Alternating current, phasors, and electrical power

Start with the energy crossing a pair of terminals

A source can maintain a sinusoidal terminal voltage while the current has the same frequency but a different phase. In the original controls, a 10 V RMS voltage and a 1 A RMS current can correspond to 10 W, 6 W, or zero average power. The product of the two RMS magnitudes alone does not tell us how much energy crosses the terminals over a complete cycle.

The difference matters even when we care about ordinary heating. A resistive wire carrying current dissipates i(t)²R at each instant and I_rms²R on average. A current component that returns energy to the source during part of a cycle can therefore still occupy current capacity and heat a resistive feeder. Zero mean energy transfer at one port does not mean that the surrounding circuit has no losses.

We will use one declared reference system throughout, solve a network whose branches cannot all be reduced by series and parallel combinations, and then examine two ways of maintaining a load voltage. The final comparisons deliberately leave the pure-sinusoid setting so that the limitations of a single phasor and a simple power triangle become visible.

Choose a first reading and an advanced return

For a first reading, follow the RMS convention, element impedances, two node equations, instantaneous and complex power, and the correction and harmonic examples. The circuit-foundations lesson supplies associated voltage/current references and Kirchhoff laws; the DC-network lesson supplies nodal analysis. The RC/RL and RLC lessons explain why a steady-state solution is not an initial-value solution.

Return afterwards to the conjugated network proof, the real-signal representation, the equality condition for power factor, and balanced three-phase power. These extend the same argument rather than introducing another set of unexplained signs. The advanced discussion of singular operators and material response states when the familiar algebra ceases to describe the physical problem.

Every waveform here is prescribed or belongs to a stated linear lumped model. The harmonic current is not a hidden rectifier calculation, and the balanced three-phase example is not an arbitrary power-grid model. Those boundaries let us investigate power carefully without assuming devices or system conditions that have not been specified.

Separate AC, DC and sinusoidal steady state

An alternating signal varies with time; it need not be a perfect sinusoid and it may have a nonzero mean. We write the mean separately as a DC component. For a periodic function x(t) with period T, the mean is T⁻¹∫₀ᵀx(t)dt. The fundamental angular frequency is ω=2π/T, measured in radians per second; f=1/T is measured in hertz.

Sinusoidal steady state means that the response being described has one settled sinusoidal frequency. A stable linear time-invariant network driven at that frequency has such a particular response after its decaying natural terms have become negligible. The phasor calculation determines that particular response. It does not reconstruct the initial state or erase undamped natural motion.

An ideal lossless system may keep an independently prepared oscillation. At a resonant forcing singularity it may have no bounded sinusoidal particular solution of the assumed form. Thus “replace derivatives by jω” is an ansatz to be checked, not a universal argument that every switched or forced circuit is already in steady state. The RLC lesson gives the corresponding natural-response and resonance analysis.

Choose one cosine, RMS and phase convention

Here j²=−1 and i(t) denotes a real current. Our RMS phasor V is defined by v(t)=√2 Re{V exp(jωt)}. If V=V_rms exp(jφ_v), then v(t)=√2 V_rms cos(ωt+φ_v). Its magnitude is the nonnegative RMS value; its argument is the phase relative to the chosen time origin. Current uses the same convention.

For a sinusoid, x_rms²=T⁻¹∫₀ᵀx(t)²dt=X_peak²/2. The peak is therefore √2 times the RMS magnitude, not the same number. RMS itself is a real nonnegative scalar; an “RMS phasor” combines that magnitude with a phase. A two-sided Fourier coefficient has yet another normalization and must not silently replace either one.

Our physical scales are V*=10 V, I*=1 A, Z*=V*/I*, and ω*=2000 rad/s. The dimensionless clock is θ=ω*t; 1/ω* is a time scale, while a full period is 2π/ω*. Keeping the radian and the factor 2π visible prevents the common confusion between frequency and angular frequency.

Reconstruct a real signal from a complex amplitude

If V=V_r+jV_i, then v(t)=√2[V_r cos(ωt)−V_i sin(ωt)]. The minus sign follows directly from multiplying by cos(ωt)+j sin(ωt) and taking the real part. For I=(3/5−j4/5) A, the real current is √2[(3/5)cos θ+(4/5)sin θ] A. Its imaginary phasor component has not become an imaginary physical current.

Changing the common time reference rotates all phasors by the same unit complex factor. A measured relative phase does not change. Our rotated-lagging control rotates both the voltage and current together; their individual Cartesian coordinates change while their complex power remains the same. Rotating only one of them changes the physical phase difference.

A zero phasor has no phase. Assigning an arbitrary angle to it may be harmless in a private algebraic placeholder, but it is not an observable phase and must not appear as a measured zero-degree result. The zero-current and zero-voltage controls keep that distinction explicit.

Keep both halves of a real sinusoid

The RMS representation can also be written x(t)=[X exp(jωt)+X* exp(−jωt)]/√2. The negative-frequency coefficient is the conjugate of the positive-frequency coefficient because x(t) is real. The two coefficients are not two independent physical sources.

This full form explains why multiplying real voltage and current produces a constant term as well as terms at twice the frequency. Multiplying only V exp(jωt) by I exp(jωt) retains the double-frequency part and loses the terms that determine average power. Taking the real part after such an incorrect multiplication does not restore what was omitted.

An alternative exp(−jωt) convention is legitimate, but the derivative rule, phasor phases and reactive-power convention must all be translated consistently. Keeping the present jωL and V I* formulas while reversing only the time exponential would change signs. We keep one convention instead of switching between them mid-calculation.

Add only quantities that belong to the same frequency

Linear time-invariant constitutive equations permit superposition. At one specified ω, voltages and currents can be added as complex phasors because the same time factor multiplies every term. Two source phasors at different frequencies cannot be added into one frequency-independent arrow that reconstructs their real sum.

For a finite Fourier series in a linear network, solve separately at each harmonic frequency and add the resulting real waveforms. The impedance must be evaluated at that harmonic, not kept at its fundamental-frequency value. A DC term is treated by the DC model and any inherited state conditions; it is not obtained by inserting infinity into ordinary finite complex arithmetic.

Linearity does not make power a linear function of independent voltage and current drives. Products generate cross terms even when the state equations are linear. A nonlinear or time-varying element can also couple frequencies, so separate linear harmonic solutions are then a small-signal approximation or an incomplete model, not a general solution.

Derive the element impedances with associated references

For a resistor with current entering the positive voltage terminal, v=Ri. Hence V=RI and Z_R=R. For a fixed linear inductor, v=L di/dt; substituting our real sinusoid gives V=jωL I and Z_L=jωL. For a fixed linear capacitor, i=C dv/dt gives I=jωC V and Z_C=1/(jωC)=−j/(ωC).

The signs encode measurable phase relations under the stated convention: the ideal inductor current lags its voltage by π/2, while the ideal capacitor current leads by π/2. Reversing a branch reference changes its assigned voltage or current signs; it does not change the physical constitutive relation when both associated references are updated together.

Dimensions provide an immediate check. H·s⁻¹ and 1/(s⁻¹F) both have units of ohms; the reciprocal has units of siemens. These ideal laws assume fixed linear parameters and a lumped port description. Winding resistance, dielectric loss, core saturation, skin effects and propagation can change the actual impedance and its frequency dependence.

Use admittance without hiding an excluded division

Admittance is Y=1/Z when that reciprocal exists. At a common voltage, parallel branch currents add, so their admittances add. In a series path carrying one current, the impedances add. These rules describe the topology as well as the algebra; they are not a license to combine two elements that do not share the required voltage or current.

With Z=R+jX, the reciprocal is Y=(R−jX)/(R²+X²). Thus the susceptance of a simple inductive impedance is negative, even though its reactance is positive. Mixing the sign of reactance with the sign of susceptance can reverse a compensation calculation.

An absent shunt capacitor has zero admittance. Its impedance need not be represented by a finite number or a floating-point infinity. Likewise, V/I is undefined for the zero-current control. At zero frequency, ideal capacitor and inductor limits require a separate DC and initial-state analysis; all frequency controls of our bridge are strictly positive.

Define the oriented AC bridge before solving it

The original bridge has datum node 0, a maintained source node s, and unknown nodes a and b. The source is V_s=10 V RMS with phase zero. Four resistors connect s→a, a→0, a→b and s→b with resistances 2Z*, 3Z*, 4Z* and 5Z*. A capacitor C=2/(ω*Z*) connects a→0; an inductor L=3Z*/ω* connects b→0.

Each arrow defines positive branch current and voltage drop from its first node to its second. The ideal source is also oriented s→0, with positive terminal at s. Its current is determined by KCL and is positive into that positive terminal. This convention allows the source to have negative absorbed power while delivering energy to the rest of the network.

The a–b resistor links two different reactive dividers. It prevents a complete reduction to independent series/parallel pairs. We evaluate x=ω/ω* at 1/2, 1 and 2 to show how the same nodal method handles changed impedances. These are three finite operating states, not a claim that they locate or characterize a resonance peak.

Solve the two node equations in complex form

Let u_a=V_a/V* and u_b=V_b/V*. Divide the node equations by V*/Z*. At node a, [(1/2+1/3+1/4)+j2x]u_a−u_b/4=1/2. At node b, −u_a/4+[(1/4+1/5)+1/(j3x)]u_b=1/5. These equations come from summing outgoing currents at each unknown node.

The normalized coefficient matrix contains both conductance and susceptance. Its determinant must be nonzero before the familiar two-by-two inverse may be used. For the stated positive resistances and x>0, a zero-source homogeneous solution would dissipate positive real power unless the node voltages vanish. The conductive connections to fixed-potential nodes therefore rule out a nontrivial homogeneous mode and give uniqueness for this bridge.

At x=1, the exact solution is V_a=(54534−j85572)/38417 V and V_b=(152550+j65460)/38417 V. Their phases differ because they sit in different branches. Node labels do not imply that either voltage is in phase with the source. Substitution into both complex equations, rather than matching one magnitude alone, checks the solution.

Check every branch and the source

For an ordinary branch, V_branch=V_from−V_to and I_branch=Y_branch V_branch. The source current is I_source=−(I_Rsa+I_Rsb). At x=1, the a→b resistor current has negative real and imaginary coordinates under its assigned reference. A negative coordinate is not an invalid current; the complete sinusoid still follows from the same phasor definition.

The following table retains exact rational phasors and signed absorbed power. All resistor active powers are nonnegative, the ideal capacitor and inductor have zero active power, and the source supplies the total resistive loss. Its reactive sign need not match its active sign. Here the complete passive network is net capacitive, so the source delivers real power while its assigned absorbed reactive power is positive.

Checking KCL at a and b is necessary but not the whole physical check. The branch voltage differences satisfy loop sums, and the sum of every branch complex power, including the source, is zero. Leaving the source out would make an otherwise correct network appear to violate conservation.

Multiply the real waveforms before interpreting power

With associated references, p(t)=v(t)i(t) is absorbed instantaneous power. Substitute v=√2 Re{V exp(jθ)} and i=√2 Re{I exp(jθ)} and expand their conjugate pairs. The result is p(t)=Re{V I*}+Re{V I exp(2jθ)}. The first term is constant; the second oscillates at twice the signal frequency.

The complete-period average is P=Re{V I*}, because the double-frequency term integrates to zero. The other product, V I, is useful: it is the complex coefficient of that double-frequency term. It is not the complex power. Both Cartesian components of the double-frequency coefficient have power units; its imaginary coordinate is not itself the reactive-power definition.

For the lagging control V=10 V and I=(3/5−j4/5) A, p(θ)=6+6cos(2θ)+8sin(2θ) W. Its mean is 6 W even though the RMS product is 10 VA. During some parts of the cycle, p is negative and energy returns across the port. Over a full cycle the net transferred energy is positive.

Keep active, reactive and apparent power distinct

Define the complex power 𝒮=V I*=P+jQ under our RMS convention. If φ=φ_v−φ_i, then P=V_rms I_rms cos φ and Q=V_rms I_rms sin φ. Apparent power is S_app=V_rms I_rms=|𝒮| for this one sinusoidal voltage/current pair. Consequently S_app²=P²+Q² in this setting.

Active power P is measured in watts, reactive power Q in vars, and apparent power in volt-amperes. Their physical dimensions agree, but the names retain different meanings. Q describes the quadrature part of the sinusoidal port relation; it is not an energy in joules or a second independently delivered average power. Do not add Q to P as if both were ordinary heating rates.

For a fixed impedance Z=R+jX, 𝒮=Z|I|² and therefore P=R|I|² and Q=X|I|². A passive positive resistor gives P≥0. The ideal inductor gives Q>0 under our convention, while the ideal capacitor gives Q<0. Those signs describe the chosen port orientation and convention, not a universal label that every source or arbitrary waveform must obey.

Read leading and lagging relative to the defined port

If the current lags the voltage, φ_v−φ_i is positive and the passive inductive control has positive Q. If current leads voltage, the corresponding capacitive control has negative Q. For a passive sinusoidal load with nonzero RMS values, the power factor is cos φ. A magnitude by itself does not distinguish the leading and lagging controls, which have equal P and S_app but opposite Q.

The generating-port control reverses the lagging current while holding the voltage reference. Its absorbed P and Q change sign; a negative absorbed P means net delivery. Some industry descriptions report an unsigned power-factor magnitude plus separate direction and leading/lagging information. Here the ratio P/S_app is signed, so that energy direction remains visible. It must not be confused with an unsigned instrument display.

Reversing both associated branch references, V′=−V and I′=−I, leaves V′I′*=VI*. Reversing only the current changes the meaning from current entering to current leaving the marked positive terminal and changes the signed product. For a zero-current or zero-voltage control, S_app=0 and PF is undefined, even though the active power is zero. We do not assign a phase or PF to an absent signal.

Follow reactive energy without erasing the current

For a fixed ideal linear inductor, w_L(t)=Li(t)²/2 and dw_L/dt=v(t)i(t). For a fixed ideal capacitor, w_C(t)=Cv(t)²/2 and dw_C/dt=v(t)i(t). A periodic steady state returns each stored energy to its initial value after a period, so the mean absorbed power of either ideal element is zero. Their instantaneous energies and currents can still be nonzero.

In RMS notation, ⟨w_L⟩=L|I|²/2 and ⟨w_C⟩=C|V|²/2. The maximum stored energy of the respective pure sinusoid is twice that mean. For these elements, Q_L=2ω⟨w_L⟩ and Q_C=−2ω⟨w_C⟩. Confusing mean and maximum energy introduces another factor-of-two error.

The purely inductive port control has P=0 and Q=10 var. Its transferred energy from θ=0 is [10/(2ω*)][1−cos(2θ)] J, returning to zero after every half-cycle of the signal. This is reversible exchange at the ideal port. A real winding or feeder can dissipate energy while that exchange occurs, and an actual magnetic material may require a more complete energy model than a fixed L.

Integrate transferred energy with the actual clock

Integrating p with respect to θ requires dt=dθ/ω. From θ₀ to θ, the transferred energy is P(θ−θ₀)/ω+Re{VI[exp(2jθ)−exp(2jθ₀)]/(2jω)}. The first term accumulates average transfer; the second is a bounded oscillatory term. It has units of joules because a power is divided by an angular frequency.

For the lagging control from θ₀=0, E_transfer=[6θ+3sin(2θ)+4(1−cos(2θ))]/ω* J. At a complete period θ=2π, the oscillatory part is zero and the energy is 12π/ω* J. It is not appropriate to call that growing quantity “energy stored in the load” without identifying the storage and dissipation mechanisms.

Every finite-window energy statement must specify its endpoints and reference direction. An average over a short window can include a nonzero boundary contribution from the oscillating part. A complete-cycle average and an instantaneous power value answer different questions; neither should be substituted for the other.

Prove the network complex-power balance

Choose a real incidence matrix A with +1 at the from-node and −1 at the to-node of every oriented branch. For a complete closed branch collection, KCL is A I=0. If e is the vector of node phasors and branch drops use those same orientations, V=Aᵀe. Since A is real, A I*=0 as well.

Therefore Σ_b V_b I_b*=VᵀI*=eᵀA I*=0. This is the complex-power form of the same topological pairing behind Tellegen’s theorem. It requires a complete and consistently referenced set of branch ports; sources and any explicitly modelled induced-emf elements belong in that set. It does not come from summing apparent-power magnitudes.

The proof at a specified frequency does not require a resistor-only circuit. In a harmonic analysis the pairing applies separately to the valid voltage/current components at each frequency. It does not authorize applying one linear impedance to a nonlinear device at all amplitudes, nor does it turn Q into an independent conserved substance. Instantaneous energy conservation still refers to real fields and real work.

Why the conjugate is essential

A common phase rotation V′=V exp(jα), I′=I exp(jα) changes neither the real network nor its average energy transfer. The correct product satisfies V′I′*=VI*. The unconjugated product instead acquires exp(2jα). A purported average power that changes when the time origin changes cannot be correct.

This provides a useful diagnostic independent of a long numeric calculation. In the rotated-lagging control, the voltage is 10(3+4j)/5 V and the current is [(3+4j)(3−4j)/25] A. The physical relative phase is the same as before the rotation, and the complex power still has P=6 W and Q=8 var.

For peak phasors, the matching definition would contain 1/2: 𝒮=V_peak I_peak*/2. Changing the normalization without changing that factor gives twice the intended power. These are convention translations, not competing physical predictions.

Interpret the reactive-power energy identity narrowly

For a sinusoidal network of fixed ideal linear inductors and capacitors, summing the element relations gives Q=2ω(Σ⟨w_L⟩−Σ⟨w_C⟩), where Q is the net absorbed reactive power of that passive collection. Its real part P is the sum of resistive dissipation if those are its only losses. The source has the opposite signed complex power when included in the closed network balance.

At ω=ω*, the bridge has ⟨w_L⟩=2391/15366800 J and ⟨w_C⟩=13401/38417000 J. The capacitor stores more cycle-mean energy in this operating state, consistent with the negative net Q of the passive network. Neither stored energy is negative; the subtraction distinguishes their opposite quadrature relations.

A nonlinear, hysteretic, moving or dispersive material does not generally admit the same fixed-parameter energy formula. Nor does an arbitrary nonsinusoidal signal have a single ω to insert into that identity. Retain the actual constitutive model, cycle or frequency decomposition, and source-work boundary instead of extending a compact sinusoidal formula beyond its derivation.

Correct reactive loading at a fixed terminal voltage

The correction model holds a load Z_load=(2+j)Z*=(20+j10) Ω at ω*. Its admittance is (2−j)/(5Z*). A parallel capacitor C=β/(ω*Z*) contributes jβ/Z*. At a maintained V_load=10 V RMS, the load branch remains unchanged while the input current is V_load(Y_load+Y_C).

Without a capacitor, the load draws I_load=2/5−j1/5 A, absorbs P=4 W and Q=2 var, and has RMS current 1/√5 A. At β=1/5, C=1/100000 F and I_C=j1/5 A. The input current becomes 2/5 A, and the input complex power is 4+j0. The source supplies the same mean load power with less RMS input current.

The capacitor has not produced the missing real energy. It exchanges reactive energy locally while the maintained source still supplies the resistor’s 4 W. Further increasing β passes through compensation and produces leading input current: the current magnitude rises again. Cancellation is a condition on the complete port admittance, not a command to add as much capacitance as possible.

Retain the feeder when the source voltage is fixed

Now keep V_s=10 V RMS at the source and insert R_f=Z*/10=1 Ω before the same parallel load. The terminal voltage becomes V_load=V_s/[1+R_fY_total], and the input current is V_loadY_total. The load is no longer held at the same voltage for every β. Its active power can change because its voltage changes, even though its component values are fixed.

At β=1/5, V_load=125/13 V and I_input=5/13 A. The input active power is 50/13 W, load active power is 625/169 W, and feeder loss is 25/169 W. Their balance is exact. Input PF is one, yet the load receives only 25/26 of the input active power. Unity power factor has not made the feeder loss or voltage drop disappear.

The table compares the two boundary conditions. Voltages and powers remain exact fractions; an irrational power factor is shown to six significant digits. The ideal-terminal statement “load P stays fixed” belongs to that maintained terminal boundary. Applying it to the finite-feeder source boundary would change the problem being solved.

Separate power factor, efficiency and regulation

Power factor compares average transfer with a product of RMS terminal magnitudes at one port. Efficiency compares useful output active power with input active power for a specified system and operating state. Voltage regulation concerns changes in terminal voltage as loading changes. These quantities can be related by a model, but none is a replacement definition for the others.

In our ideal-terminal correction, the ideal capacitor absorbs zero average power and input PF changes while the load branch’s mean power is fixed. In the finite-feeder correction, both the voltage boundary and the feeder dissipation matter. At compensation the load still consumes reactive power internally even though the capacitor supplies its opposing reactive requirement locally.

Real capacitors have losses and ratings; magnetic loads and connecting conductors have frequency-dependent effects. A correction designed for one linear frequency can interact with harmonics and other impedances. The present model establishes the mechanism and its boundary conditions, not a universal efficiency gain, installation prescription or electricity-tariff rule.

Use RMS beyond a single sinusoid

For any real periodic signal with finite mean square, x_rms=√[T⁻¹∫₀ᵀx(t)²dt]. This definition does not require a linear device or a pure sinusoid. It gives the equivalent constant current for mean heating in a fixed resistor. The familiar peak/√2 shortcut is a consequence for a sinusoid, not the definition for every waveform.

Write a finite current series as i(t)=I_dc+√2 Re{ΣₙIₙ exp(jnωt)}, with Iₙ the RMS phasor of harmonic n. Over a complete common period, distinct harmonics and DC are orthogonal. Hence I_rms²=I_dc²+Σₙ|Iₙ|². The DC term is not divided by two, and the RMS harmonic amplitudes are not divided by two again.

The result is a squared-norm sum, not an ordinary sum of magnitudes. Two harmonic components can have different phases and peaks yet contribute the same amount to RMS. Conversely, an instrument that omits DC or a high harmonic measures a different quantity from the complete RMS specified here. Its bandwidth and coupling mode are part of the measurement.

Distinguish true and displacement power factor

Our prescribed harmonic family has v=√2V* cos θ and i=I*{d+√2[a cos θ+b sin θ+h cos(3θ+φ)]}, with a=3/5 and b=4/5 in its ordinary controls. Its fundamental current phasor is I₁=I*(a−jb). The third harmonic and DC are specified independently; this is not a claim about the current drawn by a particular diode or converter.

Orthogonality gives I_rms²=I*²(d²+a²+b²+h²), P=V*I*a, and Q₁=V*I*b. Only the fundamental current component in phase with the sinusoidal voltage contributes to P. The other terms can still raise RMS current. Therefore true PF=a/√(d²+a²+b²+h²), while displacement PF is a/√(a²+b²) when the fundamental is nonzero.

Equivalently, PF=(I₁,rms/I_rms)cos φ₁ for this sinusoidal-voltage boundary. A current can have its fundamental perfectly in phase and still have PF below one because of harmonics or a DC offset. The decomposition does not assume that every distorted current is caused by a linear circuit. It uses the actual real waveform and its matching-frequency product.

For the zero-DC subset, the stated current THD is the RMS harmonic content above the fundamental divided by fundamental RMS; here it is h/√(a²+b²). We report DC separately. With zero fundamental, displacement angle and this THD ratio are undefined, but true PF can still be defined if total RMS current is nonzero. With zero total RMS, PF itself is undefined.

Include matching harmonics when both waveforms are distorted

If both voltage and current contain harmonics, the pure-voltage simplification no longer applies. Using RMS Fourier phasors over their complete common period, P=V_dcI_dc+ΣₙRe{VₙIₙ*}. Voltage and current at the same harmonic can transfer average energy. Cross-frequency products oscillate and have zero complete-period mean, while still contributing to instantaneous power.

Our second prescribed case is v=√2V*[cos θ+(1/2)cos(3θ)] and i=√2I*[(3/5)cos θ+(4/5)sin θ+(1/3)cos(3θ)+(1/4)sin(3θ)]. Here V_rms=5√5 V and I_rms=13/12 A. The fundamental transfers 6 W and the third harmonic transfers 5/3 W, so P=23/3 W. Omitting the third-frequency voltage would miss a real contribution.

Apparent power remains V_rmsI_rms=65√5/12 VA, and true PF is 92/(65√5). A single fundamental complex power cannot represent that whole waveform. In particular, S_app²−P²−Q₁²=3445/144 VA², not zero. Per-harmonic complex powers remain useful bookkeeping, but calling one chosen sum or residual a universal nonsinusoidal reactive power would require an additional definition and measurement convention.

Nonlinear devices can mix frequencies and change those Fourier coefficients when connected to a different source. The present calculations begin from the specified waveforms. They establish energy and RMS relations for those waveforms and a warning about an unjustified shortcut; they do not solve the device physics that might produce them.

See power factor as an inner-product ratio

Over the same averaging interval, define ⟨v,i⟩=T⁻¹∫vi dt and the norms ||v||=V_rms, ||i||=I_rms. Then PF=⟨v,i⟩/(||v||||i||) when both norms are nonzero. The Cauchy–Schwarz inequality gives |PF|≤1. This argument applies to real square-integrable waveforms, not only to single-frequency sinusoids.

Equality requires i(t)=k v(t) almost everywhere over the interval for a real constant k. Positive k gives PF=1 with absorbed mean power; negative k gives PF=−1 under the same reference convention. The waveforms can contain harmonics and still attain equality if their shapes and relative amplitudes match. Matching only the fundamental phase is not enough.

For a sinusoidal pair, the ratio reduces to the cosine of their phase difference. For distorted signals, it is an angle in a function-space sense, not necessarily the displacement of two peaks on a time plot. When one norm is zero the normalization is excluded; the inequality does not manufacture an angle or a numeric PF for that case.

Connect balanced phases without confusing line and phase quantities

The advanced control uses three RMS source phasors V_a=V*, V_b=V*exp(−j2π/3), and V_c=V*exp(+j2π/3), each feeding the identical impedance (2+j)Z*. The ideal neutral is the datum. Each phase absorbs P_φ=4 W and Q_φ=2 var, while its instantaneous power still oscillates.

For a phase with voltage angle α, p_α=P_φ[1+cos(2θ+2α)]+Q_φ sin(2θ+2α). The doubled angles are 0, −4π/3 and +4π/3 for a, b and c. Their sine and cosine sums vanish, leaving p_a+p_b+p_c=3P_φ=12 W at every instant. The neutral current is zero because the three current phasors also sum to zero.

The line voltage V_ab=V_a−V_b=15+j5√3 V has magnitude 10√3 V and angle π/6 relative to V_a. That is a difference between phase voltages, not a new phase-to-neutral amplitude. Under the stated balanced sinusoidal wye conditions, total active power can be written 3V_phase,rms I_phase,rms cos φ or √3V_line,rms I_line,rms cos φ.

Identical loads, the declared phase order and sinusoidal balance are essential. Unbalance can create neutral current and power pulsation; triplen harmonic currents can reinforce rather than cancel in a neutral. Neither a zero neutral current nor a constant total power should be inferred merely from the label “three phase”. The present control does not include a machine, transformer saturation or a general distribution network.

Know when the algebra needs a larger physical model

A lumped circuit treats chosen branch variables as meaningful terminal quantities with negligible unmodelled propagation and electromagnetic coupling. A useful scale test is ωℓ/v_prop≪1 for a characteristic size ℓ and propagation speed, together with a consistent return path and the relevant parasitics. Small geometric size alone does not erase a substantial unmodelled inductance, capacitance or induced emf.

An external changing magnetic flux can make a voltage measurement depend on the lead path. The induction and circuit-foundations lessons explain how to include that effect as a defined emf element and use the completed circuit consistently. Writing an ordinary zero loop sum while omitting that element is a modelling error, not a failure of phasor arithmetic.

A real impedance can contain resistive loss, skin and proximity effects, dielectric dispersion, radiation, and magnetic hysteresis. A small-signal impedance is defined around an operating condition; it need not predict large-signal saturation or switching. A nonnegative real part at one frequency is useful for sinusoidal average passivity at that port, but does not by itself establish causality, stability or passivity at every frequency.

The invertible bridge here is deliberately dissipative. An ideal lossless matrix can instead be singular, underdetermined or inconsistent for a specified drive. Active sources can produce growing modes. Frequency-domain algebra must be interpreted alongside the dynamical and constitutive problem; it cannot supply missing initial conditions or guarantee the decay of a natural response.

Measure the quantity at the defined port

A voltage and current pair must refer to the same port, reference directions and time interval. The wattmeter quantity is the average of their real product. A current RMS measurement alone cannot determine active power or its direction. Instrument loading, channel delay and bandwidth can matter when the phase or waveform contains significant high-frequency content.

A complete-period Fourier identity assumes the stated common period or an integer number of periods. On a finite non-integer window, cross terms and endpoint contributions need not vanish. Sampling must also resolve the relevant harmonics; otherwise aliasing changes the inferred spectrum, RMS and power. A plotted sequence of samples is a representation of the specified functions, not a proof that a numerical quadrature is their exact mean.

Our waveform records use θ/π=n/64 over one complete period and retain both endpoint representations. Means, RMS and power balances come from the analytic equations, not from treating the duplicate endpoint as another independent observation. The distinction is useful when reading any instrument or simulation trace: ask which quantity was averaged, over what interval, and with which model of the input signal.

Use the connections without extending the claim

The phasor method joins the linear-system ideas of superposition with circuit topology and the energy definition p=vi. The same conjugation that makes average power independent of time origin appears in inner products and spectral energy calculations. The different roles of DC, matching harmonics and cross terms provide a direct preparation for signals and Fourier analysis.

The correction comparison joins port definitions to physical loading: a maintained source voltage and a maintained load-terminal voltage are different boundary conditions. The balanced example joins phase geometry to time-domain power cancellation. Each connection is most useful when its assumptions travel with the formula.

For a new circuit, first state the real signals, frequency content, reference directions, constitutive laws and boundary conditions. Then solve the valid phasor or harmonic problem, reconstruct the quantities requested, and check the complete signed power balance. If the model or measurement lies outside those conditions, enlarge the model rather than hiding the discrepancy in a power-factor label.

Worked examples

Try it yourself

1. What peak voltage corresponds to the original 10 V RMS sinusoid? What is its period at ω*=2000 rad/s?

   Show the answer and reasoning

   The peak is 10√2 V and the period is π/1000 s.

   The RMS-to-peak factor is √2 for a sinusoid. The period is 2π/ω*, not 1/ω*. The latter is the time for one radian of phase increase, not a complete cycle.

2. Reconstruct the lagging control I=(3/5−j4/5) A as a real current.

   Show the answer and reasoning

   i(t)=√2[(3/5)cos θ+(4/5)sin θ] A.

   Multiply the complex amplitude by cos θ+j sin θ and take its real part. The minus sign in the imaginary phasor coefficient produces a positive sine term; dropping that sign reverses the inferred phase.

3. Why does the rotated-lagging control retain P=6 W and Q=8 var?

   Show the answer and reasoning

   Both phasors acquire the same unit rotation, which cancels in V I*.

   For V′=V exp(jα) and I′=I exp(jα), V′I′*=VI*. A change of time origin changes individual phasor coordinates but cannot change average transferred power. The unconjugated product would acquire twice the rotation.

4. At ω*, find the admittance of the correction load 20+j10 Ω and identify the susceptance sign.

   Show the answer and reasoning

   Y_load=(2−j)/50 S; its susceptance is −1/50 S.

   Multiply numerator and denominator of 1/(20+j10) by 20−j10. A positive inductive reactance becomes a negative susceptance on inversion. Admittance has units A/V, not V/A.

5. Write the normalized node-a equation for the bridge at x=1.

   Show the answer and reasoning

   [(13/12)+j2]u_a−u_b/4=1/2.

   The s→a resistor contributes (u_a−1)/2 to outgoing current, the a→0 resistor contributes u_a/3, the a→b resistor contributes (u_a−u_b)/4, and the capacitor contributes j2u_a. Their sum is zero.

6. What is the bridge source’s absorbed active power at ω*, and why is it negative?

   Show the answer and reasoning

   P_source=−211142/38417 W under the assigned s→0 source-current reference.

   Current is defined into the source’s positive terminal. The source supplies the resistor losses, so its mean absorbed power is negative. Defining a delivered-power number would reverse that sign without changing the physical circuit.

7. For the leading control I=3/5+j4/5 A with V=10 V, find P,Q and the signed PF.

   Show the answer and reasoning

   P=6 W, Q=−8 var and PF=3/5.

   Conjugating the leading current changes the sign of its imaginary component in V I*. Its RMS magnitude remains 1 A, so S_app=10 VA. Equal leading and lagging PF magnitudes do not imply equal Q signs.

8. Should either zero-current or zero-voltage control be assigned PF=0?

   Show the answer and reasoning

   No. Their RMS product is zero, so the ratio P/S_app is undefined.

   A nonzero reactive sinusoidal port can have P=0 with S_app>0 and therefore PF=0. That is different from dividing zero by zero. The phase of the zero-amplitude phasor is also undefined.

9. For the pure-inductive control, compare maximum and cycle-mean stored energy.

   Show the answer and reasoning

   The maximum is 1/200 J and the mean is 1/400 J.

   The compatible ideal inductance is 1/200 H and i=√2sin θ A, giving w=(1/200)sin²θ J. The mean of sin² over a complete cycle is 1/2. Q=10 var equals 2ω* times the mean stored energy, not 2ω* times the maximum.

10. Which prescribed β cancels the input susceptance of the fixed-terminal correction load, and what C does it give?

    Show the answer and reasoning

    β=1/5 and C=1/100000 F=10 µF.

    The load susceptance is−1/(5Z*) and the capacitor contributes β/Z*. Cancellation requires their sum to be zero. At a fixed load voltage this changes input current but leaves the load resistor’s 4 W unchanged.

11. The finite-feeder input PF is one at compensation. Does that give 100% efficiency?

    Show the answer and reasoning

    No. The model gives load/input active-power efficiency 25/26.

    Input power 50/13 W equals load power 625/169 W plus feeder heat 25/169 W. Unity PF refers to the input waveform relation; it does not remove the 1 Ω resistance or the voltage drop.

12. At h=1,d=0,φ=0 with sinusoidal voltage, why is PF below 3/5?

    Show the answer and reasoning

    Total RMS current is √2 A, so true PF=3/(5√2), while displacement PF remains 3/5.

    The third harmonic adds one ampere squared to the current mean square. Its product with the purely fundamental voltage has zero complete-period mean. It raises the RMS denominator without changing the 6 W active transfer.

13. For h=0,d=1/2 in the same family, find total RMS current and true PF.

    Show the answer and reasoning

    I_rms=√5/2 A and PF=6/(5√5).

    The DC contribution to mean square is d²=1/4 A², added to the fundamental 1 A². It is not divided by two. The zero-mean sinusoidal voltage has no average product with that DC current, so P remains 6 W; the DC offset is reported separately from the stated zero-DC THD controls.

14. In the no-fundamental control, which power-factor quantities are meaningful?

    Show the answer and reasoning

    True PF is 0 because I_rms=1/2 A and P=0. Fundamental phase, displacement PF and the stated THD ratio are undefined.

    The denominator of true PF is 10 V times 1/2 A, which is nonzero. The denominators defining fundamental phase and harmonic-to-fundamental ratios are absent. A finite total current is not evidence of a nonzero fundamental current.

15. Why is 6 W an incomplete mean-power result for the both-distorted case?

    Show the answer and reasoning

    The third-frequency pair contributes another 5/3 W, giving 23/3 W in total.

    The voltage includes a third harmonic of 5 V RMS. It has a nonzero in-phase product with the third current phasor 1/3−j1/4 A. Only cross-frequency terms average to zero; matching nonfundamental harmonics need not.

16. What is constant in the balanced control, and does the pure-inductor control have zero mean power over every shorter interval?

    Show the answer and reasoning

    Balanced total instantaneous power is 12 W and its neutral current is zero. The pure-inductor mean over θ=0 to π/2 is 20/π W, despite its zero complete-cycle mean.

    The balanced cancellation uses three equally shifted double-frequency terms and identical loads. For the inductor, the interval transfers 1/200 J in time π/4000 s. The ratio is 20/π W; a complete-cycle averaging identity does not apply to every finite window.

What to carry forward

• An RMS phasor includes magnitude and phase; the physical waveform is √2 Re{X exp(jωt)} under the stated convention.
• Use jωL and 1/(jωC) with associated references, and evaluate each harmonic at its own frequency.
• For a sinusoidal RMS pair, 𝒮=VI*=P+jQ; the unconjugated VI controls the double-frequency power term.
• P is average absorbed power, Q is the stated quadrature quantity, and S_app is an RMS magnitude product; their meanings differ even when dimensions agree.
• Include the source and every oriented branch when checking the complex-power sum; apparent magnitudes do not add in the same way.
• Reactive exchange can have zero mean at an ideal element while still carrying current and causing real loss elsewhere.
• Fixed source voltage and fixed load-terminal voltage are different boundaries. Unity PF does not imply zero feeder loss or unit efficiency.
• For distorted current under sinusoidal voltage, true PF includes the fundamental-to-total RMS factor; a displacement angle alone is incomplete.
• When both waveforms contain harmonics, mean power includes every matching-frequency product; no single fundamental triangle describes the whole waveform.
• Zero RMS excludes normalized PF or phase claims. Balanced three-phase and lumped linear formulas retain their stated conditions.