Power in AC circuits and power factor

Every AC electricity bill you have ever seen is a small monument to a single Greek letter: $\phi$ , the phase angle between voltage and current. Plug a heater into the wall and every watt the meter counts is real, usable heat — voltage and current rise and fall together, in step. Plug in a large motor or a bank of fluorescent tubes instead, and something odd happens: the current drawn can be large, the wires can get warm, and yet a chunk of that current does no useful work at all. It sloshes into the coil's magnetic field and back out again, cycle after cycle, contributing nothing to the shaft that turns or the light that shines. Power engineers gave this a name worth knowing before you ever open a physics exam: wattless current. The bridge between "current flows" and "work gets done" is the power factor, $\cos\phi$ — a single dimensionless number, always between $0$ and $1$ , that tells you what fraction of the volt-amperes a circuit draws actually turns into watts. A power factor of $1$ means every joule pulled from the socket does something; a power factor of $0$ means the circuit is all bluster, drawing current and returning it every quarter cycle without ever converting it to heat, light, or motion. Real household and industrial loads sit somewhere in between, and that "in between" costs money: utilities size their generators, transformers, and copper cables for the apparent power (volt-amperes), not the real power (watts), so a poor power factor means paying for capacity that never gets used productively. This article picks up exactly where the phasor diagram for the series LCR circuit left off. You already know how to draw $V_R$ , $V_L$ , $V_C$ as phasors and combine them into $V$ and $Z$ . Now you multiply voltage by current, average that product over a cycle, and discover that the innocent-looking formula $P = VI$ from DC circuits needs one crucial correction the moment the circuit becomes reactive — the factor $\cos\phi$ — and that correction is the whole story of why power factor matters, on your exam and on the national grid alike.

Key takeaways

Average AC power needs a phase correction. For $v = V_0\sin\omega t$ and $i = I_0\sin(\omega t-\phi)$ , the average power over one cycle is $P_{avg} = V_{rms}I_{rms}\cos\phi$ , not simply $V_{rms}I_{rms}$ .
The power factor $\cos\phi$ is dimensionless, between 0 and 1. It measures what fraction of the apparent power ( $V_{rms}I_{rms}$ , in volt-amperes) becomes real power (in watts).
Pure R gives $\cos\phi = 1$ ; pure L or pure C gives $\cos\phi = 0$ . In a purely reactive element voltage and current are $90°$ out of phase, so the time-averaged power is exactly zero even though a real, measurable current flows.
That zero-average current in a pure L or C is called wattless current — it does no net work over a cycle, sloshing energy between source and field instead of dissipating it.
In a series LCR circuit, $\cos\phi = R/Z$ , so $P_{avg} = I_{rms}^2R$ — power is dissipated only in the resistor; $L$ and $C$ never consume energy on average, they only shift the phase.
A low power factor forces a larger current for the same real power ( $I_{rms} = P_{avg}/(V_{rms}\cos\phi)$ ), which raises $I^2R$ losses in the supply wiring even though nothing extra is "wasted" inside an ideal reactive load itself.
Key idea: average AC power is $P_{avg}=V_{rms}I_{rms}\cos\phi$ , where the power factor $\cos\phi=R/Z$ in a series LCR circuit measures how much of the apparent power is real, watt-consuming power, with $\cos\phi=0$ marking pure wattless (reactive) current and $\cos\phi=1$ marking pure resistive current.

The core insight

Average power delivered to any AC circuit is $P_{avg}=V_{rms}I_{rms}\cos\phi$ , so the power factor $\cos\phi$ (equal to $R/Z$ in a series LCR circuit) is the fraction of the apparent power $V_{rms}I_{rms}$ that is actually consumed as real power, with $\cos\phi=0$ describing current that is $90°$ out of phase and therefore wattless.

Where it fits in the syllabus

This topic sits inside the Alternating Current unit, right after the series LCR circuit and its phasor diagram, and it is fully within the CBSE 2025-26 prescribed scope — the syllabus explicitly lists "power in AC circuits, power factor, wattless current" as an examinable outcome, to be reached purely through the phasor picture, not through the differential-equation solution of the circuit (that route is explicitly out of scope). Everything below leans only on averaging $v\times i$ over one cycle using a trigonometric identity, plus the impedance triangle you already built for phasors. This topic also sets up the very next one: resonance in the LCR circuit, where the power factor and the current both peak sharply as the applied frequency approaches the natural frequency $\omega_0 = 1/\sqrt{LC}$ .

The physical picture

In a DC circuit, power is simply $P = VI$ because voltage and current never change with time. In an AC circuit both $v$ and $i$ oscillate, and in a reactive circuit they are not even in step — current can lag or lead voltage by a phase angle $\phi$ . The instantaneous power $p = vi$ therefore also oscillates, sometimes positive (energy flowing into the circuit) and sometimes negative (energy flowing back out, from the reactive elements to the source). What we actually want — and what any wattmeter or electricity meter reports — is the average of $p$ over a full cycle, not its instantaneous value.

v = V_0\sin\omega t, \qquad i = I_0\sin(\omega t - \phi)

where: $V_0, I_0$ are the peak voltage and current, $\omega$ is the angular frequency of the AC supply, $t$ is time, and $\phi$ is the constant phase angle by which the current lags the voltage (a positive $\phi$ means an inductive circuit; $\phi$ can also be negative for a capacitive circuit, where current leads). Multiplying these and averaging over one full cycle — using the identity for $\sin\theta\sin(\theta-\phi)$ , whose average over a cycle picks out only the $\cos\phi$ term because the oscillating part averages to zero — gives the central result of this article:

P_{avg} = \frac{V_0 I_0}{2}\cos\phi = V_{rms}I_{rms}\cos\phi

where: $P_{avg}$ is the average (real) power in watts, $V_{rms} = V_0/\sqrt2$ and $I_{rms} = I_0/\sqrt2$ are the RMS voltage and current (the quantities your multimeter actually displays), and $\cos\phi$ is the power factor. Physically, $\cos\phi$ answers the question "how much of what the source pushes out actually lands as delivered power, and how much just bounces back?" A circuit with $\phi$ close to $90°$ (nearly pure $L$ or $C$ ) has $\cos\phi$ close to zero: current flows, the wires carry it, but almost none of the corresponding volt-amperes converts to watts. Two useful companion quantities are the apparent power $S = V_{rms}I_{rms}$ , measured in volt-amperes (VA) — this is what the source "sees" itself supplying, and what transformers, cables, and generators must be rated for — and the real power $P_{avg} = S\cos\phi$ , measured in watts (W), which is what actually does useful work. The ratio $P_{avg}/S = \cos\phi$ is exactly the power factor, so an electricity bill (which charges for real energy delivered, in kWh) can be much smaller than the apparent capacity a low-power-factor industrial customer forces the utility to hold in reserve — which is why many supply contracts specifically penalise a poor power factor.

Figure

A right triangle showing apparent power as the hypotenuse, real power along the horizontal side, and reactive power along the vertical side, with the angle phi between apparent power and real power equal to the same phase angle from the impedance triangle

Special cases you should recognise on sight

The three familiar circuit elements sit at the extremes and midpoint of the power-factor scale, and the syllabus expects you to identify each instantly. For a pure resistor, current and voltage are exactly in phase, $\phi = 0$ , so $\cos\phi = 1$ — the maximum possible power factor, and $P_{avg} = V_{rms}I_{rms}$ , the familiar DC-looking formula, is recovered exactly. All of the apparent power becomes real power; nothing is wattless. For a pure inductor or a pure capacitor, current and voltage are exactly $90°$ out of phase, $\phi = 90°$ , so $\cos\phi = 0$ and $P_{avg} = 0$ regardless of how large $V_0$ and $I_0$ are. The current is perfectly real and measurable — an ammeter reads a nonzero value — but averaged over a full cycle it delivers exactly zero net energy, because for one quarter-cycle energy flows from the source into the magnetic (or electric) field and for the next quarter-cycle it flows straight back out. This is wattless current: a genuine oscillating current that transports energy back and forth without ever consuming any of it. For a series LCR circuit, the phase angle comes from the impedance triangle you already drew for the phasor diagram: $\tan\phi = (X_L - X_C)/R$ , where $X_L = \omega L$ and $X_C = 1/\omega C$ are the reactances and $Z = \sqrt{R^2 + (X_L-X_C)^2}$ is the impedance. Reading directly off that same right triangle, the adjacent side over the hypotenuse gives

\cos\phi = \frac{R}{Z}

so that $P_{avg} = V_{rms}I_{rms}\cdot\dfrac{R}{Z} = I_{rms}^2 R$ . This final equivalent form is the most physically telling: power is dissipated only in the resistive part of the circuit. However elaborate the $L$ and $C$ combination, they never remove net energy from the source on average — they only rotate the phase angle $\phi$ and thereby determine how much of $V_{rms}I_{rms}$ counts as real power.

Why utilities care: the cost of a low power factor

A low power factor is not a purely academic inconvenience — it is a genuine engineering and economic cost. To deliver a fixed real power $P_{avg}$ to a load, the RMS current the source must supply is

I_{rms} = \frac{P_{avg}}{V_{rms}\cos\phi}

where: for a fixed target $P_{avg}$ and fixed supply voltage $V_{rms}$ , a smaller $\cos\phi$ in the denominator forces a larger $I_{rms}$ . That larger current produces larger $I_{rms}^2R_{\text{wire}}$ losses in the transmission wiring, and requires thicker cables and larger-capacity transformers rated for the higher current. This is why large industrial consumers with reactive loads — banks of induction motors, fluorescent lighting ballasts — are commonly required to install capacitor banks that partially cancel an inductive phase lag and push $\cos\phi$ back toward $1$ , even though those capacitors do no useful work themselves; their sole job is to correct the phase.

Derivation ADeriving average AC power from first principles

The goal here is to show, using only the trigonometric product-to-sum identity and the definition of a time average, exactly why the familiar DC formula $P=VI$ gains a $\cos\phi$ correction the moment current and voltage fall out of step — and to confirm that this correction reduces correctly to the three special cases (pure R, pure L or C, and general LCR) already stated above, with no assumption beyond sinusoidal steady-state signals.

Step 1. Write the instantaneous voltage and current with an explicit phase lag: $v = V_0\sin\omega t$ and $i = I_0\sin(\omega t - \phi)$ . The instantaneous power delivered to the circuit at any instant is simply their product, $p = vi = V_0 I_0 \sin\omega t\,\sin(\omega t-\phi)$ .

Step 2. Expand the product using the identity $\sin A\sin B = \tfrac12[\cos(A-B)-\cos(A+B)]$ , with $A=\omega t$ and $B=\omega t-\phi$ . This gives $p = \dfrac{V_0I_0}{2}\big[\cos\phi - \cos(2\omega t-\phi)\big]$ . The instantaneous power has split cleanly into two pieces: a constant term $\cos\phi$ that does not depend on time at all, and an oscillating term $\cos(2\omega t-\phi)$ that swings through a full cycle twice for every cycle of the original AC supply (frequency $2\omega$ , not $\omega$ ).

Step 3. Average over one full cycle of the supply. The oscillating cosine term $\cos(2\omega t-\phi)$ is positive for exactly as long as it is negative over a complete cycle, so its time average is zero — it represents energy sloshing into and back out of any reactive element, contributing nothing net. Only the constant term survives averaging: $P_{avg} = \dfrac{V_0I_0}{2}\cos\phi$ .

Step 4. Rewrite using RMS values. Since $V_{rms}=V_0/\sqrt2$ and $I_{rms}=I_0/\sqrt2$ , we have $V_0I_0 = 2V_{rms}I_{rms}$ , so $P_{avg} = \dfrac{2V_{rms}I_{rms}}{2}\cos\phi = V_{rms}I_{rms}\cos\phi$ — the result stated above. Checking the special cases confirms consistency: setting $\phi=0$ (pure R) gives $P_{avg}=V_{rms}I_{rms}$ , the DC-like maximum; setting $\phi=90°$ (pure L or C) gives $\cos\phi=0$ and $P_{avg}=0$ , the wattless case; and substituting the LCR phasor result $\cos\phi=R/Z$ gives $P_{avg}=I_{rms}^2R$ , matching the impedance-triangle picture exactly.

Worked examples

Board level - apparent power, power factor and real power from a stated phase angle

A series LCR circuit is connected to a $220\ \text{V}$ (RMS) AC source and draws an RMS current of $4\ \text{A}$ . The phase angle between voltage and current is $\phi = 60°$ . Find (a) the apparent power, (b) the power factor, and (c) the average (real) power consumed.

Step 1. Apparent power is simply the product of RMS voltage and RMS current, with no phase correction: $S = V_{rms}I_{rms} = 220\times4 = 880\ \text{VA}$ . This is the volt-ampere capacity the source must be able to supply, whatever the phase angle turns out to be.

Step 2. Power factor follows directly from the stated phase angle: $\cos\phi = \cos60° = 0.5$ . Since this is well below $1$ , we already expect roughly half the apparent power to be real, watt-consuming power.

Step 3. Average (real) power uses the central formula of this article: $P_{avg} = V_{rms}I_{rms}\cos\phi = 880\times0.5 = 440\ \text{W}$ .

The apparent power is $S=880\ \text{VA}$ , the power factor is $\cos\phi=0.5$ , and the average power is $P_{avg}=440\ \text{W}$ — although the circuit "handles" 880 VA of apparent power, only 440 W is actually real, usable power. The wattless part is the reactive power $Q = V_{rms}I_{rms}\sin\phi = 880 \times \sin 60° = 762\ \text{VAR}$ , sloshing between the source and the circuit's inductive/capacitive elements. Note that real and reactive power do not subtract from the apparent power — they add in quadrature, $S^2 = P^2 + Q^2$ (check: $\sqrt{880^2 - 440^2} = 762\ \text{VAR}$ ).

Competitive level - working backward from impedance data, plus a resistor-swap follow-up

A series LCR circuit has resistance $R = 60\ \Omega$ and total impedance $Z = 100\ \Omega$ at its operating frequency, carrying an RMS current of $1.5\ \text{A}$ . (a) Find the power factor and the average power dissipated. (b) State whether the circuit is closer to purely resistive or purely reactive behaviour. (c) If an identical source were instead connected across the resistor alone (removing $L$ and $C$ from the circuit, holding $V_{rms}$ fixed), by what factor would the real power increase?

Step 1. Power factor from the impedance triangle: $\cos\phi = \dfrac{R}{Z} = \dfrac{60}{100} = 0.6$ .

Step 2. Average power using $P_{avg} = I_{rms}^2R$ — the most direct route here, since $R$ is already known and this form sidesteps needing $V_{rms}$ at all: $P_{avg} = (1.5)^2\times60 = 2.25\times60 = 135\ \text{W}$ .

Step 3. Interpret the power factor geometrically. $\cos\phi=0.6$ gives $\sin\phi = \sqrt{1-0.6^2} = 0.8$ , so the net reactance is $X_L-X_C = Z\sin\phi = 100\times0.8 = 80\ \Omega$ , which is larger than $R = 60\ \Omega$ . So the circuit is actually somewhat more reactive than resistive — the reactance dominates the resistance, even though $\cos\phi=0.6$ is closer to the resistive extreme ( $1$ ) than to the purely reactive extreme ( $0$ ) on a linear scale; this is a reminder that $\cos\phi$ compresses angles nonlinearly, so always cross-check with the actual $X_L-X_C$ versus $R$ comparison when a question asks which behaviour "dominates."

Step 4. Compare with the resistor alone. First recover $V_{rms}$ from the original circuit: $V_{rms} = I_{rms}Z = 1.5\times100 = 150\ \text{V}$ . Connecting this same $150\ \text{V}$ (RMS) directly across just the $60\ \Omega$ resistor (no reactive elements now, so $\phi=0$ and $\cos\phi=1$ ) draws a new current $I'_{rms}=V_{rms}/R = 150/60 = 2.5\ \text{A}$ , giving a new real power $P'_{avg} = V_{rms}I'_{rms} = 150\times2.5 = 375\ \text{W}$ (equivalently $I_{rms}'^2R = 2.5^2\times60=375\ \text{W}$ ).

Removing the reactive elements and keeping the same source voltage raises the real power from 135 W to 375 W, an increase by a factor of 375/135 ≈ 2.78 — the reactance was not just "wasting" apparent power, it was actively limiting how much current (and hence real power) the same source voltage could drive through the resistor. Timing note: in a competitive exam, do not re-derive $\cos\phi = R/Z$ or $P_{avg}=I_{rms}^2R$ from scratch — quote them directly from the impedance triangle, exactly as done in Step 1 and Step 2 above, and spend your time instead on the numeric bookkeeping (recovering $V_{rms}$ , comparing the two scenarios). A full problem like this one, with the resistor-swap follow-up, should take about 3 minutes if the formulas are already fluent.

Common traps

Students consistently lose marks in three specific ways on this topic. First, applying $P_{avg}=V_{rms}I_{rms}$ to any AC circuit — this shortcut is only valid when $\phi=0$ (a pure resistor, or a series LCR circuit exactly at resonance where $X_L=X_C$ ); in every other case the $\cos\phi$ factor is mandatory. Second, misreading wattless current as meaning "no current actually flows" — a wattless current is a perfectly real, ammeter-measurable current; it merely does zero average work because it is $90°$ out of phase with the voltage. Third, assuming a low power factor means energy is being wasted inside an ideal reactive load — it is not: an ideal $L$ or $C$ dissipates nothing internally over a full cycle. The genuine cost of a low power factor is extra $I^2R$ heating in the supply and transmission wiring, which must carry a larger current to deliver the same real power at reduced $\cos\phi$ . A fourth, more mechanical trap is forgetting that $\cos\phi$ can be reached by several equivalent routes — a stated phase angle, $R/Z$ from impedance-triangle data, or $P_{avg}/(V_{rms}I_{rms})$ if real and apparent power are both given — and mixing up which route the given data actually supplies.

You should be able to…

State and derive, by averaging $v i = V_0\sin\omega t\cdot I_0\sin(\omega t-\phi)$ over one cycle, the general AC power formula $P_{avg}=V_{rms}I_{rms}\cos\phi$ .
Identify the power factor instantly for a pure resistor ( $\cos\phi=1$ ), a pure inductor or capacitor ( $\cos\phi=0$ ), and a general series LCR circuit ( $\cos\phi=R/Z$ ).
Explain what "wattless current" means physically, in terms of the $90°$ phase relationship between $v$ and $i$ , and state clearly that it is a real, non-zero current.
Compute apparent power ( $S=V_{rms}I_{rms}$ , in VA), real power ( $P_{avg}=S\cos\phi$ , in W), and power factor from given circuit data, using whichever of the equivalent routes ( $R/Z$ , a stated $\phi$ , or $P_{avg}/S$ ) the problem's data actually supplies.
Explain qualitatively, using $I_{rms}=P_{avg}/(V_{rms}\cos\phi)$ , why utilities and large industrial consumers actively work to keep the power factor close to $1$ .

Ask the tutor: if you would like the full trigonometric averaging worked out line by line with your own choice of numbers, or more practice converting between apparent power, real power, and power factor, ask the AI tutor to build a few extra worked problems with you.

If wattless current does no work, why do power companies care about it at all?

Because "does no net work in the load" is not the same as "costs nothing to supply." The wattless component still flows through every wire, transformer, and switch between the generating station and the load, heating that copper via ordinary $I^2R$ resistive loss exactly like any other current would. A factory with a poor power factor forces the utility to build thicker cables and larger transformers than the factory's real (watt) consumption alone would need, purely to carry the extra reactive current back and forth — which is why low power factor is commercially penalised even though the reactive energy itself is never "consumed."

Why does cos phi equal R over Z specifically, and not some other ratio?

It comes straight from the impedance triangle you already drew for the phasor diagram. In a series LCR circuit, $R$ is the horizontal (in-phase) side, $X_L-X_C$ is the vertical (quadrature) side, and $Z=\sqrt{R^2+(X_L-X_C)^2}$ is the hypotenuse, with $\phi$ the angle between $Z$ and $R$ . By plain right-triangle trigonometry, the side adjacent to $\phi$ divided by the hypotenuse is $\cos\phi$ , and that adjacent side is $R$ — so $\cos\phi=R/Z$ falls straight out of the same diagram that already gives you $Z$ and $\tan\phi$ .

Is there a maximum power factor a circuit can have, and can it ever be exactly zero in practice?

The maximum is $\cos\phi=1$ , reached only when $\phi=0$ exactly — a pure resistor, or a series LCR circuit tuned so $X_L=X_C$ (precisely the resonance condition of the next topic). A power factor of exactly zero needs $\phi=90°$ exactly, which in practice only an idealised pure inductor or capacitor with genuinely zero resistance achieves; every real coil has some winding resistance, so real inductors always sit slightly above zero, never exactly at it. Exam questions treat "pure L" and "pure C" as ideal limiting cases precisely so you can use $\cos\phi=0$ cleanly.

How is power factor different from efficiency — aren't they basically the same thing?

They measure different things. Efficiency compares energy output to energy input for a device converting energy from one form to another, and inefficiency there usually means real heat losses inside the device. Power factor instead compares the apparent electrical power a source supplies (volt-amperes) to the real electrical power actually delivered (watts), and a low power factor by itself does not mean energy is wasted as heat inside an ideal reactive component — it means the supply system carries more current than the real power alone would require. A device can have both issues at once, but they are separate bookkeeping questions.