Power in AC circuits

Plug in a fluorescent tube fitted with a choke coil and it lights up perfectly well, yet the electricity meter barely notices the coil is there — despite a substantial current flowing through it every instant. How can current flow steadily without the meter charging for it? The answer is that in AC circuits, voltage and current are usually out of step with each other, and the power actually delivered depends not just on their sizes but on the phase angle $\phi$ between them, through a factor called the power factor, $\cos\phi$ . When $\phi = 90°$ — as in an ideal inductor or capacitor — current flows but does no net work at all. This article builds the average-power formula from scratch, applies it to a series LCR circuit, and uses it to explain wattless current and the choke coil as a practical energy-saving device.

The core insight

Average AC power is $\bar P = V_{rms}I_{rms}\cos\phi$ , so only the in-phase (resistive) part of the current does any net work — reactive current flows without dissipating energy.

The core idea

In a DC circuit, power is just $P = VI$ . In an AC circuit this fails whenever current lags or leads voltage, because the instantaneous product $p = vi$ swings positive and negative over a cycle, and only its time average corresponds to real energy delivered. For a series AC circuit driven by $v = V_0\sin\omega t$ with resulting current $i = I_0\sin(\omega t - \phi)$ , averaging the instantaneous power $p = vi$ over a full cycle gives the average power:

\bar P = \frac12 V_0 I_0\cos\phi = V_{rms}I_{rms}\cos\phi

where: $V_{rms} = V_0/\sqrt2$ and $I_{rms} = I_0/\sqrt2$ are the RMS voltage and current, and $\cos\phi$ is the power factor — the cosine of the phase angle between voltage and current. For a series LCR circuit, the phasor triangle gives $\cos\phi = R/Z$ (R is the adjacent side of the triangle, Z the hypotenuse), so the average power can equally be written

\bar P = I_{rms}^2 R

This second form is the more physical one: only resistance dissipates energy as heat. An ideal inductor or capacitor stores energy in its field during part of the cycle and returns all of it during the rest, with zero net loss.

Two limiting cases worth memorising. A pure resistor has $\phi = 0$ , so $\cos\phi = 1$ (the maximum possible power factor) and $\bar P = V_{rms}I_{rms}$ — every AC circuit's best case. A pure inductor or pure capacitor has $\phi = \pm 90°$ , so $\cos\phi = 0$ and $\bar P = 0$ — substantial current can flow while zero net power is transferred, because the instantaneous power is exactly as often negative (energy flowing back to the source) as positive. The current component responsible for this non-dissipative flow, $I_{rms}\sin\phi$ , is called the wattless (idle) current. It does no useful work by itself, but it still causes real $I^2R$ heating losses in the non-ideal wires and equipment it passes through elsewhere in the supply network — which is exactly why power utilities penalise industrial consumers for running circuits at a low power factor, and why they encourage keeping $\cos\phi$ close to 1.

Figure

Phasor triangle showing impedance Z as hypotenuse, resistance R as the adjacent side and reactance X as the opposite side, with phase angle phi between them, illustrating cos phi equals R over Z

Choke coil. A choke coil is a near-ideal inductor — very large reactance $X_L = \omega L$ , very small winding resistance $r$ — used to limit AC current (e.g. in tube-light fittings) without wasting power as heat, unlike a resistor achieving the same current limit. Since $r \ll X_L$ , the phase angle $\phi \approx 90°$ and $\cos\phi = r/Z \approx 0$ ; current still flows and is limited by the reactance, but the average power drawn from the supply is tiny, because almost all of it is exchanged with the coil's magnetic field and handed back, not converted to heat.

Derivation AAverage power from the instantaneous power p = vi

The goal is to show, starting only from $v = V_0\sin\omega t$ and $i = I_0\sin(\omega t - \phi)$ , that time-averaging the instantaneous power $p = vi$ over one full cycle collapses the product of two oscillating quantities into the compact result $\bar P = \tfrac12 V_0 I_0\cos\phi$ , with no calculus beyond the standard cycle-average of $\sin^2$ and $\sin\cdot\cos$ terms that a student already knows from RMS work.

Step 1. Write the instantaneous power as a product of the two signals: $p = vi = V_0 I_0 \sin\omega t \sin(\omega t - \phi)$ . This is a single number at each instant — sometimes positive (energy flows into the circuit), sometimes negative (energy flows back out) — and we want its average over one period.

Step 2. Expand $\sin(\omega t - \phi) = \sin\omega t\cos\phi - \cos\omega t\sin\phi$ and substitute: $p = V_0I_0\left[\sin^2\omega t\cos\phi - \sin\omega t\cos\omega t\sin\phi\right]$ . The product has split into two separate oscillating terms, one multiplying $\cos\phi$ and one multiplying $\sin\phi$ .

Step 3. Average each term over a full cycle. The cycle-average of $\sin^2\omega t$ is $\tfrac12$ (the standard RMS result), while the cycle-average of $\sin\omega t\cos\omega t$ is exactly zero, because it equals $\tfrac12\sin2\omega t$ , which is equally positive and negative over a period. So the second term vanishes entirely and only the first survives.

Step 4. Collect the surviving term: $\bar P = V_0I_0\cos\phi\times\tfrac12 = \tfrac12V_0I_0\cos\phi$ . Writing $V_0 = \sqrt2\,V_{rms}$ and $I_0 = \sqrt2\,I_{rms}$ gives the usual form $\bar P = V_{rms}I_{rms}\cos\phi$ . The physical content is now visible directly in the algebra: the $\cos\phi$ term is what survives averaging (net energy transfer), and the $\sin\phi$ term is what cancels to zero (energy sloshing back and forth, contributing nothing net) — this cancelling term is exactly the wattless current's contribution.

Worked example

A series LCR circuit has $R = 25\ \Omega$ and impedance $Z = 50\ \Omega$ at its operating frequency. It draws an RMS current of 3 A from a 150 V (RMS) supply. Find (a) the power factor, (b) the average power consumed, (c) the wattless current.

Step 1. Power factor from the impedance triangle: $\cos\phi = R/Z = 25/50 = 0.5$ , so $\phi = 60°$ .

Step 2. Average power using $\bar P = V_{rms}I_{rms}\cos\phi = 150 \times 3 \times 0.5 = 225\ \text{W}$ . Cross-check with the resistive form: $\bar P = I_{rms}^2R = 3^2\times25 = 225\ \text{W}$ — the two formulas agree, confirming the value.

Step 3. Wattless current uses $\sin\phi = \sin60° = 0.866$ : $I_{rms}\sin\phi = 3\times0.866 = 2.6\ \text{A}$ . Answer: power factor 0.5, average power 225 W, wattless current 2.6 A.

Competitive stretch

Extend the choke-coil case from Pass-1: a choke of $L = 1.5\ \text{H}$ and $r \approx 2\ \Omega$ on a 200 V, 50 Hz supply gives $X_L = \omega L = 314.16\times1.5 = 471.2\ \Omega$ , $Z \approx 471.2\ \Omega$ (r negligible), $I_{rms} = 200/471.2 = 0.424\ \text{A}$ , $\cos\phi = r/Z = 2/471.2 = 0.00424 \approx 0$ , and $\bar P = 200\times0.424\times0.00424 = 0.36\ \text{W}$ . A plain resistor limiting the same current would instead dissipate $\bar P = I_{rms}^2R = (0.424)^2\times471.2 \approx 84.7\ \text{W}$ — nearly 235 times more power lost as heat for the same current-limiting job, which is exactly why chokes, not resistors, are used in ballasts.

Common traps

Using $\bar P = V_{rms}I_{rms}$ (dropping $\cos\phi$ ) for any AC circuit — valid only for a purely resistive load; every other circuit needs the power factor.
Assuming zero net power means zero current — a purely reactive circuit ( $\phi = 90°$ ) still carries substantial current; it is the time-averaged product $vi$ that vanishes, not the current itself.
Treating a choke coil as just another current-limiting resistor — a choke limits current through reactance while dissipating far less power than a resistor giving the same current limit.
Applying $\cos\phi = R/Z$ outside a series circuit — this specific ratio assumes R is the net series resistance; it is not a universal formula for every circuit configuration.

Ask the tutor: Want to see how power factor correction (adding a capacitor to cancel an inductive load's lagging phase) works in practice?

If a choke coil dissipates almost no power, why does the tube light still glow using real energy?

The choke does not power the tube — it merely limits the current flowing through the whole circuit using reactance rather than resistance. The tube itself (a resistive load, or one with its own resistance) is where the actual power delivered by the supply gets converted into light and heat, governed by $\bar P = I_{rms}^2R_{tube}$ . The choke's job is only to keep the current within a safe, steady value while wasting as little of the supply's power as possible in doing so — that is the entire point of using a choke instead of a plain resistor for current limiting.

Why does averaging sin(omega t) times cos(omega t) over a cycle give zero?

Because $\sin\omega t\cos\omega t = \tfrac12\sin2\omega t$ , and $\sin2\omega t$ is a symmetric oscillation that spends exactly half of every cycle positive and half negative, with equal magnitude in each half. Averaged over a whole number of cycles, the positive and negative areas cancel exactly, leaving zero. This is the same reason the reactive (wattless) part of the current contributes nothing to net average power — it represents energy sloshing into the field and back out, never a one-way transfer.

Is the power factor the same thing as efficiency?

No, though they are often confused. Efficiency compares useful output power to total input power in an energy-conversion device (like a motor or transformer) and is always less than or equal to 1 due to internal losses. The power factor $\cos\phi$ describes how well the current is phase-aligned with the voltage in an AC circuit — a low power factor does not mean energy is "wasted" as heat inside the reactive component itself (an ideal inductor loses nothing), but it does mean larger currents are needed to deliver the same real power, which increases $I^2R$ losses elsewhere in the supply lines and forces power stations and cables to be oversized.

Power in AC circuits

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads