Single-Phase AC Power

The product, using \(\cos A\cos B = \tfrac12\left[\cos(A-B) + \cos(A+B)\right]\):

Two terms: a constant, and one at twice the supply frequency. The second averages to zero over any whole number of cycles, so

The three cases, with \(\tfrac12V_mI_m = 250\ \text{W}\):

The resistive case is \(p = 250(1+\cos2\omega t)\) — always non-negative, touching zero twice per cycle, and pulsating at 100 Hz on a 50 Hz supply. Energy flows one way only.

This is why incandescent lamps and transformer cores hum at twice the supply frequency: the power, not the voltage, drives the mechanical excitation.

The purely reactive case is the striking one:

The average is exactly zero, yet the instantaneous power reaches \(\pm250\ \text{W}\). Energy flows into the element for a quarter cycle, then back out for the next quarter, indefinitely. Nothing is consumed — but a real current of 5 A amplitude flows in the conductors the whole time.

What negative power means. The load is returning energy to the source. A capacitor discharging or an inductor collapsing its field pushes current back against the supply voltage. Over a cycle the returns exactly cancel the deliveries for a pure reactance, and partly cancel them for a mixed load.

The quantity that has appeared is the amplitude of the circulating flow. Separating the \(2\omega\) term into components in phase and in quadrature with the constant part gives

where \(t'\) is measured from a convenient origin. \(P\) rides on a pulsation that never goes negative; \(Q\) is a pure oscillation with zero mean. Problem 2 names them.

Complex power packages both quantities into one number:

The conjugate on the current is essential: it makes the angle the \(v\!-\!i\) difference, which is the impedance angle. Without it the angle would be \(\theta_v+\theta_i\), which depends on the arbitrary time origin and is therefore meaningless.

Its rectangular parts:

Three different units for quantities all dimensionally equal to power. The distinction is deliberate: it makes clear from the unit alone which quantity is meant.

The power triangle. Since \(P\) and \(Q\) are the legs and \(S\) the hypotenuse:

Why it is similar to the impedance triangle. Writing \(\mathbf{S} = |\mathbf{I}|^2\mathbf{Z}\):

The power triangle is the impedance triangle scaled by \(|\mathbf{I}|^2\) — same angle, same shape. Set 20, Problem 14's voltage triangle is a third member of the family, scaled by \(|\mathbf{I}|\).

The three triangles together:

All three share the angle \(\theta\), so knowing any one gives the others immediately.

A caution on "power factor". It is \(\cos\theta\), which is positive for any \(|\theta| < 90°\) and therefore cannot distinguish inductive from capacitive. The words lagging and leading must always be attached — they carry the sign of \(Q\), which \(\cos\theta\) throws away.

Convert amplitudes to RMS phasors, following this set's convention:

1Complex and apparent power:

The two \(\sqrt2\) factors combine into the familiar \(\tfrac12\) — which is exactly what the RMS convention absorbs, so it need never be written again.

2Real and reactive powers from the rectangular form:

Negative \(Q\) means capacitive. Note that \(|Q| > P\) here — most of the volt-amperes are doing no work at all.

3Power factor and impedance:

The \(\sqrt2\) factors cancel in the ratio, so impedance is the same whichever convention is used — the point made in the convention block.

Cross-check by the alternative formula:

What a pf of 0.5 costs. The load consumes 22.5 W but the supply must deliver 45 VA — twice the current a resistive load of the same wattage would need, and hence four times the \(I^2R\) loss in the cables feeding it.

Reduce to a single impedance:

The power factor is the cosine of the impedance angle:

Leading, because the net reactance is capacitive. The sign of the angle is the only thing distinguishing this from a lagging 0.9734.

The source current:

The average power, computed two ways.

aFrom the definition:

bFrom the resistance, since only resistance dissipates:

The second route is usually faster and is immune to power-factor sign errors — it uses only the real part of the impedance, which cannot be mistaken for anything else.

Where the power goes. The 6.8 Ω is not a physical resistor — it is 6 Ω plus the 0.8 Ω contributed by the parallel pair. Splitting it:

The capacitor dissipates nothing, as it must — it contributes only to \(Q\).

The reactive power:

Confirming the capacitive character, and giving \(S = \sqrt{125.4^2 + 29.5^2} = 128.8\ \text{VA}\).

Recover the angle from the power factor, taking it positive because the load is lagging:

"Lagging" is what fixes the sign — \(\cos^{-1}\) alone gives no information about it.

1Real and reactive powers:

Check: \(\sqrt{10.272^2 + 6.204^2} = 12.00\) kVA ✓.

2The current, from the definition of complex power:

Note the conjugation at the end — forgetting it reverses the phase and turns a lagging load into a leading one.

The peak current:

This is what matters for saturation, semiconductor ratings and insulation stress; the RMS value is what matters for heating.

3The load impedance:

Cross-check: \(P = |\mathbf{I}|^2R = (100)^2(1.027) = 10.27\) kW ✓, and \(Q = (100)^2(0.620) = 6.20\) kvar ✓.

A shortcut worth knowing:

obtained by combining \(\mathbf{S} = \mathbf{V}\mathbf{I}^{*}\) with \(\mathbf{V} = \mathbf{Z}\mathbf{I}\). It skips the current entirely.

1Apparent power is the product of the RMS values:

Power factor from the phase difference:

Leading, because the current's phase exceeds the voltage's — the current reaches its peak first.

2The impedance fixes the elements:

Negative reactance confirms a capacitor, consistent with the leading power factor.

The resistance is read directly:

The capacitance, from \(X_C = 1/\omega C\) with \(\omega = 100\pi = 314.16\ \text{rad/s}\), i.e. 50 Hz:

Cross-check the powers:

Matching the direct calculation, and confirming \(\text{pf} = 207.8/240 = 0.866\).

Note the ambiguity this resolves. A power factor of 0.866 alone is consistent with \(\pm30°\) and therefore with either a capacitive or inductive load — and hence with a capacitor of 212 µF or an inductor of \(15/(100\pi) = 47.7\ \text{mH}\). Only the waveforms, or the word "leading", settles which.

Take a pure inductor carrying \(I_{rms}\) at frequency \(\omega\). Its reactive power is

Its average stored energy. The instantaneous energy is \(w = \tfrac12Li^2\), and averaging \(i^2\) over a cycle gives \(I_{rms}^2\) by definition:

Compare the two:

Reactive power is the average stored energy, multiplied by \(2\omega\). The factor of two appears because the energy completes two full store-and-return cycles per period of the supply — as Problem 1's \(2\omega\) term showed.

A numerical instance. With \(L = 0.1\ \text{H}\), \(I_{rms} = 5\ \text{A}\) at 50 Hz:

For a capacitor the sign reverses, because it stores in the electric field while the current leads:

Which is why an inductor and a capacitor can exchange reactive power directly, each supplying what the other absorbs — the principle behind every correction capacitor.

Why it costs money despite consuming nothing. The reactive current is a real current in real conductors:

The energy is not lost in the reactive element — but transporting it back and forth wastes energy in everything between.

The load's complex power:

The load current:

The same current flows in the line, since load and line are in series.

aMethod 1 — power balance. The line absorbs

Complex power is conserved, so the source must supply the sum:

bMethod 2 — KVL. The line drop is

Note what conservation does and does not say. Real and reactive powers each balance independently:

Apparent power does not add. \(P\) and \(Q\) are conserved; \(S = |P+jQ|\) is a magnitude and obeys no conservation law. Adding kVA figures is one of the commonest errors in power calculations.

The engineering reading. A 13% voltage rise is needed at the sending end to hold 220 V at the load — and note that most of the drop comes from \(Q\) flowing through the line's reactance, not from \(P\) through its resistance. Correcting the power factor would reduce the current, the loss and the voltage drop together.

Branch currents, all driven by the same voltage:

Total current by KCL:

Note that \(\mathbf{I}_1\) is capacitive (positive imaginary) and \(\mathbf{I}_2\) inductive, so they largely cancel — leaving a small net capacitive component.

The total complex power:

The input power factor:

Leading because \(Q < 0\) overall — the capacitive branch slightly outweighs the inductive one.

Cross-check branch by branch, using \(\mathbf{S}_k = |\mathbf{V}|^2/\mathbf{Z}_k^{*}\):

Agreeing with the current-based calculation. Adding complex powers is usually the faster route for parallel loads, since each branch is independent.

The practical point. The near-unity power factor here is accidental — the capacitive load happens to cancel the inductive one. In a real installation dominated by motors the result would be strongly lagging, and deliberate correction would be needed.

aRoute 1 — complex power per branch, using \(\mathbf{S} = |\mathbf{V}|^2/\mathbf{Z}^{*}\):

Note the conjugate in the denominator flips each angle's sign, so \(\mathbf{S}\) ends up with the same angle as \(\mathbf{Z}\) — as it must, since both share the impedance angle.

Add:

bRoute 2 — via the total current. The branch currents are

Identical, and note \(481.6\angle16.22° = 462.4 + j134.6\) exactly.

Observe the individual magnitudes:

but the total is only 481.6 VA. Apparent powers do not add, because the branches have different phase angles — the same point as Problem 8's table, seen here as a 20% discrepancy.

Which route to prefer. Route 1 avoids computing currents at all and is immediate for any number of parallel branches. Route 2 gives the supply current, which is needed for cable sizing and for the \(I^2R\) losses upstream. Doing both, as here, is a free check.

The key observation: a shunt capacitor changes \(Q\) but not \(P\). It dissipates nothing, and it does not alter the load's own operation — the load still sees 230 V and draws exactly what it did.

Reactive power before and after, from \(Q = P\tan\theta\):

The capacitor supplies the difference:

"Supplies" is the right word: a capacitor's \(Q\) is negative, so adding it to the load's positive \(Q\) reduces the total. The inductive load absorbs reactive power; the capacitor provides it locally instead of the generator providing it down the line.

The capacitance, from \(Q_C = V^2/X_C = \omega CV^2\):

The general formula, worth having directly:

The current reduction. The same real power is now drawn at a better power factor:

A 16% reduction — and since cable losses go as \(I^2\), they fall by 29%.

Why stop at 0.95 rather than correcting to unity? Three reasons:

The three operating points, with \(P\) fixed at 500 kW:

The real power never changes — the factory does the same work. Everything in the table is about what the supply system must carry to deliver it.

1Apparent power released:

A transformer sized at 715 kVA now has 188 kVA spare — enough to add roughly 180 kW of further corrected load without replacing it. That deferred capital cost is usually the largest single benefit.

2The capacitor bank required:

3Line losses. Loss goes as \(I^2\), and current goes as \(1/\text{pf}\) at fixed \(P\):

Losses fall to 54% of their previous value — a 46% saving on every watt wasted in cables and transformer windings between the correction point and the supply.

4Where the capacitor goes. This is the part most often got wrong:

A capacitor relieves only the conductors between itself and the generator. Everything downstream still carries the full reactive current.

The tariff mechanism. Utilities recover the cost in one of three ways — a maximum-demand charge in kVA rather than kW, an explicit kvarh charge, or a power-factor penalty multiplier. All three make the same point: the customer's \(Q\) occupies capacity that could have carried someone else's \(P\).

The conjugate match, derived in Set 22, Problem 14 and unaffected by the convention change:

The reactances cancel:

Purely resistive, so the current is in phase with the Thévenin voltage — the matched circuit is at resonance.

The maximum power, with no factor of \(\tfrac12\) because the phasors are RMS:

Compare with Set 22, Problem 19, where the same numbers were used with the amplitude convention:

Both correct under their own convention. A 20 V RMS source genuinely delivers twice the power of a 20 V peak source, because its RMS value is \(\sqrt2\) times larger and power goes as the square.

The complete power picture at the match:

The reactive powers cancel exactly, so the source supplies pure real power at unity power factor — while 24 var circulates internally between the two reactances. Efficiency is 50%, as always at maximum power transfer.

Note the convention. This problem is stated with an amplitude source, so the factor \(\tfrac12\) must be retained throughout — a deliberate exception, kept so both conventions appear in one set.

The Thévenin impedance:

The Thévenin voltage, by division onto the \(j20\):

The load is restricted to a resistor, so Set 22, Problem 16's result applies rather than the conjugate match:

Not \(R_{Th} = 9.41\ \Omega\). Because the reactance cannot be cancelled, the optimum resistance is the impedance magnitude.

The load current and power:

Compare with the unrestricted case. Had a complex load been permitted:

So the restriction costs 44% of the available power — much more than Set 22, Problem 16's 22%, because here \(|X_{Th}|\) is more than twice \(R_{Th}\). The more reactive the source, the more a resistive-only load loses.

The four forms, all equivalent with RMS phasors:

Why the conjugates sit where they do. Start from \(\mathbf{S} = \mathbf{V}\mathbf{I}^{*}\) and substitute:

The conjugate follows the substituted quantity. The reliable check is that \(\mathbf{S}\) must end up with the same angle as \(\mathbf{Z}\) — if it comes out with the opposite sign, a conjugate is misplaced.

Choosing in practice:

For real power alone, the safest expression of all is

It has no conjugate, no angle, no power factor and no sign convention — only the real part of the impedance the current flows through. Whenever the current is available, this should be the default.

A worked comparison. Problem 3's load, three ways:

The convention follows from \(\mathbf{S} = |\mathbf{I}|^2\mathbf{Z}\), so \(Q\) takes the sign of the reactance:

"Absorbs" and "supplies" are the useful words. An inductor takes reactive power from the system; a capacitor gives it back. They can therefore trade directly, which is the whole basis of correction (Problem 11).

Why "pf = 0.8" is incomplete. The power factor is a cosine, and

so it cannot distinguish the two cases. Two loads at 0.8 pf with the same \(P\) differ entirely:

Applying a capacitor to a leading load makes matters worse, not better — the specification must always carry the word.

How to determine it in practice. Three equivalent tests:

All three test the same thing, and the first is usually available with least work.

A note on the sign of \(\mathbf{Y}\). Susceptance reverses relative to reactance — an inductor has \(X > 0\) but \(B < 0\) (Set 20, Problem 11). Since \(\mathbf{S} = |\mathbf{V}|^2\mathbf{Y}^{*}\), the conjugate restores the expected sign:

The construction. A wattmeter has two coils:

Why it reads \(P\). The deflecting torque is proportional to the product of the two coil currents:

The pointer's inertia cannot follow the \(2\omega\) component of Problem 1, so it settles at the average:

The instrument performs the multiplication and the time-averaging mechanically — which is why it reads watts and not volt-amperes, and why it does so correctly for any waveform.

Contrast with separate meters. A voltmeter and ammeter give \(V_{rms}\) and \(I_{rms}\), whose product is \(S\), not \(P\):

The two agree only at unity power factor. Three instruments — voltmeter, ammeter and wattmeter — determine \(P\), \(S\) and hence \(\text{pf} = P/S\) and \(|Q| = \sqrt{S^2-P^2}\). What they cannot give is the sign of \(Q\).

Reversing the current coil reverses the torque, so the reading becomes \(-P\) and the pointer drives backwards against its stop. Both coils have a marked \(\pm\) terminal for this reason; connecting them consistently gives a positive reading for power absorbed by the load.

A wattmeter can legitimately read negative — when the element it measures is supplying power. Set 25's two-wattmeter method depends on this, where one meter routinely reads negative at low power factors.

When can a wattmeter read zero with current flowing? When \(\theta = 90°\) — a purely reactive load. The needle sits at zero while the current coil carries full current and may overheat, which is why a wattmeter's current and voltage ratings must be respected independently of its watts range.

Cross terms vanish. The average of a product of sinusoids at different frequencies is zero:

This orthogonality — noted in Set 20, Problem 18 — means each harmonic contributes power independently.

Power harmonic by harmonic:

Note the third-harmonic angle is \(30° - (-20°) = 50°\), its own phase difference — unrelated to the fundamental's.

RMS values, where mean squares add:

Apparent power and power factor:

Why this is not \(\cos\theta\) for any \(\theta\). There is no single phase difference — the fundamental is displaced by 30° and the harmonic by 50°. The power factor splits into two factors:

approximately reproducing 0.837 — the small difference is the harmonic's own real-power contribution, which the two-factor form neglects. When the harmonic carries no power the identity is exact.

The engineering significance. Distortion degrades power factor even when every harmonic is perfectly in phase:

Worse, a correction capacitor presents a low impedance to harmonics and may resonate with the supply inductance, amplifying them. This is why modern installations full of switched-mode power supplies need harmonic filters, not merely capacitor banks.

Individual powers:

The actual power. Superpose the voltages first, as Set 22, Problem 12 requires:

Not 520 W. The discrepancy is 240 W — 46% of the naive answer.

Identify the cross term. Expanding \(|\mathbf{V}_A + \mathbf{V}_B|^2\):

When the cross term vanishes. It is proportional to \(\cos(\theta_A - \theta_B)\), so:

Set 22, Problem 12's contributions happened to be 90° apart, which is why its powers appeared to add. That was a coincidence of that circuit, not a rule.

The extreme cases are worth noting. With \(\theta_A = \theta_B\) the voltages add fully:

and with them 180° apart:

— far less than either source alone would give. So the error can go in either direction, and can be arbitrarily large.

What phase does. In DC there is one power. In AC there are three quantities where one would do, and the angle decides how the volt-amperes divide:

At \(\theta = 0\) the split is entirely into \(P\); at \(90°\) entirely into \(Q\) and nothing is delivered at all.

The consequences established:

The one irreducible cost. Even at unity power factor, single-phase power pulsates:

from 0 to \(2P\), twice per cycle. No amount of correction removes this — it is inherent in multiplying two sinusoids at the same frequency. A single-phase motor therefore receives pulsating torque, vibrates at \(2\omega\), and cannot start unaided.

What three phases achieve. Take three loads of equal power fed by voltages 120° apart. Their instantaneous powers are

and the three cosines are themselves 120° apart in \(2\omega t\), so they sum to zero:

The pulsation cancels exactly. That is the fundamental reason three-phase power exists, and it is why essentially all generation and transmission uses it.

What Sets 24 and 25 must supply:

All of it rests on this set: \(P\), \(Q\), \(S\) and the power factor mean exactly the same things per phase.