Power (PIE) Formula

The power formula, often remembered as PIE, says that electric power equals current times voltage: P = I × E. It answers everyday questions like "how much current does a 60 W bulb draw?", "how hot will this resistor get?" and "what does it cost to leave this on?" This page explains each letter from the start, shows the three ways to write the formula, and gives you a live calculator to try your own numbers.

1. What power is

Start with energy: the ability to make something happen, such as heating water, lighting a room or turning a motor. Energy is never used up; it only changes form. A battery turns chemical energy into electrical energy, and a lamp turns electrical energy into light and heat.^[1] The SI unit of energy (SI is the International System of Units, the modern metric system) is the joule (J).^[2] One joule is small: roughly the energy needed to lift an apple one meter.

Power is the rate at which energy is converted or moved: how many joules per second.^[3] Its unit is the watt (W), and one watt is exactly one joule per second.^[2] So a 60 W lamp converts 60 joules of electrical energy into light and heat every second it is on.

Energy versus power

The two words are easy to mix up, so it is worth being precise:

• Energy is an amount, like the number of gallons of water in a bucket. It is measured in joules, or in watt-hours and kilowatt-hours (section 7).
• Power is a rate, like how many gallons per minute flow from a tap. It is measured in watts.

The link between them is time: energy = power × time.^[3] A 60 W lamp left on for twice as long uses twice the energy, but its power is still 60 W. Your electric bill charges for energy, not power; the power rating tells you how fast the meter runs.^[3]

Watts often carry a prefix, a letter in front that multiplies the unit by a power of ten: a milliwatt (mW) is one-thousandth of a watt and a kilowatt (kW) is a thousand watts.^[2] SparkFun gives a sense of scale: microcontrollers such as the Arduino run in the microwatt-to-milliwatt range, laptops and desktop computers in watts, and a whole house in kilowatts.^[1]

2. The PIE formula

Electric power comes from two things working together. Current is the rate at which electric charge flows past a point, measured in amperes (A, "amps"); one ampere is one coulomb of charge per second, where a coulomb (C) is the SI unit of charge.^[4] Voltage is the energy each coulomb of charge gains or gives up between two points, measured in volts (V); one volt is one joule per coulomb.^[5] Multiply them and the coulombs cancel:

joules per coulomb × coulombs per second = joules per second = watts

That is the whole formula. Power is current times voltage:^[3]​^[6]

P = I × E

So 1 W = 1 A × 1 V. NIST, the U.S. standards agency, defines the volt the same way from the other direction: one volt is one watt per ampere.^[2] OpenStax puts it simply: "1 A · V = 1 W."^[3]

Notice what the formula says: neither voltage nor current alone is power. A battery sitting on a shelf has voltage but no current flowing, so it delivers zero power; anything multiplied by zero is zero.^[6] A small current at a high voltage can carry the same power as a large current at a low voltage.

Why "E" for voltage?

E stands for electromotive force, often shortened to emf. Despite the name, emf is not a force at all; it is a voltage. Strictly, it is the voltage a source such as a battery or generator produces when no current is flowing.^[7] Older books and many technicians use E for voltage, which is how the memorable "PIE" came about. Kuphaldt notes that E and V are interchangeable for the most part, although some texts keep E for the voltage of a source and V for the voltage across everything else.^[6] NIST's table of SI units lists "electric potential difference" and "electromotive force" side by side, both measured in volts.^[2]

So P = I × E and P = V × I (the form OpenStax and the Ohm's Law helper use) are the same formula with a different letter for voltage. Be careful not to confuse the letter V for the quantity voltage with the symbol V for the unit volt: "V = 12 V" means "the voltage is twelve volts." That is one reason some people prefer E. And I? It is usually said to stand for the "intensity" of the current, though Kuphaldt notes the origin is disputed.^[6]

3. The PIE circle

PIE is one equation, but you can solve it for whichever quantity you do not know. Here is the algebra, step by step, starting from P = I × E:

1. To find power, use it as is: P = I × E.
2. To find current, divide both sides by E. On the right, E ÷ E cancels to 1, leaving I = P ÷ E.
3. To find voltage, divide both sides by I instead, leaving E = P ÷ I.

If algebra is rusty, the circle in Figure 1 is a memory aid. P sits on top, with I and E side by side underneath. Cover the letter you want with a finger, and what is left shows the formula: two letters side by side multiply, and a letter above another divides by it. The same layout drawn as a triangle works exactly the same way.

Before you calculate, convert values to plain watts, amperes and volts. Mixing units is the most common mistake: 500 mA is 0.5 A, and 2 kW is 2,000 W. Watts, amps and volts fit together directly, and so do milliwatts, milliamperes and volts (because mA × V = mW, calculated from the prefix values).

4. The calculator

Type values into any two of the three boxes and the third is calculated for you. The line under the boxes shows which form of the formula was used, with your numbers filled in. A few details:

• Which box is calculated? The highlighted one, marked "Calculated." Use the Solve for buttons to choose it, or simply type in it: the box you used least recently then becomes the calculated one.
• Units. Pick mW, mA, kV and so on from the list beside each box instead of typing letters. The calculated box picks a sensible unit by itself; change it if you prefer another.
• Equivalent resistance. The calculator also shows R = E ÷ I, the resistance that would draw this current at this voltage (section 6).
• Energy cost. Enter how many hours a day the device runs and your electricity price, and it estimates the daily and monthly cost (section 7). The price starts at an example rate; change it to the rate on your own bill.
• Checks. Values cannot be negative, boxes must hold plain numbers, and you cannot divide by zero. If something is off, a message explains it instead of showing a nonsense answer.
• The drawing is the PIE circle from Figure 1. The letter being calculated lights up, and each part shows its current value.

The calculator treats P, I and E as steady values. For household AC, enter the voltage and current printed on the label (these are already the right kind of average; see section 8).

5. Worked examples

Each example follows the same three steps: write down what you know in plain units, pick the form of the formula that gives the unknown, then calculate. All results here are calculated.

Example 1: a 60 W bulb on 120 V (find the current)

Most U.S. household outlets supply 120 V.^[8] Known: P = 60 W, E = 120 V. Unknown: I, so cover I in the circle and use I = P ÷ E.

I = 60 W ÷ 120 V = 0.5 A = 500 mA

A bulb rated for more power at the same voltage must draw more current. OpenStax uses this comparison: a 60 W bulb draws more current than a 25 W bulb on the same supply, so its filament has the lower resistance.^[3] (Checking with numbers: 25 W ÷ 120 V ≈ 0.21 A.)

Example 2: a 5 V USB device drawing 2 A (find the power)

A phone charger's label might read "Output: 5V ⎓ 2A" (the symbol ⎓ means DC, direct current, which flows one way only). Known: E = 5 V, I = 2 A. Use P = I × E.

P = 2 A × 5 V = 10 W

So that charger can deliver up to 10 W. A charger rated 5 V and 1 A gives only 5 W, which is why it charges more slowly. Many chargers print the wattage directly; if yours shows only volts and amps, PIE turns them into watts.

Example 3: a 12 V car accessory drawing 3 A (find the power)

A car's electrical system runs at about 12 V.^[3] Known: E = 12 V, I = 3 A. Use P = I × E.

P = 3 A × 12 V = 36 W

Turn it around to check whether an accessory will fit an outlet. OpenStax's example: a car's auxiliary outlet rated 20 A can deliver at most 20 A × 12 V = 240 W.^[3] A 120 W accessory would draw I = 120 W ÷ 12 V = 10 A, inside that limit; a 300 W one would need 25 A, more than the outlet is rated for.

Example 4: 24 W at 2 A (find the voltage)

Known: P = 24 W, I = 2 A. Unknown: E, so use E = P ÷ I.

E = 24 W ÷ 2 A = 12 V

Checking it the other way: 2 A × 12 V = 24 W. Running a result back through another form of the formula is a quick way to catch mistakes.

Example 5: is a ¼ W resistor big enough? (a resistor power check)

A resistor is a part made to have a definite resistance, and all the power it takes in becomes heat.^[1] Every resistor has a power rating, the most heat it can safely shed (section 8). Suppose a 100 Ω resistor sits directly across a 5 V supply. (To read a resistor's value from its colored stripes, see the Resistor Color Codes helper.)

1. Find the current with Ohm's law: I = E ÷ R = 5 V ÷ 100 Ω = 0.05 A (50 mA).
2. Find the power with PIE: P = I × E = 0.05 A × 5 V = 0.25 W.
3. Compare: 0.25 W is exactly the rating of a common ¼ W resistor, with no margin at all. It would run hot. The twice-the-power habit from section 8 points to a ½ W part.

Section 6 shows a shortcut that skips step 1: P = E² ÷ R = 25 ÷ 100 = 0.25 W.

6. PIE meets Ohm's law

Resistance (R, measured in ohms, Ω) is how strongly something opposes current. Ohm's law links it to voltage and current: E = I × R, or I = E ÷ R.^[6] The Ohm's Law helper explains it from the start and has its own calculator.

When you know only the resistance and one other quantity, you can substitute Ohm's law into PIE to get two more forms of the power formula:^[3]​^[6]

• Know current and resistance? Replace E with I × R: P = I × (I × R), so P = I²R (current squared, times resistance).
• Know voltage and resistance? Replace I with E ÷ R: P = (E ÷ R) × E, so P = E² ÷ R (voltage squared, divided by resistance).

All three give the same answer for a resistor. Kuphaldt adds a historical note: it was James Prescott Joule, not Ohm, who first found the I²R relationship, published in 1841, so it is properly called Joule's law.^[6] There is one important difference between the forms. OpenStax points out that P = I × E is always valid, for any part, while the other two apply only to resistors.^[3] For a motor, an LED (light-emitting diode) or a charger, use PIE with the measured current and voltage.

The squares matter. Double the voltage across a resistor and the current doubles too, so the power goes up four times.^[3] OpenStax's example: a 25 W bulb made for 120 V, connected to 240 V, glows very brightly for a moment and then burns out.^[3]

The calculator's equivalent resistance

For any device, E ÷ I gives the resistance a plain resistor would need to draw the same current at the same voltage. For the 60 W bulb in Example 1, that is 120 V ÷ 0.5 A = 240 Ω. This is the bulb's resistance only while it is hot and lit. A filament's resistance rises as it heats, so a cold filament has a much lower resistance and draws a surge of current at the instant it is switched on; OpenStax's car headlight draws about 34 A when cold against 2.5 A when hot.^[3]

The combined wheel

Put PIE and Ohm's law together and you get twelve formulas, three for each of P, I, E and R. Many reference cards print them as a wheel (Figure 2). You do not need to memorize it: every slice comes from P = I × E and E = I × R by the same substitutions shown above. The square-root (√) forms simply undo a square: if P = I²R, then I² = P ÷ R and I = √(P ÷ R).

7. Energy, kilowatt-hours and cost

Because energy = power × time, any power unit times any time unit is an energy unit. Utilities bill in the kilowatt-hour (kWh): the energy used by a 1,000 W load running for one hour.^[3] One kilowatt-hour is 3.6 million joules (1,000 J per second × 3,600 seconds).^[3] A smaller cousin, the watt-hour (Wh), is one-thousandth of that; battery packs are often rated in watt-hours.

To estimate what something costs to run, Virginia Tech's extension service gives a three-step recipe:^[9]

1. Daily energy in kWh = watts × hours used per day ÷ 1,000.
2. Multiply by the number of days it is used.
3. Multiply by your utility's price per kWh.

Example: the 60 W bulb, 5 hours a day

This example uses a rate of $0.18 per kWh. That is an example rate, close to the U.S. average residential price of 18.31 cents per kWh that the U.S. Energy Information Administration reported for July 2026; prices vary widely by state, from under 13 cents in Louisiana to 48 cents in Hawaii that month.^[10] Use the price on your own bill.

• Daily energy: 60 W × 5 h ÷ 1,000 = 0.3 kWh.
• Daily cost: 0.3 kWh × $0.18 = $0.054, about 5 cents.
• Monthly energy (30 days): 0.3 kWh × 30 = 9 kWh; monthly cost: 9 kWh × $0.18 = $1.62.

OpenStax notes that a 15 W compact fluorescent bulb gives about the same light as a 60 W incandescent.^[3] At a quarter of the power, the same 5 hours a day costs a quarter as much: 2.25 kWh a month, about $0.41 at the example rate. The calculator's cost row does this arithmetic for you.

Two cautions. A nameplate wattage is not always what a device draws: many appliances have a range of settings, and the actual power depends on the setting in use.^[9] Also, a refrigerator's compressor cycles on and off, so Virginia Tech suggests dividing its plugged-in hours by three. And many devices keep drawing a small "phantom load" when switched off.^[9]

8. Power ratings, heat, fuses and wire

Most of the power that does not become light, sound or motion becomes heat. Every part has a limit to how much heat it can shed, so PIE is the first check in almost any design.^[1]

Choosing a resistor's wattage

Every resistor has a power rating, the most heat it can shed continuously without damage. SparkFun notes that most ordinary resistors are rated from ⅛ W to ½ W, and that larger power resistors are built for watts or more.^[1] Exceed the rating and the resistor overheats; Ohmite lists drifting value, cracking, discoloration and, in severe cases, burning open as results.^[11]

Even inside its rating, a resistor can be too hot to touch. KOA's datasheet for its metal oxide resistors shows the surface rising roughly 100 °C above the surrounding air for the 1 W size at full rated power, and the full rating applies only up to 70 °C ambient; above that the allowed power falls, which is called derating.^[12] Ohmite advises choosing a rating with margin above the calculated power.^[11] A simple habit, used on this site, is to pick a rating at least twice the calculated power. That is a rule of thumb, not a standard; the datasheet is the authority. Article 3, Resistors, covers this in more depth.

Fuses

A fuse is a short piece of wire designed to melt and open the circuit when the current is too high. It goes in series with what it protects, so when it opens, the current stops.^[13]​^[14] Fuses are rated in amperes, so PIE is how you get from a device's wattage to the current a fuse must carry: a 36 W, 12 V accessory draws 36 ÷ 12 = 3 A (Example 3).

A fuse should not run right at its rating. Article 17, Fuses and Circuit Protection, walks through the manufacturer's checklist, including running a fuse at no more than 75% of its rating: a 3 A load points to a fuse rated at least 3 ÷ 0.75 = 4 A (calculated). The fuse's voltage rating must also be at least the circuit voltage. And never replace a blown fuse with a bigger one; find out why it blew.

Wire size

Wires have a little resistance too, so current through them makes heat, at a rate of P = I²R.^[13]​^[14] A thinner wire has more resistance per foot, so it heats more at the same current. The most current a wire can safely carry is its ampacity, set mainly by how hot its insulation can get before it softens or melts.^[13] Because the heat goes with the square of the current, a load that draws twice the current makes four times the heat in the same wire.

This is why the same power is much harder on wiring at low voltage. A 120 W load draws 1 A at 120 V but 10 A at 12 V (calculated), so low-voltage, high-power projects like car accessories need thicker wire. At a beginner level: find the current with I = P ÷ E, choose wire rated for more than that current, and put a fuse that will open before the wire overheats close to the supply. Article 9, DC Circuits, shows how to add up a power budget for a whole project, and its safety section gives example wire ratings.

A note on household AC

Wall outlets supply alternating current (AC), which reverses direction many times a second. The "120 V" and the amps printed on a label are rms (root-mean-square) values, a special kind of average chosen so that, for a simple resistive load such as a heater or a filament bulb, power works out exactly as it would with DC: P = I × E.^[8] For motors and many electronic power supplies, volts times amps (called volt-amperes, VA) can be more than the true power in watts; Article 10, AC Circuits, explains why and introduces the power factor.

9. Safety

• Check the power before you build. Work out P for every resistor and any other part that gets warm. A part pushed past its rating can get hot enough to burn skin, scorch a board or fail.^[1]​^[12] Allow margin.
• Short circuits release enormous power. A short is an accidental low-resistance path across a source. Because P = E² ÷ R, a tiny resistance means a huge power: OpenStax's example is 120 V across a 0.100 Ω short, which works out to 144 kW, enough to melt or ignite nearby material very quickly.^[14] Never short a battery or supply, even a small one.
• Do not overload wires or outlets. Too much current heats the supply wires (P = I²R) and is a fire hazard; this is the job fuses and circuit breakers do.^[14] Add up the watts of everything on one outlet, divide by the voltage, and compare the current with the rating.
• Watch out for switch-on surges. Lamps and motors can draw several times their running current for a moment when switched on.^[3] Fuses for such loads are usually time-delay ("slow-blow") types; see Fuses and Circuit Protection.
• Keep to low voltage while learning. The power formula works the same at 5 V and 120 V, but wall (mains) voltage can drive a dangerous current through your body. Leave mains wiring to qualified people.^[14]

The whole story in brief

• Power (P, watts) is the rate of energy conversion: 1 W = 1 joule per second. Energy is power × time, billed in kilowatt-hours.
• PIE: P = I × E, so I = P ÷ E and E = P ÷ I. E (electromotive force) and V both mean voltage.
• With Ohm's law: P = I²R = E² ÷ R for resistors; PIE itself works for any part.
• Cost: watts × hours ÷ 1,000 = kWh; kWh × your rate = cost.
• Heat: choose resistors rated well above the calculated power, size fuses and wire from the current, and treat short circuits with respect.

For the parts behind these numbers, see Resistors, DC Circuits and Fuses and Circuit Protection; for the other half of the math, see the Ohm's Law helper.

Sources

All links were checked in October 2026. External links open in a new tab.

1. SparkFun Learn, "Electric Power" tutorial (all pages; power ratings): https://learn.sparkfun.com/tutorials/electric-power/all
2. NIST, Guide for the Use of the International System of Units (SI) (SP 811), chapter 4, "The Two Classes of SI Units and the SI Prefixes" (Tables 3 and 5): https://www.nist.gov/pml/special-publication-811/nist-guide-si-chapter-4-two-classes-si-units-and-si-prefixes
3. OpenStax, College Physics 2e, section 20.4, "Electric Power and Energy": https://openstax.org/books/college-physics-2e/pages/20-4-electric-power-and-energy
4. OpenStax, College Physics 2e, section 20.1, "Current": https://openstax.org/books/college-physics-2e/pages/20-1-current
5. OpenStax, College Physics 2e, section 19.1, "Electric Potential Energy: Potential Difference": https://openstax.org/books/college-physics-2e/pages/19-1-electric-potential-energy-potential-difference
6. Tony R. Kuphaldt, Lessons In Electric Circuits, Volume I (DC), chapter 2, "Ohm's Law" (including "Power in electric circuits" and "Calculating electric power"): https://www.ibiblio.org/kuphaldt/electricCircuits/DC/DC_2.html
7. OpenStax, College Physics 2e, section 21.2, "Electromotive Force: Terminal Voltage": https://openstax.org/books/college-physics-2e/pages/21-2-electromotive-force-terminal-voltage
8. OpenStax, College Physics 2e, section 20.5, "Alternating Current versus Direct Current" (rms values and average power): https://openstax.org/books/college-physics-2e/pages/20-5-alternating-current-versus-direct-current
9. Virginia Cooperative Extension (Virginia Tech), "Energy Series: Estimating Appliance and Home Electronic Energy Use" (BSE-301, PDF): https://www.vt.edu/content/dam/pubs_ext_vt_edu/2901/2901-9014/BSE-301.pdf
10. U.S. Energy Information Administration, Electric Power Monthly, Table 5.6.A, "Average Price of Electricity to Ultimate Customers by End-Use Sector, by State" (July 2026 data, released September 24, 2026): https://www.eia.gov/electricity/monthly/epm_table_grapher.php?t=epmt_5_6_a
11. Ohmite, "Common Causes of Resistor Failure and How to Prevent Them" (blog, August 2026): https://www.ohmite.com/blog/2026/08/24/common-causes-of-resistor-failure-and-how-to-prevent-them
12. KOA Speer, MOS/MOSX "Reduced Size Metal Oxide Power Type Leaded Resistor" datasheet (derating curve and surface temperature rise chart; PDF): https://www.koaspeer.com/pdfs/MOS.pdf
13. Tony R. Kuphaldt, Lessons In Electric Circuits, Volume I (DC), chapter 12, "Physics of Conductors and Insulators" (sections "Conductor ampacity" and "Fuses"): https://www.ibiblio.org/kuphaldt/electricCircuits/DC/DC_12.html
14. OpenStax, College Physics 2e, section 20.6, "Electric Hazards and the Human Body": https://openstax.org/books/college-physics-2e/pages/20-6-electric-hazards-and-the-human-body