Every symbol has to belong to the same element

Put a $4\ \Omega$ resistor in series with an $8\ \Omega$ resistor across $12$ V, and ask for the power in the $4\ \Omega$ resistor. The number $12$ is on the page, so $(\Delta V)^2/R = 144/4 = 36$ W is easy to write and badly wrong: $12$ V is across the pair, not across the $4\ \Omega$ resistor.

Work from something that resistor owns:

1. Series current: $I = 12/12 = 1$ A. That passes through the $4\ \Omega$ resistor, so it is genuinely its own.
2. $P = I^2R = (1)^2(4) = 4$ W.
3. Cross-check with its own potential difference: $\Delta V = IR = 4$ V, so $(\Delta V)^2/R = 16/4 = 4$ W. Same answer.

The rule is short: match every symbol in a power formula to one element before any numbers go in. If a quantity belongs to the whole circuit, it does not belong in that element's formula.

Ranking brightness by resistance alone gets half of all circuits wrong, because the answer flips depending on the connection.

1. In series, the elements share a current. Use $P = I^2R$, and the larger resistance dissipates more.
2. In parallel, the elements share a potential difference. Use $P = (\Delta V)^2/R$, and the smaller resistance dissipates more.

The classic demonstration: wire a $60$ W lamp and a $100$ W lamp in series across one source, and the $60$ W lamp is the brighter one. A higher wattage rating at a fixed household voltage means a smaller resistance, and in series that is the loser.

Which brings up the other half. A wattage printed on a bulb applies only at the potential difference it was rated for. Move it to another circuit and the rating is a fact about its resistance, not about its power there.

Power is energy per unit time, so a $60$ W lamp converts $60$ J each second. To get an amount of energy, multiply:

$$E = P\,\Delta t = (60\ \text{W})(7200\ \text{s}) = 4.32\times10^{5}\ \text{J}.$$

The per-second is already built into the watt, so handing it a duration has to be a multiplication. Dividing by the time is the reflex to watch for, and answering an energy question in watts is the other.

The kilowatt-hour is the same idea in units built for a utility bill: a kilowatt sustained for an hour, which is

$$1\ \text{kWh} = (1000\ \text{W})(3600\ \text{s}) = 3.6\times10^{6}\ \text{J}.$$

Despite the "per hour" sound of the name, it is an energy, not a rate. Checking the unit against what the question asked for catches every version of this.

It is worth naming the mechanism, because it ties this topic back to the first one in the unit. A charge $q$ crossing a potential difference $\Delta V$ loses energy $q\,\Delta V$. Divide by the time and you get the rate:

$$P = \frac{q\,\Delta V}{\Delta t} = I\,\Delta V.$$

In a resistor, that energy goes into the lattice as internal energy and the component warms. In a motor, most of it becomes mechanical work. In a battery being charged, it goes into chemical energy. The formula is the same; where the energy ends up depends on the device.

One consequence worth keeping: a real battery has internal resistance $r$, and $I^2r$ is dissipated inside the battery. That is energy the external circuit never sees, and it is why a battery under heavy load gets warm.