Electrical Power, Energy and Heating

Electrical power (rate of energy use), in watts: $P = VI = I^2 R = \dfrac{V^2}{R}$ — three equivalent forms via Ohm's law $V = IR$. Use $P = VI$ always; $P = I^2R$ when current and resistance are known; $P = V^2/R$ when voltage and resistance are known. Expressions like $IR^2$ or $I^2/R$ are NOT power (wrong dimensions).

An appliance rated $(V_0, P_0)$ has a fixed resistance $R = V_0^2 / P_0$. Run at a different voltage $V$, its actual power is $P = V^2 / R$, so $P \propto V^2$ (resistance fixed): halving the voltage gives one-quarter the power.

Electrical energy = power × time. The commercial unit is the kilowatt-hour (kWh): $1\text{ kWh} = 1\text{ kW} \times 1\text{ h}$ = one 'unit'. Energy in kWh = (power in kW) × (time in hours). Cost = (number of units) × (rate per unit).

Joule's law of heating: heat produced $H = I^2 R t = VIt = \dfrac{V^2}{R}\,t$ (joules). It depends on the current, the resistance/voltage, AND the time — all three. For a heating coil on a fixed supply, the temperature rise grows with the supply voltage, the current, and the time the voltage is applied.

At a fixed voltage, power dissipated $P = V^2/R$ — so SMALLER equivalent resistance ⟹ MORE heat. Parallel gives a smaller equivalent than series, so a parallel combination dissipates more. For two equal resistors at the same applied voltage: $R_\parallel = R/2$, $R_\text{series} = 2R$, so $\dfrac{P_\parallel}{P_\text{series}} = \dfrac{R_\text{series}}{R_\parallel} = \dfrac{2R}{R/2} = 4$.