Electrical Power (P = V²/R)

This is the same power as $P = VI$, written for the situation you actually meet at a wall outlet: the voltage is fixed and the resistance is what you choose. Substitute $I = V/R$ into $P = VI$ and the current disappears, leaving $P = V^2/R$. Read it carefully, because the two variables behave in opposite directions. Power rises with the square of the voltage but falls inversely with resistance, so halving the resistance doubles the power, while doubling the voltage quadruples it.

A 1500 W kettle on a 120 V supply must therefore be built with $R = V^2/P = 14\,400/1500 = 9.6\ \Omega$ of element. The same element plugged into 240 V would try to deliver $57\,600/9.6 = 6000\ \text{W}$, four times its rating, which is why travel appliances fail spectacularly rather than gradually. Run the arithmetic the other way for a 1000 W element on 230 V mains and you need about 53 Ω. Notice that the element's resistance is a design choice made to hit a wattage at one particular voltage; the wattage on the label is not a property of the element, it is a property of the element and the supply it was designed for.

The three power forms, $P = VI$, $P = I^2R$ and $P = V^2/R$: are one equation seen from three sides, and picking the right one is mostly about which quantity is being held constant. Devices in parallel across a fixed supply are the $V^2/R$ case: every extra appliance you plug in adds power, because it adds a path, and lowering the effective resistance raises the total draw. Devices in series carrying a common current are the $I^2R$ case. Only $P = VI$ makes no assumption at all, which is why it is the one that still works for motors, LEDs and batteries where Ohm's law does not.

Where this goes wrong. The resistance to use is the resistance at operating temperature, and for anything that glows that is not what an ohmmeter reads on the bench. A tungsten filament's resistance climbs by roughly a factor of fifteen between room temperature and incandescence, so a 60 W lamp measuring 20 Ω cold does not draw 720 W: it draws a large inrush for a few milliseconds and then settles near 240 Ω. The second error is the mains one: use RMS voltage, always. Plugging the 170 V peak of a 120 V circuit into this formula doubles the answer. The third is applying it to a load that is not resistive. A motor at 120 V drawing 5 A is not a 24 Ω resistor; most of that opposition is reactance, which stores and returns energy rather than turning it into heat, and $V^2/R$ with an impedance in the denominator will overstate the watts by exactly the power factor.