Electrical Power

P = VI is the general statement; the other two forms carry conditions

Move a charge q through a potential difference V and the energy exchanged is qV. Divide by the time taken and you have divided the charge by time as well, which is current. So the power delivered to a two-terminal element is P = VI. Nothing in that derivation assumes anything about what the element is. It holds for a resistor, a motor, an LED, a battery being charged, or an entire appliance treated as a black box. What it does demand is bookkeeping discipline: V is the voltage across the two terminals of the element you are asking about, and I is the current through that same element.

If the element is a resistance obeying Ohm's law, substitute V = IR to get P = I squared R, or substitute I = V/R to get P = V squared over R. Both are algebraic consequences of P = VI, and both inherit the condition that came with them: there has to be a resistance R, and the V and I in the expression have to be that resistance's own voltage and current.

Which form to reach for is a question about what you already know. Use I squared R when current is the quantity you know and share, which is the case along a series path or down a transmission line. Use V squared over R when you know the voltage sitting directly across the element, which is the case for something wired straight across a supply. Use P = VI whenever the element is not a resistance at all, because then there is no R to put into the other two.

The standard error is feeding in the wrong V or the wrong I. Take a 12 V source driving a 10 ohm resistor in series with a 30 ohm resistor. The current is 12/40 = 0.3 A, so the 10 ohm resistor dissipates I squared R = 0.9 W and the whole circuit draws VI = 3.6 W. Reach for V squared over R using the source voltage and you get 144/10 = 14.4 W: sixteen times too large, and larger than the power the entire circuit is drawing. The formula did not fail. It was handed a voltage that never appeared across that resistor.

Power is a rate; energy is the accumulated total

The watt is the SI coherent derived unit of power and equals one joule per second. The joule is the unit of energy. That single sentence settles most of the confusion in this topic: a watt answers how fast, a joule answers how much.

The kilowatt-hour is where students lose the thread, because it sounds like a rate and is not one. It is a kilowatt sustained for an hour, so it is power multiplied by time, which is energy. Executing the conversion makes the point concrete: 1 kW multiplied by 1 h is 1000 W times 3600 s, and since a watt is a joule per second the seconds cancel to leave 3.6 million joules, or 3.6 MJ. Utility meters accumulate energy, which is why the bill is in kilowatt-hours and not kilowatts. The expression 'kilowatts per hour' is not a smaller version of this; it is a rate of change of power, and it is almost never what someone writing it means.

A space heater whose nameplate reads 1,500 W has a power of 1.5 kW while it is running. That number alone does not tell you what it costs. Run it four hours a day for thirty days and the energy is 1,500 W times 4 h times 30, which is 180,000 W-h, or 180 kW-h, or 6.48 times 10 to the 8 joules.

Resistive heating, and why bulk transmission runs at high voltage

Charge moving through a resistance dissipates energy as heat. That is Joule heating, and in conductors it is the loss mechanism that dominates. A transmission conductor is not there to make heat, but it has resistance and it carries current, so it makes heat anyway, and the amount is I squared R.

The squared dependence on current is what drives the design of the grid. For a fixed power delivered to a load, current and voltage trade off directly: I = P/V. Double the transmission voltage and the current halves; halve the current and the I squared R loss falls to a quarter. The loss at fixed delivered power therefore scales as one over V squared, which is a much steeper lever than intuition suggests.

Put numbers on it. Suppose a line must deliver 5.0 MW and its conductors have a total resistance of 2.0 ohm. At 15 kV the current is 5.0 x 10^6 / 15,000 = 333.333 A, and the loss is 333.333 squared times 2.0 = 222,222 W, about 222 kW, or 4.4 percent of the delivered power. At 150 kV the current is 33.3333 A and the loss is 2,222 W, about 2.2 kW, or 0.044 percent. Ten times the voltage, one tenth the current, one hundredth the loss. Note the idealization being made: the 15 kV case means the sending end must actually supply about 5.22 MW, the line resistance was given rather than computed from conductor geometry, and reactance has been ignored entirely.

As a reality check on the scale of grid losses, EIA estimates that annual United States transmission and distribution losses averaged about 5 percent of the electricity transmitted and distributed from 2018 through 2022. That published figure is an aggregate. It is not broken down by mechanism, and transformer core losses and other equipment losses sit inside it alongside conductor heating, so do not read it as the I squared R share.

This is educational material about why the physics points the way it does. It is not guidance for electrical installations. Real power systems and real wiring are governed by codes and require a licensed professional; nothing here is a procedure.

Ratings, derating, and where the missing watts go

A component's power rating is a thermal limit wearing electrical clothing. A 0.25 W resistor is not forbidden from dissipating more than a quarter of a watt by any law of physics; it is telling you that at a stated ambient temperature it can shed that much heat without its own temperature climbing past what its materials tolerate. Above that ambient temperature the allowable dissipation falls, which is why datasheets carry a derating curve rather than a single number.

Derating means deliberately operating a part below its nameplate maximum so that stress, and therefore degradation rate, stays low. NASA Goddard's EEE-INST-002 is one documented example of how this gets specified: it requires the designer to derate parts against a table for each part category, and its resistor table applies a power derating factor of 0.6 to most listed styles below a stated rated-power temperature, with linear derating to zero allowable power at a higher temperature. On that rule a 0.25 W part would be held to 0.15 W. Do not generalize the number. Derating practice varies by industry, by mission, and by what happens when the part fails; spaceflight practice is deliberately conservative and is not a universal design rule.

Efficiency is the useful output compared with the input. Feed 1,000 W into a device that is 92 percent efficient and 920 W leaves as the output you wanted, while 80 W leaves as heat. Energy is not lost in the sense of ceasing to exist; 'loss' is shorthand for converted into something you were not trying to make. In a motor that missing power is winding I squared R heating, core losses, and friction. Because the number depends entirely on where the boundary is drawn and what is counted as useful, an efficiency figure is only meaningful next to a statement of what went in and what came out.

Batteries: ampere-hours are charge, watt-hours are energy

An ampere-hour is one ampere flowing for one hour, which is 3,600 coulombs. It is a quantity of charge. On its own it says nothing about how much energy a battery holds, because energy requires the voltage that charge falls through. The FAA states the conversion directly in its packing guidance: watt-hours are the battery voltage multiplied by the amp hours.

The distinction has teeth. A 4,500 mAh cell at a nominal 3.85 V stores 4.5 A-h times 3.85 V = 17.325 W-h. The same 4,500 mAh at a nominal 3.7 V stores 16.65 W-h. Identical capacity label, different energy, purely because the voltage differs. A 20,000 mAh pack built from cells at a nominal 3.7 V stores 20 A-h times 3.7 V = 74 W-h, which is why it sits under the FAA's 100 W-h ceiling for a spare lithium-ion battery in carry-on baggage even though the milliamp-hour number on the case looks enormous. Regulators write the limit in watt-hours precisely because energy, not charge, is what can be released in a failure.

Two honest caveats. A nominal voltage is a label, not a measurement: real cell voltage sags under load and drifts with state of charge, so the watt-hour figure computed this way is a rating rather than a guaranteed delivery. And this is educational material about units, not guidance for building, charging, or handling battery systems, which is specialized safety-critical work.

AC at survey depth: RMS, and real versus apparent power

In an AC circuit the voltage and current vary through every cycle, so the instantaneous power varies too, and a single number has to be defined before P = VI means anything. That number is the root mean square: square the waveform, average the square over a cycle, take the square root. RMS is defined that way for a reason. For a resistive load, using RMS values in the DC expressions returns the correct average power, so a 120 V rms supply heats a resistor exactly as a 120 V DC supply would.

For a sinusoid the RMS value is the peak divided by the square root of two. A 120 V rms supply therefore peaks at 120 times 1.41421 = 169.706 V, which is why insulation and semiconductor ratings are set by the peak while power arithmetic uses the RMS. A 1,500 W resistive heater on that supply draws I = P/V = 12.5 A rms and presents R = V squared over P = 14,400/1,500 = 9.6 ohm. Both derived forms agree: 12.5 squared times 9.6 = 1,500 W, and 14,400/9.6 = 1,500 W. The instantaneous power still swings between zero and 3,000 W, twice the average.

Once the load is not purely resistive, current and voltage stop peaking together and the accounting splits three ways. Apparent power is the plain product of RMS voltage and RMS current, expressed in volt-amperes; it comprises both of the parts below. Real, or active, power is the part that actually gets converted, measured in watts, and it is what an energy meter accumulates. Reactive power, in volt-amperes reactive, is the portion that establishes and sustains the electric and magnetic fields in equipment such as motors and transformers; it flows back and forth rather than being consumed. The power factor is the ratio of real to apparent power, written as cos of the phase angle for the sinusoidal case.

Why engineers care: apparent power sets the current, and current sets the I squared R heating in every conductor between the generator and the load. A load pulling 20 A at 240 V presents 4,800 VA; at a power factor of 0.80 it is doing 3,840 W of real work. Delivering that same 3,840 W at unity power factor would need only 16 A, cutting the loss in a 0.5 ohm feeder from 200 W to 128 W. State this carefully: the cos-phi picture assumes sinusoidal voltage and current, and with the distorted currents drawn by many electronic loads the full accounting needs more machinery than a single phase angle. That belongs to a circuits course, not here.