Worked examples: LED strip and a resistor that won't burn

Two problems, two forms of the power law. First a question about how much a running LED strip actually costs you; second, a question about whether a resistor you picked will survive being switched on.

Example 1 — LED strip and the weekly bill

A 12 V LED strip draws 0.8 A. You run it six hours a night. At £0.30 per kWh, what does a week of use cost?

1. Given. $V = 12\,\text{V}$, $I = 0.8\,\text{A}$, runtime $t = 6 \cdot 7 = 42\,\text{h}$, tariff $\pounds 0.30/\text{kWh}$.
2. Equations. $P = V \cdot I$ for the instantaneous draw, then $E = P \cdot t$ for weekly energy.
3. Substitute.
   $$P = 12\,\text{V} \cdot 0.8\,\text{A} = 9.6\,\text{W}$$
   $$E = 9.6\,\text{W} \cdot 42\,\text{h} \approx 0.40\,\text{kWh}$$
4. Result. The strip draws $9.6\,\text{W}$ and costs roughly $0.40 \cdot \pounds 0.30 = \pounds 0.12$ per week — pennies. Tiny wattages stay tiny; the kWh unit is designed for exactly this kind of everyday arithmetic.

Example 2 — picking a resistor that won't burn

You drop a 6 V rail across a 100 Ω resistor. You have a ¼-watt part on hand. Will it survive?

1. Given. $V = 6\,\text{V}$, $R = 100\,\Omega$. Current isn't stated — use the form of the power law that skips it.
2. Equation. $P = V^{2} / R$ gives power from voltage and resistance alone.
3. Substitute.
   $$P = \frac{(6\,\text{V})^{2}}{100\,\Omega} = \frac{36}{100} = 0.36\,\text{W}$$
4. Result. $0.36\,\text{W} > 0.25\,\text{W}$, so the ¼-watt part will overheat and eventually fail. Use a ½-watt resistor (double the headroom) or pick a larger resistance so less current flows for the same voltage.