonic is still3A. RMS source current is $\sqrt{3.2^2+3^2}\approx 4.39\mathrm{A}$, so total factor is about0.73 even though the displacement factor has become one. This illustrates a limitation of phase-only power factor correction, not a universal model of a correcting device.

The cosine also loses lead/lag sign: phases $+{\phi }_1$ and $-{\phi }_1$ have the same displacement factor but opposite sinusoidal reactive power. Retain the phase or lead/lag convention when that distinction matters. If the potential difference itself contains harmonics, total active power needs the contributions of the corresponding harmonic pairs; this simple fundamental-only reduction is no longer sufficient.

Electric power transmission transfers bulk electrical power between generating areas, major network junctions and receiving systems. Transmission lines carry electric current; electrical transformers change the operating electric potential difference in alternating current networks. Raising that potential can reduce the electric current required for a specified transferred power, and therefore reduce resistive heating. It also changes insulation and equipment requirements.

For an original fixed-output direct current comparison, deliver $P_L=10\mathrm{MW}$ through a line with total conductor electrical resistance $R=2\Omega$. At receiving-end electric potential difference $V_L$, useAt $V_L=10\mathrm{kV}$, the electric current is $1000\mathrm{A}$, conductor heating is $2\mathrm{MW}$, and the source must supply $12\mathrm{MW}$ at $12\mathrm{kV}$. At $100\mathrm{kV}$, the corresponding values are $100\mathrm{A}$, $20\mathrm{kW}$ and $10.02\mathrm{MW}$ at $100.2\mathrm{kV}$. This compares the same receiving load and line electrical resistance; it does not assume equal source power in the two cases.

The resulting loss reduction by a factor of one hundred follows from squaring the tenfold electric current reduction. It does not imply lossless transfer: electrical transformers, converters and conductors have their own losses. The DOE treatment explicitly separates winding copper loss from useful transformer output.^[ref]

Long alternating current lines also have distributed inductance and capacitance, so a resistive balance alone cannot determine their phase relationships or operating limits. An electric power transmission calculation must specify whether its electric potential differences are sending-end or receiving-end values, and whether stated power is active power or apparent power. Those conventions keep conductor-current calculations consistent with power factor.

The harmonic active power sum expresses average electrical power as the sum of contributions from matching components of periodic terminal electric potential difference and electric current. With a common fundamental frequency and root mean square magnitudes for nonzero harmonic components, writeHere $V_0,I_0$ are the signed DC components. The passive sign convention makes the average product absorbed active power. Averaging over a complete fundamental period givesFor finite sums this follows from the cosine-product identity: products of different integer harmonic components average to zero, and matching components give half the cosine of their relative phase. For square-integrable periodic waveforms, the corresponding Fourier series identity follows by continuity of the inner product in the mean-square norm. The orthogonality is a property of the averaging interval and basis, not an assumption that the physical sources are independent.

For an original example, set $V_0=10\mathrm{V}$, $I_0=2\mathrm{A}$, $V_1=100\mathrm{V}$, $I_1=3\mathrm{A}$, and fundamental relative phase $\pi /3$. Add third-harmonic magnitudes $V_3=20\mathrm{V}$, $I_3=1\mathrm{A}$ with relative phase $\pi$. ThenThe third-harmonic contribution is $-20\mathrm{W}$, even though the total absorbed active power is positive. The root mean square values satisfy $V_{\mathrm{rms}}^2=10500{\mathrm{V}}^2$ and $I_{\mathrm{rms}}^2=14{\mathrm{A}}^2$, so apparent power is about $383.41\mathrm{VA}$ and total power factor about $0.3912$.

For a purely sinusoidal supply with no DC component, only $V_1$ survives, giving $P=V_1{I}_1\cos ({\alpha }_1-{\varphi }_1)$ even if the electric current is distorted. Current in other harmonic components can still increase conductor heating. The usual one-frequency complex electrical power construction is derived in the primary MIT treatment; the sum above applies that average-product reasoning separately to orthogonal components.^[ref]

A proposed measurement records the two waveforms synchronously and compares their directly averaged product with the resolved sum. A non-integer-period window, unresolved harmonic components, channel timing error or changing waveform can spoil the simple finite-record equality. The formula assumes a shared periodic model; multiplying separately measured RMS magnitudes gives apparent power instead.

James L. Kirtley Jr. is an electrical engineer whose work concerns electric machines and electric power systems. MIT's departmental profile identifies him as a professor emeritus of electrical engineering, with energy and robotics among his research areas.^[ref]

The engineering problem is to match a motor's behaviour to its application. MIT reported in 2016 that Kirtley worked on customized electric motors for the MIT Cheetah robot, where brief peaks of torque serve running and jumping. The same interview describes his interest in electric motors interacting with microgrids, particularly their dynamic behaviour during supply disturbances.^[ref] A short burst and sustained operation place different demands on the system; the largest instantaneous output alone does not describe its operating role.

His AC Power Flow in Linear Networks by James L. Kirtley Jr. connects the phasor representation of electric potential difference and electric current to physical energy transfer. In its peak-amplitude convention, complex electrical power is $S=V_{\mathrm{pk}}{I}_{\mathrm{pk}}^{\ast }/2$. With RMS phasors $V=V_{\mathrm{pk}}/\sqrt{2}$ and $I=I_{\mathrm{pk}}/\sqrt{2}$, this becomes $S=V{I}^{\ast }$. The complex conjugate ensures that the relative phase enters the active power.^[ref]

For an original load example in sinusoidal steady state, take $V=20\mathrm{V}$ and electrical impedance $Z=(3+4i)\Omega$, using RMS phasors and the passive sign convention: electric current enters the positive-potential terminal. A series resistor and inductor realize this model at the specified angular frequency when $R=3\Omega$ and $\omega L=4\Omega$. ThenThus active power is 48 watts, reactive power is 64 vars, apparent power is 80 volt-amperes, and power factor is $48/80=0.6$. Write $\varphi =\arctan (4/3)$ and $x=\omega t$. The corresponding electric potential difference and electric current areMultiplying and applying a trigonometric identity gives instantaneous powerIts cycle average is 48 watts, while its range is $[-32,128]$ watts. The negative portions represent temporary return of stored energy from this ideal load; the electrical resistance still dissipates a positive average. A synchronized product of the two waveforms reveals this behaviour, whereas multiplying their root mean square magnitudes gives only the apparent power.

John D. Ramboz is the first author, with Oskars Petersons, of NIST's 1991 account of a laboratory current transformer calibration service. The report describes ratio and phase measurements, their uncertainties and quality-control procedures for electrical measurement work.^[ref] It is a primary technical record of the service and its measurement conventions.

The separation of magnitude and phase has a direct consequence for electric power measurements. Consider an original sinusoidal model with root-mean-square electric potential difference $V$, secondary current magnitude $I_s$, true current ratio $K$ and primary-current lag $\varphi$. True active power isIf the secondary leads the primary by $\beta$, simply multiplying the secondary signal by nominal ratio $K_n$ givesprovided $\cos \varphi \ne 0$. This uses the lead-positive current-transformer phase displacement convention stated in the report.^[ref]

With exact magnitude conversion $K_n=K$ and small $\beta$ in radians, the relative power error is approximately $\beta \tan \varphi$. For stipulated $\beta =0.001$ rad and $\varphi =60^{\circ }$, that term is about $0.173\%$. At $\varphi =0$, its first-order value vanishes. Thus the same phase error can have different effects at different power factors.

This derivation is an explanatory example, not a reported performance test or an attribution of the calculation to Ramboz. It explains why a calibration report needs operating conditions, a sign convention and measurement uncertainty, in addition to a nominal transformer ratio. Precision in the current magnitude does not remove phase-dependent power error.

A leading power factor describes an absorbing sinusoidal electrical load whose electric current leads its terminal electric potential difference. Under the passive sign convention, writeThe current maximum occurs earlier by $\theta /\omega$. The complex electrical power isThe negative reactive power is consistent with a capacitive terminal response. It does not imply negative mean absorbed active power.

For an original $100\mathrm{V}$, $2\mathrm{A}$ example with $\theta =\pi /3$, the power factor magnitude is $0.5$, active power is $100\mathrm{W}$, and reactive power is $-100\sqrt{3}\approx -173.21\mathrm{var}$. At $50\mathrm{Hz}$ the current leads by $1/(6f)\approx 3.33\mathrm{ms}$. A load with the same RMS magnitudes and the opposite phase displacement has identical active power but a lagging power factor and opposite reactive power.

For an ideal capacitor, $i=Cdv/dt$ gives a quarter-cycle current lead. Its mean active power is zero, although instantaneous power alternates as electrical energy enters and leaves the field. The primary one-frequency power derivation separates this quadrature exchange from net energy transfer.^[ref]

An inductive load can acquire a leading power factor at the source when power factor correction supplies more capacitive reactive power than the load absorbs. The current reduction interval for shunt reactive compensation shows why mildly leading operation can still draw less source electric current than no correction, while excessive correction reverses that benefit.

Record phase direction or signed reactive power together with the numerical factor. The same positive cosine can describe leading and lagging loads, and a meter may assign a separate sign according to its display convention.^[ref] For nonsinusoidal signals, state whether leading describes the fundamental displacement power factor; the complete power factor also depends on waveform distortion.

Optimal shunt capacitance for sinusoidal power factor correction minimizes the source's RMS electric current for an ideal fixed sinusoidal supply and a fixed load with active power $P>0$ and inductive reactive power $Q_L>0$. A lossless parallel capacitor supplies signed reactive power $-\omega C{V}^2$, where $V$ is the supply RMS electric potential difference and $\omega >0$ the angular frequency.

The squared source electric current isDifferentiating givesThus the unique minimum isThe fundamental quadrature electric current vanishes there. Strict convexity proves this is a global optimum in the stated model, rather than merely a stationary point. The capacitive sign and one-frequency complex electrical power relation follow from the primary MIT derivation.^[ref]

For an original example, $P=800\mathrm{W}$, $Q_L=600\mathrm{var}$, $V=100\mathrm{V}$ and $f=50\mathrm{Hz}$ give $C_{\ast }\approx 190.99\mu \mathrm{F}$. Source electric current falls from $10\mathrm{A}$ to $8\mathrm{A}$. A capacitance 20% below or above $C_{\ast }$ leaves $Q_{\mathrm{net}}=+120$ or $-120\mathrm{var}$, respectively, and both draw about $8.0895\mathrm{A}$. They have the same power factor magnitude, about $0.9889$, but opposite lead/lag direction.

The exact excess above the optimal squared electric current isSmall sizing errors therefore affect this loss objective quadratically around the optimum, while residual reactive power changes linearly and changes sign. An allowed interval $0\le C\le C_{\max }$ clips the optimum to its nearest feasible point. If the original load has $Q_L\le 0$, the constrained optimum for added nonnegative capacitance is instead $C=0$.

The result assumes a fixed load, ideal sinusoidal supply, lossless correction and no additional network effects. It does not optimize a system with waveform distortion, component heating, variable terminal electric potential difference or discrete stages. The current reduction interval for shunt reactive compensation describes a wider set of beneficial choices, while automatic power factor correction follows a changing load with available stages.

The power factor is the ratio of active power to the product of RMS terminal electric potential difference and electric current:when the denominator is positive. It measures how effectively those terminal magnitudes support net average transfer. With signed absorbed power it can be negative for net generation; for an absorbing load the usual range is zero to one. The Cauchy-Schwarz inequality applied to the time average of $v(t)i(t)$ proves $|P|\le V_{\mathrm{rms}}{I}_{\mathrm{rms}}$.

For one-frequency sinusoids, $P=V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \phi$, so the power factor is $\cos \phi$. The phase-only quantity for the fundamental is also called displacement power factor. The primary treatments make the RMS convention explicit.^[ref][ref]

A distorted current can give another result. For purely sinusoidal supply potential difference and a zero-mean current with fundamental RMS magnitude $I_1$,Only the in-phase fundamental contributes to the average product with that supply waveform. For an original example with fundamental3A in phase and third-harmonic RMS magnitude4A, total RMS current is5A. The displacement power factor is one but total power factor is0.6. Changing the fundamental phase alone would not remove this harmonic-current contribution.

At fixed absorbed active power and supply RMS potential difference, $I_{\mathrm{rms}}=P/(V_{\mathrm{rms}}\mathrm{pf})$. For an original900W,230V example, changing the factor from0.6 to0.9 changes source current from about6.52 to4.35A. In the same upstream resistance, the corresponding $I^{2}R$ loss becomes $(0.6/0.9{)}^2=4/9$ of its former value. This is a stipulated circuit calculation, not an assertion about a particular tariff or installation.

Power factor correction addresses the mechanism causing the low ratio. An ideal capacitive branch can offset inductive reactive power at one frequency; it does not automatically correct an arbitrary distorted current. State the waveform conditions and whether the reported number is total or displacement factor when comparing measurements.

Power factor correction changes the current drawn from a source so that a specified active power is supplied with a more favorable power factor. For a sinusoidal inductive load, a parallel capacitor can supply an opposite reactive power component. It changes the source current's quadrature component while an ideal lossless correction branch leaves the load's active power unchanged.

Let the original load absorb $P+j{Q}_L$ at RMS potential difference $V$ and angular frequency $\omega$. A parallel ideal capacitance contributes $Q_C=-\omega C{V}^2$, soTo change a lagging phase angle from ${\phi }_1$ to a smaller lagging angle ${\phi }_2$ at fixed $P>0$,These follow from complex electrical power and the capacitive phase relationship.^[ref]

For an original ideal example, take $P=1000\mathrm{W}$, $V=200\mathrm{V}$ and50Hz. An initial factor0.8 corresponds to750var. A target factor0.95 corresponds to about328.68var, so a capacitance about33.53µF supplies the difference. RMS source current changes from6.25A to about5.26A. Resistive feeder loss at fixed resistance changes by $(0.8/0.95{)}^2\approx 0.709$, while the1000W load conversion remains the same in this model.

Reactive power overcompensation occurs if the capacitive contribution exceeds the intended inductive component. The net $Q$ then changes sign; continuing to add capacitance does not keep improving the magnitude of the factor. At a different load or frequency the same component can produce another result. Component ratings and nonideal circuit behavior require their own model rather than being determined by this one-frequency algebra.

Correction of displacement power factor is also distinct from correction of waveform distortion. Harmonic current can remain after the fundamental quadrature component is canceled. A useful measurement compares synchronized source potential-difference/current waveforms, RMS magnitudes and their average product before and after a change, retaining the operating condition. Separately measuring only RMS magnitudes cannot establish the resulting active power.