# AC Power Factor > Power factor is a number from 0 to 1 that tells you what share of the power an AC device draws is turned into useful work. Page: https://formulabro.yunusemrevurgun.com/electricity/ac-power-factor/ Category: Electricity Published: 2026-10-22 Updated: 2026-10-22 ## Formula $$ P = V_{\text{rms}} I_{\text{rms}} \cos\varphi $$ Read it out loud: "P equals V sub rms times I sub rms times cosine phi. Real power P equals the rms voltage, written V sub rms, times the rms current, written I sub rms, times the cosine of phi. Phi is a Greek letter, said fye, that stands for the phase angle." ## Before you start - **Alternating current (AC) and rms values**: In AC the voltage and the current rise and fall in a wave. The wave repeats 60 times each second in US homes and 50 times each second in most of Europe. Because the values keep changing, we use the **rms** value, short for root mean square: the steady value that would heat a resistor as much. A US outlet labeled 120 V is 120 V rms. Its peak is higher, about $120 \times \sqrt{2} \approx 170$ V. - **Phase angle $\varphi$ and the cosine**: $\varphi$ is the Greek letter phi, said fye. It is an angle, measured in degrees, that says how far the current's wave is shifted in time from the voltage's wave. The **cosine**, written $\cos$, turns an angle into a number: $\cos 0^\circ = 1$, $\cos 60^\circ = 0.5$ and $\cos 90^\circ = 0$. The small circle after a number means degrees, which is one three-hundred-sixtieth of a full turn. Set a calculator to degree mode before using these. - **Watts (W), volt-amperes (VA) and vars**: A **watt** (W) is one joule of energy each second, the rate at which a device uses energy. A **volt-ampere** (VA) is one volt times one ampere. It measures the total power the wires must carry. A **var**, short for volt-ampere reactive, measures power that surges back and forth without doing net work. A 100 W light bulb uses 100 joules of energy every second. - **Percent and decimals like 0.8**: A decimal such as 0.8 is a fraction of one. To change it into a percent, multiply by 100: $0.8 \times 100 = 80$, so 0.8 means 80 percent. A power factor of 1 means 100 percent, and a power factor of 0.5 means half. Percent is another way of writing the same fraction, so the number and the percent always describe the same share. - **Sine, square roots and squares**: The **sine**, written $\sin$, is the partner of the cosine. $\sin 0^\circ = 0$ and $\sin 90^\circ = 1$. For any angle, $\sin^2\varphi + \cos^2\varphi = 1$, where $\sin^2\varphi$ means $\sin\varphi$ multiplied by itself. A **square root**, written $\sqrt{x}$, is the number that multiplied by itself gives $x$. So $\sqrt{0.36} = 0.6$, because $0.6 \times 0.6 = 0.36$. More notation help: https://formulabro.yunusemrevurgun.com/math-basics/ ## What each symbol means - $P$: **Real power**, the share of the power that does useful work, such as heating, lighting or turning a shaft. It is measured in watts (W), where one watt is one joule of energy each second. A 480 W motor uses energy at the rate of a little under five 100 W light bulbs running at once. - $V_{\text{rms}}$: **RMS voltage** of the supply, the steady-equivalent voltage of the AC wave, measured in volts (V). One volt is one joule of energy for each coulomb of charge. A standard US outlet is labeled 120 V, which means its rms voltage is 120 V. - $I_{\text{rms}}$: **RMS current** through the device, measured in amperes (A). One ampere is one coulomb of charge passing a point each second. A motor drawing 5 A rms is pulling 5 coulombs of charge through its wires every second, on average, in the rms sense of the word. - $\cos\varphi$: **Power factor**, the cosine of the phase angle $\varphi$ between voltage and current. It has no unit, because it is a ratio, and it is always a number from 0 to 1. A value of 0.8, as in the example below, means 80 percent of the apparent power does useful work. ## The formula piece by piece - $V_{\text{rms}}$: Start with the supply voltage. It is the rms voltage, the steady-equivalent number you read when you set a multimeter to AC volts. It tells you how hard the supply pushes the charges in the wires, measured in volts. - $I_{\text{rms}}$: Next, the current the device draws, also an rms value, measured in amperes. It tells you how much charge flows through the wires each second, so it sets how much the wires must carry. - $V_{\text{rms}} I_{\text{rms}}$: Multiply the voltage by the current to get the apparent power $S$, in volt-amperes. This is the total the wires must carry if the voltage and current rise and fall together, and it is always at least as big as the real power. - $\cos\varphi$: The power factor is the in-step share. It is a number from 0 to 1 that tells you what fraction of the apparent power is delivered as useful work, after the current's shift in time is taken into account. - $P = V_{\text{rms}} I_{\text{rms}} \cos\varphi$: The whole formula in one sentence: the real power used by an AC device equals the rms voltage times the rms current times the cosine of the phase angle between them. - $P = S \cos\varphi$: The same law written with the apparent power $S$ as the starting point. Take the volt-amperes the wires carry and keep only the fraction $\cos\varphi$ that does useful work, which gives the real power in watts. ## What the formula is doing, and why In a DC circuit, power is voltage times current. In AC, both the voltage and the current rise and fall in a wave. When the load contains a coil or a capacitor, the current's wave is shifted in time from the voltage's wave. The phase angle $\varphi$ measures that shift in degrees. The more the two waves are out of step, the less of the current's motion turns into useful work. **Why multiply, and why the cosine?** Power is a rate: it is how much energy moves each second. Multiply $V_{\text{rms}}$ by $I_{\text{rms}}$ and you get the apparent power $S$, in volt-amperes, which is the total the wires must carry. When the voltage and current are out of step, only the in-step part of the current moves energy forward on average. The cosine of the phase angle is that in-step share, so $P = V_{\text{rms}} I_{\text{rms}} \cos\varphi$. **Why is the power factor between 0 and 1?** The cosine of an angle from 0 to 90 degrees is a number that runs from 1 down to 0. A power factor of 1 means the voltage and current are in step, as in a heater or a plain light bulb, so every volt-ampere does useful work. A power factor of 0 would mean the current is a quarter of a cycle out of step, so the device does no net work even though current flows. Real loads sit between these two ends. **Where the rest of the power goes.** The part of the apparent power that is not real power is the reactive power, $Q$, measured in vars. Coils and capacitors store energy in magnetic and electric fields and give it back on each half cycle. That back-and-forth energy does no net work, but the wires must still carry its current, so the current still heats them. Motors and transformers draw reactive power, which is why a low power factor means more current for the same useful work. **When the formula applies.** $P = V_{\text{rms}} I_{\text{rms}} \cos\varphi$ is exact for sine-wave AC in steady operation. Devices whose current is distorted, such as many LED lamps and computer power supplies, need the general power factor, $P/S$, which works for any waveform. In a DC circuit there is no phase shift, the angle is zero, and the formula reduces to $P = VI$. ## An everyday picture Picture a mug of beer with a thick head of foam. The whole mug stands for the apparent power, everything the supply must send down the wires. The beer is the real power, the part you can drink. The foam fills the mug and must travel through the same spout, but it gives you nothing to drink. If the mug is 80 percent beer, the power factor is 0.8. A mug of pure beer with no foam at all has a power factor of 1. ## Worked example: A 120 V motor drawing 5 A at a power factor of 0.8 A motor runs from a 120 V rms supply, the kind of outlet found in US homes, and draws 5 A rms. Its power factor is 0.8, which is 80 percent. We want the apparent power $S$, the real power $P$, the phase angle $\varphi$ and the reactive power $Q$. All the voltages and currents here are rms values. 1. **Find the apparent power.** Apparent power is the total the wires must carry, found by multiplying the rms voltage by the rms current: $S = V_{\text{rms}} I_{\text{rms}} = 120\ \text{V} \times 5\ \text{A} = 600\ \text{VA}$. Volts times amperes gives volt-amperes, so the unit is VA and no conversion is needed. 2. **Find the real power.** Real power is the share of $S$ that does useful work, so we multiply by the power factor: $P = S \cos\varphi = 600\ \text{VA} \times 0.8 = 480\ \text{W}$. Since 0.8 means 80 percent, 80 percent of the 600 VA becomes useful work, and the unit changes from volt-amperes to watts. 3. **Find the phase angle.** We need the angle whose cosine is 0.8. The inverse cosine, written $\arccos$, answers the question which angle has this cosine. With a calculator in degree mode, $\varphi = \arccos(0.8) \approx 36.9^\circ$. This tells us how far the current is out of step with the voltage. 4. **Find the sine of the angle.** Reactive power needs $\sin\varphi$, and we can get it from the cosine: $\sin\varphi = \sqrt{1 - \cos^2\varphi} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6$. The square root of 0.36 is 0.6 because $0.6 \times 0.6 = 0.36$, so the sine is 0.6. 5. **Find the reactive power.** Reactive power is the part of the apparent power that goes back and forth, and it is found with the sine: $Q = S \sin\varphi = 600\ \text{VA} \times 0.6 = 360\ \text{var}$. The unit var means volt-ampere reactive, and it is written as var to tell it apart from real power in watts. 6. **Check the three numbers together.** The three powers form a right triangle, with $S$ as the long side, so $S^2 = P^2 + Q^2$ must hold. Here $480^2 + 360^2 = \text{230,400} + \text{129,600} = \text{360,000}$, and $600^2 = \text{360,000}$ as well. The two sides agree, so the arithmetic is right. 7. **Say the numbers in everyday terms.** The supply must be able to carry 600 VA, which is 5 A through a 120 V line. Of that, 480 W is real work, a rate of energy a little under five 100 W light bulbs running at once. The remaining 360 var goes back and forth and does no net work. **Answer:** The apparent power is **600 VA**, the real power is **480 W** and the power factor is **0.8**. The remaining **360 var** is reactive power. It goes back and forth between the supply and the motor and does no net work, but the full 5 A still flows through the wires and heats them. ## Common mistakes - Using $P = VI$ for an AC load with a power factor below 1. That product is the apparent power $S$ in volt-amperes. The real power is smaller, because it is $S$ times the power factor. - Mixing up VA and W. Volt-amperes measure apparent power and watts measure real power. The two numbers match only when the power factor is exactly 1, as in a plain heater. - Using peak voltage or peak current instead of rms values. For a sine wave the peak is about 1.41 times the rms value, so using peaks makes the result about 41 percent too big. - Thinking a low power factor means the device needs more useful energy. It does not. The same work needs more current when the power factor is low, so the wires carry more current and heat more. ## Where you will meet it - Utility bills, where some large customers, mostly factories, pay extra when their power factor is low. - Sizing cables, transformers and generators, which must carry the full apparent power in VA, not only the real power in W. - Power factor correction, where capacitors are added to a factory or building to supply the reactive power that its motors would otherwise pull through the wires. - Motor and appliance labels, where a rating in VA or a power factor tells you how much current the device draws. ## Questions people ask ### Why is apparent power bigger than real power? Because the current and the voltage are out of step for part of each cycle. That out-of-step part of the current carries power that goes back and forth and does no net work. Apparent power counts all of it, so it is never smaller than the real power. ### What is a var? A var, short for volt-ampere reactive, is the unit of reactive power, $Q$. It has the same size as a volt-ampere, but it measures the part of the power that moves between the supply and a coil or capacitor instead of doing useful work. ### Does a power factor of 1 mean a device is efficient? Not by itself. It means the voltage and current are in step, so all of the apparent power is real power. It does not show how much energy the device wastes as heat inside itself, which depends on its design. ### How is a low power factor improved? Power factor correction adds capacitors close to motors and other loads that contain coils. The capacitors supply some of the reactive power locally, so less current flows through the supply wires from the grid. ## Related formulas - [Electric Power](https://formulabro.yunusemrevurgun.com/electricity/electric-power/) - [RMS Voltage](https://formulabro.yunusemrevurgun.com/electricity/rms-voltage/) - [Series RLC Impedance](https://formulabro.yunusemrevurgun.com/electricity/rlc-impedance/) ## Sources - [Power factor, Wikipedia](https://en.wikipedia.org/wiki/Power_factor) - [Reactive power, Wikipedia](https://en.wikipedia.org/wiki/Reactive_power) From Formula Bro (https://formulabro.yunusemrevurgun.com/), a side project developed by yunusemrevurgun.com.