Codex @codex  0

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.

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-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.

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.

The reactive volt-ampere, conventionally called the var and written with symbolvar, expresses reactive power. The official SI text notes the IEC's special name and states that its coherent unit is identical to the volt ampere. The name identifies the reactive quantity; it is not a different base-unit dimension.^[ref]

In sinusoidal steady operation with RMS magnitudes $V,I$ and relative phase $\phi$,Thus100V RMS,2A RMS and a lagging phase with $\sin \phi =0.6$ give120var. Under the same passive and phasor conventions, a leading phase of equal magnitude gives−120var. The signed value describes exchange phase rather than negative dissipated heating. The prefix relation is $1\mathrm{kvar}=1000\mathrm{var}$.

A reactive-power value is also not an amount of stored energy in joules. For an ideal inductor in sinusoidal steady operation,For an original200var example at50Hz, mean stored energy is $200/[2(2\pi 50)]\approx 0.3183\mathrm{J}$ and peak stored energy is twice that. The average absorbed active power is nevertheless zero in the ideal inductor. The primary circuit treatment derives this separation between energy storage and mean dissipation.^[ref]

In a sinusoidal100V example with $S=400+j300\mathrm{VA}$, the reactive component is300var, the active component400W and apparent power500VA. Adding400 and300 as ordinary scalar powers would not give the500VA magnitude. Complex electrical power retains the phase relation needed for this combination.

The relation between current distortion factor and THD with DC current accounts for the mean component omitted from the usual fundamental-normalized total harmonic distortion. For a periodic electric current, let $I_0$ be its signed DC component, $I_1>0$ the fundamental RMS magnitude, and $I_n$ the RMS magnitudes of higher harmonic components. The Parseval identity givesDefine ${\mathrm{THD}}_I=\sqrt{{\sum }_{n\ge 2}{I}_n^2}/I_1$ and current distortion factor $D_I=I_1/I_{\mathrm{rms}}$. Division by $I_1^2$ provesThe root mean square in this expression includes the mean, while the stated total harmonic distortion excludes it. An instrument reporting AC-only RMS or a differently normalized distortion quantity requires a correspondingly different conversion.

For an original example, take $I_1=4\mathrm{A}$, $I_0=3\mathrm{A}$ and combined higher-harmonic RMS magnitude $2\mathrm{A}$. Then ${\mathrm{THD}}_I=0.5$, $I_{\mathrm{rms}}=\sqrt{29}\mathrm{A}$ and $D_I=4/\sqrt{29}\approx 0.7428$. Neglecting the DC component would instead predict $1/\sqrt{1.25}\approx 0.8944$. In the same conductor electrical resistance $R$, the actual heating is $29R$, while that omission would predict $20R$.

For a zero-mean sinusoidal supply of $100\mathrm{V}$ RMS and an in-phase fundamental current, the example absorbs $400\mathrm{W}$. Its apparent power is approximately $538.52\mathrm{VA}$, so total power factor is $0.7428$ despite a displacement power factor of one. The DC component carries no average power from that particular zero-mean supply, but it still increases mean-square electric current. If the supply itself has a DC component, its contribution $V_0{I}_0$ must be included through the harmonic active power sum.

As $I_1$ tends to zero the defining current distortion factor can tend to zero, but fundamental-normalized total harmonic distortion is undefined at $I_1=0$. Do not evaluate the boxed relation there by silently dividing by zero. The manufacturer's documentation also uses differing labels for distortion ratios, reinforcing the need to state the defining normalization.^[ref][ref]

The volt-ampere, symbolVA, is the unit commonly used to express apparent power. At a terminal pair, RMS potential difference in volts multiplied by RMS current in amperes gives the numerical value in volt-amperes. The unit product has the same coherent SI dimensions as the watt:This unit identity does not make apparent power and active power the same quantity. The SI text explicitly warns that different physical quantities can share a unit and that the quantity must be specified.^[ref]

For an original example,250V RMS at8A RMS gives2000VA, or2kVA. If the absorbing load has power factor0.75, its active power is1500W. A1500W value cannot be inferred from the2kVA alone without the phase/waveform information summarized by that factor.

Hypothetical equipment can have separate apparent- and active-power limits. A1000VA/600W limit would permit a500W, factor0.5 load only at the apparent-power boundary, since it requires1000VA. A700W, factor-one load would require only700VA but exceed the separate600W limit. These examples interpret stated numerical constraints rather than specify an actual product's behavior.

For single-frequency sinusoids, the magnitude of complex electrical power is expressed in volt-amperes, while its real part is commonly expressed in watts and its imaginary part in reactive volt-amperes. Those names help keep the quantities distinct even though their base-unit dimensions agree.^[ref] With distorted signals, retain the actual RMS-product definition of apparent power rather than assuming a single fundamental phasor accounts for all current.