Power factor and AC power

Plug a heater into the mains and every watt you pay for turns into heat — voltage and current rise and fall together, $P = VI$ , and that is the whole story. Plug in a motor instead, and the story gets a twist: current lags voltage, and only part of the $VI$ product does useful work. The rest sloshes back and forth between the supply and the motor's magnetic field, doing nothing except making the cables hotter than they need to be. The fraction of $VI$ that actually counts is set by a single number — the power factor, $\cos\theta$ — and once you have it, three power quantities (active, reactive, apparent) fall out of the same phase angle $\theta$ that you already met in phasor and impedance work. This article turns that phase angle into watts, VARs, and VA, and shows why utilities obsess over keeping $\cos\theta$ close to 1.

The core insight

In an AC circuit the phase angle $\theta$ between voltage and current splits the product $VI$ into active power $P=VI\cos\theta$ (useful, watts), reactive power $Q=VI\sin\theta$ (oscillating, VAR) and apparent power $S=VI$ (VA), related by $S=\sqrt{P^2+Q^2}$ .

The core idea

From phasors to power. For an RMS voltage $V$ and RMS current $I$ separated by phase angle $\theta$ (current relative to voltage), power factor is defined as $\cos\theta$ . For a simple series R-L-C circuit this is exactly the ratio from the impedance triangle you already built:

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

where: $R$ is the circuit resistance, $Z=\sqrt{R^2+(X_L-X_C)^2}$ is the impedance magnitude, and $\theta$ is the same angle by which current leads or lags voltage. Power factor is quoted as a magnitude between 0 and 1, with the sign of $\theta$ stated separately as lagging (inductive, current lags — $X_L>X_C$ , the common case with motors and transformers) or leading (capacitive, current leads). A purely resistive circuit gives $\theta=0$ , unity power factor.

Three powers, one triangle. Multiplying $V$ and $I$ two different ways with $\theta$ gives the three quantities that matter in practice:

P = VI\cos\theta \qquad Q = VI\sin\theta \qquad S = VI

where: $P$ is active (real) power in watts (W) — the power actually converted to heat, light, or mechanical work; $Q$ is reactive power in volt-amperes-reactive (VAR) — energy that shuttles between the source and the circuit's inductors/capacitors without net transfer; $S$ is apparent power in volt-amperes (VA) — simply $V$ times $I$ , ignoring phase. These three sides form a right triangle, the power triangle, a direct echo of the impedance triangle ( $R,X,Z$ scaled by current):

S = \sqrt{P^2+Q^2}, \qquad \cos\theta = \frac{P}{S}

The svg figure below lays the two triangles side by side so the correspondence is visible at a glance: $R\to P$ , $X\to Q$ , $Z\to S$ , same angle $\theta$ in both.

Why it matters for the supply system. A low power factor means a fixed useful power $P$ needs a large current, since $I = S/V = P/(V\cos\theta)$ : as $\cos\theta$ falls, $I$ rises for the same $P$ . That extra current does not deliver more work — it just raises $I^2R$ losses in the supply cables and forces generators, transformers, and cables to be rated for a bigger apparent power $S$ than the useful power $P$ they actually deliver. This is why industrial consumers with lots of motors (low, lagging power factor) are penalised on their electricity bill unless they correct their power factor toward unity.

Derivation ADeriving the active, reactive, and apparent power from V, I, and theta

The goal is to show, starting only from the instantaneous power $p(t) = v(t)i(t)$ of a sinusoidal voltage and current separated by phase angle $\theta$ , that its time-average splits cleanly into the active power $P = VI\cos\theta$ plus a purely oscillating part whose amplitude is the reactive power $Q = VI\sin\theta$ , so that the "useful" and "sloshing" parts of AC power are not assumptions but a direct consequence of averaging a product of two sinusoids over one cycle.

Step 1. Write the instantaneous voltage and current as $v(t) = V_m\sin(\omega t)$ and $i(t) = I_m\sin(\omega t - \theta)$ , where $\theta$ is how much current lags voltage (negative $\theta$ for leading current), and $V_m, I_m$ are peak values.

Step 2. Instantaneous power is the product, $p(t) = v(t)i(t) = V_mI_m\sin(\omega t)\sin(\omega t - \theta)$ . Using the product-to-sum identity $\sin A\sin B = \tfrac12[\cos(A-B)-\cos(A+B)]$ with $A=\omega t$ , $B=\omega t-\theta$ gives $p(t) = \dfrac{V_mI_m}{2}\big[\cos\theta - \cos(2\omega t - \theta)\big]$ .

Step 3. Split this into a constant term and a term that oscillates at twice the supply frequency: $p(t) = \dfrac{V_mI_m}{2}\cos\theta \;-\; \dfrac{V_mI_m}{2}\cos(2\omega t-\theta)$ . The first term never changes with time; the second averages to zero over a full cycle because a cosine integrates to zero over $2\pi$ .

Step 4. Average $p(t)$ over one full cycle. Only the constant term survives: $P = \left\langle p(t)\right\rangle = \dfrac{V_mI_m}{2}\cos\theta$ . Writing peak values in terms of RMS values, $V_m = \sqrt2\,V$ and $I_m=\sqrt2\,I$ , this becomes $P = \dfrac{(\sqrt2 V)(\sqrt2 I)}{2}\cos\theta = VI\cos\theta$ — the active power formula, now derived rather than assumed.

Step 5. The oscillating term never vanishes on its own (it is only zero on average); its amplitude, $\dfrac{V_mI_m}{2}\sin\theta = VI\sin\theta$ , measures how strongly power surges back and forth each half-cycle without net transfer — this amplitude is exactly the reactive power $Q$ . Apparent power $S=VI$ is simply the product of RMS magnitudes with no reference to $\theta$ at all, so $P=S\cos\theta$ and $Q=S\sin\theta$ automatically satisfy $S=\sqrt{P^2+Q^2}$ by the Pythagorean identity $\cos^2\theta+\sin^2\theta=1$ .

Worked example

A single-phase load draws 15 A from a 230 V, 50 Hz supply at a lagging power factor of 0.8. Extend Pass-1's result by also correcting this load to unity power factor conceptually, and check what reactive power a capacitor would need to cancel.

Step 1 — apparent power. $S = VI = 230\times15 = 3450\ \text{VA} = 3.45\ \text{kVA}$ .

Step 2 — phase angle. $\cos\theta = 0.8 \Rightarrow \theta = \cos^{-1}(0.8) = 36.87^\circ$ lagging, so $\sin\theta = 0.6$ .

Step 3 — active and reactive power. $P = VI\cos\theta = 3450\times0.8 = 2760\ \text{W} = 2.76\ \text{kW}$ , and $Q = VI\sin\theta = 3450\times0.6 = 2070\ \text{VAR} = 2.07\ \text{kVAR}$ , absorbed by the inductive load.

Step 4 — check the power triangle. $S = \sqrt{P^2+Q^2} = \sqrt{2760^2+2070^2} = \sqrt{7.6176\times10^6 + 4.2849\times10^6} = \sqrt{11.9\times10^6} = 3450\ \text{VA}$ , matching Step 1.

Step 5 — what correcting to unity power factor would need. To bring $\cos\theta$ from 0.8 to 1 at the same $P=2760\ \text{W}$ , a capacitor must supply reactive power equal and opposite to the load's $Q$ , i.e. $Q_C = 2070\ \text{VAR}$ leading, so the source only ever sees $Q_{\text{net}} = Q - Q_C = 0$ . At unity power factor the same 2760 W would need only $I' = P/V = 2760/230 = 12\ \text{A}$ instead of 15 A — a 20 percent drop in supply current for identical useful power, which is the concrete payoff of power-factor correction.

Common traps

Confusing "leading" and "lagging": lagging means current lags voltage (inductive load, e.g. motors — the common industrial case); leading means current leads voltage (capacitive load).
Using peak values of $V$ and $I$ instead of RMS values in $P=VI\cos\theta$ — the formula is built from RMS phasor magnitudes, as the derivation above shows explicitly.
Treating apparent power $S=VI$ as the real power delivered — only $P=VI\cos\theta$ is useful (watt) power, which is why generators and transformers are rated in kVA but energy is billed in kWh.
Swapping cosine and sine: active power always uses $\cos\theta$ , reactive power always uses $\sin\theta$ , of the same voltage-current phase angle.

Ask the tutor: if you want to see how a capacitor is actually sized to correct power factor toward unity, ask the AI tutor to walk through the power-factor-correction calculation building on this article's reactive power.

Why does a low power factor cost the utility money even though the customer's watt-hour meter reads the same?

Because the watt-hour meter only bills active power $P$ , but the cables, transformers, and generators supplying that customer must be sized for the larger apparent power $S = P/\cos\theta$ . A lagging load at $\cos\theta = 0.5$ needs twice the current of the same load at unity power factor to deliver identical watts, doubling $I^2R$ losses in the wires and forcing the utility to install thicker cables and bigger transformers for no extra revenue. That mismatch between what is billed and what is supplied is exactly why many utilities add a power-factor penalty clause for large industrial consumers.

Can power factor ever be negative, and what would that mean?

The cosine of an angle can be negative for $\theta$ beyond 90 degrees, but in normal single-phase loads (a resistor with inductors and/or capacitors) the phase angle stays between minus 90 and plus 90 degrees, so $\cos\theta$ stays between 0 and 1 and is always quoted as a positive magnitude with "leading" or "lagging" attached separately. A negative sign convention is reserved for cases like a source actually absorbing power (feeding energy back into the grid, as a generator does) rather than describing an ordinary passive load, which is beyond this article's scope.

How is this different from the impedance triangle we already learned?

It is the same triangle, scaled. The impedance triangle has sides $R$ , $X$ , $Z$ in ohms; multiply every side by the current magnitude $I$ (or more precisely by $VI$ using $V=IZ$ ) and $R\to P$ , $X\to Q$ , $Z\to S$ , with the angle $\theta$ unchanged. That is why $\cos\theta = R/Z$ from the impedance triangle and $\cos\theta = P/S$ from the power triangle are two routes to the same number — whichever quantities a problem gives you, you can compute $\theta$ from either triangle and use it in the other.

Power factor and AC power

The core idea

Worked example

Common traps

Before this

Where this leads