How this instrument works
Power is the rate at which a circuit converts electrical energy into something else — heat, light, motion — measured in watts, one watt being one joule per second. This instrument computes it the most direct way there is: read the voltage across a load, read the current flowing through it, multiply the two. No resistance value has to enter the calculation at all, which matters more than it sounds, because plenty of real loads do not hold a fixed resistance to plug in. A motor's winding resistance drifts as it spins up; an LED driver holds current constant while its forward voltage wanders with temperature. For those, a voltmeter and an ammeter tell the truth that a single resistance figure cannot.
P = V × I is one identity in a family of three, and Ohm's law is the hinge between them. Swap V for I·R inside the power formula and the current appears twice: P = I²R, the form an electrician reaches for when a wire's resistance and its carried current are both known and the question is how many watts that copper sheds as heat. Swap I for V⁄R instead and voltage appears twice: P = V²⁄R, useful when a component's resistance and its supply voltage are the fixed knowns — an incandescent filament's rated hot resistance at its rated voltage, for instance. All three forms describe the same watts; they differ only in which two quantities you happened to measure first.
The substituted forms carry a hidden assumption the direct one does not: that Ohm's law actually holds for the component in question, meaning its resistance stays put while voltage and current move. That is a fair bet for a copper conductor or a wirewound heater element, and a bad one for a diode, a battery under load, or a switch-mode charger, where resistance is not a constant but a consequence of whatever the device happens to be doing at that instant. Measuring V and I directly, as this page does, sidesteps the question entirely — it returns the correct wattage whether or not the load is well-behaved enough to call ohmic.
- Enter Voltage — the potential difference actually measured across the load, not a battery's printed nominal figure. Millivolts and volts share the unit menu.
- Enter Current — the amperage the load draws while running, read from a clamp meter or an inline ammeter. Switch to milliamps for small electronics.
- Read Power in watts, updated instantly; switch its unit to milliwatts for low-draw sensors and indicator circuits.
- Before trusting the result on a live circuit, check the Current field against your fuse or breaker rating — watts alone will not tell you whether a cable can carry the amperage that produced them.
Worked example — a 12 V accessory drawing 3 A
A 12-volt driving-light bar on a van or utility trailer is rated at 3 amps. Enter 12 into Voltage and 3 into Current: Power returns 36, in watts. That is 12 × 3 = 36.0 W exactly — the figure a spec sheet would call it, and the number that actually decides whether a 5-amp fuse and 16-gauge wire are enough, since a fuse is rated in amps but a cable's heating and a battery's drain are best judged in watts.
That wattage translates directly into runtime against a battery's stored energy: a 50 Ah battery at 12 V nominal holds roughly 600 Wh, so a steady 36 W draw would run it down in about 16.7 hours before any allowance for depth of discharge. Move the same fixture to a 24-volt system and, because its internal driver holds current at a constant 3 A regardless of supply, power rises with voltage alone: 24 × 3 = 72 W, double the draw for the identical current reading — proof that watts track voltage even when amperage does not budge.
Questions
Why enter voltage and current instead of resistance?
Because plenty of real loads don't have one fixed resistance to enter. A voltmeter and an ammeter measure what a circuit is actually doing right now, while a single resistance figure only describes components — a fixed resistor, a length of wire — whose resistance genuinely stays constant. Motors, LED drivers, and batteries under load all shift their effective resistance as conditions change, so multiplying two live meter readings gives the correct wattage every time; computing from an assumed resistance does not.
How is P = V × I related to P = I²R and P = V²/R?
They are the same physical quantity reached three ways. Ohm's law says V = I·R, so substituting that into P = V·I in place of V gives P = I²R, and substituting I = V⁄R in place of I gives P = V²⁄R. Which form is convenient depends only on which two quantities you already know — this instrument assumes you know voltage and current directly, the common case when working from meter readings rather than a component datasheet.
My 12 V device drew more power than its nameplate wattage. Why?
Check the actual voltage first — a lead-acid battery under load commonly sags to 11.8–12.2 V rather than holding a clean 12.0 V, and current can spike above a nameplate's steady-state figure during motor start-up or an LED driver's inrush. Enter the values a meter shows in the moment, not the labels printed on the case, and a 36 W-style result will match reality rather than the spec sheet's best-case number.
What happens if current is zero?
Power returns zero, regardless of voltage. A load sitting at 12 V with no current flowing — a switch left open, a blown fuse — is converting no energy at all, because power needs both a voltage difference and a moving charge to mean anything; either one alone does no work.
Does this formula work for AC as well as DC?
As an instantaneous relationship, yes — but for alternating current you need RMS voltage and RMS current, and the result is only true watts if the load is purely resistive. A motor, ballast, or switch-mode supply pulls current out of step with voltage, so V × I there gives apparent power in volt-amperes, not real watts; that gap is what a separate power-factor calculation accounts for.
If I only know resistance and one of voltage or current, can I still use this page?
Find the missing value first with Ohm's law — V = I·R or I = V⁄R — then bring both V and I here. This page deliberately keeps to the direct multiplication so the working stays visible: one multiplication, one answer, with nothing hidden inside a resistance substitution.