Electrical Power (P = I²R)

This is the heat form of electrical power, and the square is the whole point of it. Start from $P = VI$, substitute Ohm's law for the voltage across the resistance, $V = IR$, and you get $P = I^2 R$. Because the current appears twice, the heat does not track the load: it tracks the square of the load. Double the current and a conductor dissipates four times the heat; triple it and nine times. Nothing else in ordinary wiring punishes a modest overload so hard, and it is the reason a circuit that runs warm at rated current runs dangerously hot at 150% of it.

Take a 30 m branch circuit run in 12 AWG copper. The conductor is about 3.31 mm², copper's resistivity is $1.68 \times 10^{-8}$ Ω·m, so each metre is roughly 5.1 mΩ, and the current has to go out and come back, so the loop is 60 m and about 0.30 Ω. At 15 A the copper dissipates $15^2 \times 0.30 = 68\ \text{W}$, spread along the run, and drops $15 \times 0.30 = 4.6\ \text{V}$ of the supply before it ever reaches the load. Drop the current to 5 A and the loss falls to 7.5 W, not to a third: that is the square doing its work.

James Joule established the law in 1841 by immersing coils in water and measuring the temperature rise, which was also his route to the mechanical equivalent of heat and thence to the first law of thermodynamics. The same relation is the reason the grid transmits at hundreds of kilovolts. Delivered power is $VI$, so raising the voltage a hundredfold cuts the current a hundredfold for the same power, and cuts the line loss by ten thousand. Every transformer between a generating station and a house exists to move a fixed quantity of watts into a lower current, purely so that this equation returns a smaller number.

Three traps, in rising order of consequence. First, $R$ is the resistance of the thing dissipating the heat, and $I$ is the current through that same thing: mixing the conductor's resistance with the load's current is fine only because they are in series, and on a branched circuit it is not fine at all. Second, on AC the current must be RMS. A peak reading gives twice the power for a sine wave, and using it is one of the most common ways to double an answer without noticing. Third, and the one that bites in the field: this equation uses resistance, and on an AC circuit the opposition to current is impedance. A long run of steel-armoured cable, a coil, or a motor feeder has reactance as well as resistance, and the voltage drop computed from DC resistance alone will understate the real drop. The heat, though, still comes only from the resistive part: reactance stores energy and hands it back, so it moves voltage around without ever warming the copper.