How this instrument works
One volt is joules per coulomb; one ampere is coulombs per second. Multiply those two and coulombs cancel, leaving joules per second — the watt. That cancellation is why P = V·I carries no fudge factor and no material property anywhere inside it: one term counts how much energy each unit of charge surrenders, another counts how much charge passes per second, and their product is the rate of energy conversion. Nothing is stored. Your 2 kW kettle does not contain 2 kW; it turns electrical energy into heat at two thousand joules every second for as long as it stays switched on, and owes you nothing once you unplug it.
James Watt never measured his own unit. He sold steam engines, and invented horsepower — 33,000 foot-pounds per minute — as sales ammunition against actual horses. Attaching his name to the SI unit of power came sixty-odd years after his death, proposed by C. William Siemens in an 1882 address to Britain's association for advancing science, and written into SI at the eleventh General Conference on Weights and Measures in 1960. Joule had already pinned down the electrical case in 1841, showing that heat raised in wire tracks I²R. Metrology later closed the loop Watt himself would have relished: the Kibble balance, devised at Britain's National Physical Laboratory in 1975, holds mechanical power mgv equal to electrical power VI, and since May 2019 that equality is how the kilogram gets realised from Planck's constant. Volts multiplied by amperes now underwrite mass itself.
P = V·I is exact for instantaneous values, a qualifier that does real work. On alternating current, multiplying RMS volts by RMS amperes yields apparent power in volt-amperes, not watts; only where voltage stays in step with current do those two figures agree. Motors, fluorescent ballasts, cheap switch-mode supplies — all drag current out of phase, pulling genuine power down to S·cos φ. Data-centre gear carries kVA ratings for precisely that reason. Treat this sheet as DC arithmetic, or as AC at unity power factor, and it will not mislead you.
- Enter Voltage — potential difference actually present across your load, not your supply's open-circuit rating. Millivolts and kilovolts sit on its unit menu.
- Enter Current, meaning what your load genuinely draws. Milliamps suit signal work; kiloamps suit welding plant and substation arithmetic.
- Read Electrical power, returned to four significant figures and switchable to kW, MW, or horsepower.
- Working on AC? If your load is reactive, read that answer as volt-amperes and scale it by power factor to get watts.
- Sanity-check against your nameplate. Figures well above what any plug, fuse, or cable is rated for usually mean one mistyped decimal point.
Worked example — one volt, one ampere, one watt
Picture your bench source trimmed to 1.000 V, feeding whatever load settles at 1.000 amperes. Enter 1 into Voltage and 1 into Current; Electrical power returns 1. Nothing rounds and nothing approximates, because the watt is defined as exactly this case — one joule of energy handed across every second.
Scale outward, then intuition follows quickly. Run that 1 W for an hour; it moves 3600 joules, billed as 0.001 kWh. Run it for one year and it swallows 8.76 kWh, roughly two pounds' or dollars' worth, which is why standby draws of one watt or two, spread across every gadget in one house, are worth arguing about. Multiply both inputs by ten instead: power climbs hundredfold, so 10 V at 10 amperes gives 100 W, one bright old-fashioned filament bulb.
Everyday hardware is that same unit case with its decimal point shifted. European kettles pulling 13 amperes from 230 V convert 2990 W. North American 15-ampere branch circuits at 120 V top out near 1800 W. USB-C chargers negotiating 20 V at 5 amperes push 100 W into your laptop. Each one is 1 × 1, relocated.
Questions
Why does multiplying volts by amperes give watts?
Because charge cancels. A volt is one joule per coulomb — energy carried by each unit of charge. An ampere is one coulomb per second — how much charge moves. Multiply (J/C) by (C/s), coulombs vanish, and J/s remains, which is a watt. No experimental constant enters that step anywhere, so it holds for any element you point it at: resistor, motor, LED, battery, or a whole building's incomer.
Does this give watts or volt-amperes on AC?
Multiply RMS voltage by RMS current on alternating current, then what you obtain is apparent power S, measured in volt-amperes. Real power in watts is S·cos φ, where φ is phase angle between them. For heaters or filament lamps φ sits near zero, so both figures agree. For motors running at 0.8 power factor, 230 V × 10 amperes reads 2300 VA yet delivers only about 1840 W. Generators, transformers, UPS units — all carry kVA ratings, because their true limit is current-driven heating, which cares nothing about phase.
Should I enter peak or RMS values for AC?
RMS, always. Root-mean-square values are defined so that resistive heating matches an equivalent DC figure, which is exactly what makes power arithmetic come out right. European mains at 230 V RMS peaks near 325 V; feeding that peak in overstates power by 41%, and multiplying peak volts by peak amperes overstates by a factor of two. Meters and nameplates quote RMS unless they explicitly say otherwise.
What does a negative power result mean?
Sign is bookkeeping convention rather than physics. Under passive sign convention — current entering at whichever terminal counts as positive for voltage — positive answers mean that element absorbs energy, negative ones mean it supplies energy. Batteries read negative while discharging, positive while charging. Rooftop solar exporting to the grid shows negative power at your meter. Enter magnitudes here, then attach sign yourself, from whichever way energy is travelling.
How do I apply this to a three-phase supply?
Insert the √3 factor when working from line-to-line voltage: P = √3 · V_LL · I_line · cos φ. Working per phase instead, take V_phase × I_phase, then multiply by three — same result, since line-to-line voltage is √3 times phase voltage in star connection. One 400 V three-phase motor drawing 10 amperes per line at 0.85 power factor comes to roughly 5.9 kW. Dropping that √3, then calling it 4 kW, is both common and expensive when sizing cable.
How do watts relate to kilowatt-hours on my bill?
Watts are a rate, kilowatt-hours a quantity — multiply power by time. A 2000 W appliance running 90 minutes uses 2 kW × 1.5 h = 3 kWh, and one kilowatt-hour is 3.6 megajoules in SI terms. Utilities bill energy because energy is what makes a generator burn fuel. Industrial tariffs usually add a separate demand charge in kW as well, which prices something different: how much capacity a network must hold in reserve for your worst quarter-hour.