Every symbol must belong to the same element

A charge $dq$ crossing an element through a potential difference $V$ gives up energy $dU = V \, dq$. Divide by $dt$:

$$P = \frac{dU}{dt} = V\frac{dq}{dt} = IV, \qquad 1 \ \text{W} = 1 \ \text{J/s} = 1 \ \text{V}\cdot\text{A}.$$

That form is completely general: it works for a resistor, a battery, a capacitor, a motor. For a resistor specifically, Ohm's law lets you eliminate whichever variable you do not know:

$$P = IV = I^2 R = \frac{V^2}{R}.$$

They are algebraically identical, so choosing between them is a matter of what you actually know locally, not of correctness. The useful habit:

1. Series chain? The current is shared, so $I$ is the known local quantity. Use $P = I^2R$.
2. Parallel branches? The voltage is shared, so $V$ is the known local quantity. Use $P = V^2/R$.
3. Both known for the element? $P = IV$ needs no assumption about Ohm's law at all, which matters for a bulb or a motor.

In a resistor this energy leaves as heat. In a motor most of it leaves as mechanical work, which is why $P = IV$ and $P = I^2R$ give different numbers for a motor: the $I^2R$ piece is only the heating in its windings.

Here is the error this topic is built around. A $12$ V battery drives $4 \ \Omega$ and $8 \ \Omega$ in series. Asked for the power in the $4 \ \Omega$ resistor, a student writes

$$P = \frac{V^2}{R} = \frac{12^2}{4} = 36 \ \text{W}. \qquad \text{Wrong.}$$

The formula is right and the $12$ is not the $4 \ \Omega$ resistor's voltage. That resistor sees only its share. Do it properly:

1. $R_{\text{eq}} = 4 + 8 = 12 \ \Omega$.
2. $I = \varepsilon/R_{\text{eq}} = 12/12 = 1$ A, shared by both resistors.
3. $P_4 = I^2R = (1)^2(4) = 4$ W, and $P_8 = (1)^2(8) = 8$ W.
4. Check: $4 + 8 = 12$ W, and the battery delivers $\varepsilon I = 12 \times 1 = 12$ W. The books balance.

Notice that $36$ W exceeds the battery's entire output of $12$ W, which is the sanity check that catches this every time. The total power dissipated can never exceed what the sources deliver.

Notice too which resistor won. In series, the current is common, so $P = I^2R$ says the larger resistance dissipates more. In parallel, the voltage is common, so $P = V^2/R$ says the smaller resistance dissipates more. Those are opposite conclusions from the same physics, and the only thing that decides which applies is which quantity the elements share.

A bulb's brightness tracks the power dissipated in that bulb. Nothing else. So a question about brightness is a question about $P$, and it has to be answered by solving the circuit, not by counting components.

Take identical bulbs of resistance $R$ on an ideal source of emf $\varepsilon$:

1. One bulb alone. $P = \varepsilon^2/R$. Call this full brightness.
2. Two in series. They share one current $I = \varepsilon/2R$, so each dissipates $I^2R = \varepsilon^2/4R$, a quarter of full brightness each. Not half.
3. Two in parallel. Each branch sees the full $\varepsilon$, so each dissipates $\varepsilon^2/R$: both at full brightness, and the battery now supplies twice the current and twice the power.

So "adding a bulb makes everything dimmer" is right for a series addition and flatly wrong for a parallel one. With an ideal source, adding a parallel branch does not touch the original branch at all, because the voltage across it has not changed. A real battery has internal resistance $r$, so its terminal voltage sags a little as the current rises and the original bulb dims slightly; that is a second-order correction, and it is not what the census reasoning was tracking.

The procedure that always works: recompute $R_{\text{eq}}$, recompute the source current, recompute each branch's voltage and current, then compute each bulb's own $P$ and compare it to its own previous $P$.

Watts already contain a per-second. Hand a wattage a duration and you get energy:

$$E = Pt, \qquad E = \int P \, dt \ \text{ if } P \text{ varies.}$$

A $60$ W bulb running for $2$ hours converts

$$E = (60 \ \text{W})(7200 \ \text{s}) = 4.32\times10^{5} \ \text{J} = 120 \ \text{W}\cdot\text{h} = 0.12 \ \text{kW}\cdot\text{h}.$$

The kilowatt-hour is a unit of energy, not power: it is a kilowatt multiplied by an hour, which is $(1000 \ \text{J/s})(3600 \ \text{s}) = 3.6\times10^{6}$ J. Utility bills quote kWh because the customer is buying energy. A battery's mA$\cdot$h rating is the same kind of construction one level down: milliamps times hours is a total charge, and multiplying that by the pack voltage gives its energy.

Two errors to refuse:

1. Dividing by the time. A $60$ W bulb over $2$ hours does not use $30$ of anything. Energy multiplies.
2. Quoting energy in watts. "The bulb used $432$ W last night" names a rate where a total was asked for. The total is $432$ kJ.