Electric Power & Heating

In a time dt a charge dq = I dt passes through an element, and every coulomb of it falls through the potential difference V, giving up energy V dq. So the energy converted is dU = V I dt, and the rate is P = dU/dt = I V. Check it in units: 1 A × 1 V = 1 C s⁻¹ × 1 J C⁻¹ = 1 J s⁻¹ = 1 W. Nothing in that argument mentioned resistance, so the relation is completely general — a 12 V motor drawing 4.0 A converts 48 W whether it is spinning freely, lifting a load, or stalled and glowing. The direction is set by which terminal the conventional current enters. If it enters at the higher potential, charge is losing energy inside the element and the element absorbs power; if it leaves at the higher potential, the element is pushing charge uphill and supplying power. That is the whole difference between a battery being discharged and the same battery being charged.

For an element that obeys V = I R you can eliminate either variable. Substituting V gives P = I (I R) = I² R; substituting I gives P = (V/R) V = V²/R. These are not two different powers competing for your attention, they are one power written twice. The apparent contradiction — I²R grows with R while V²/R falls with R — is settled by asking what the circuit holds fixed. Drive 2.0 A through 6.0 Ω and it dissipates 24 W; drive that same 2.0 A through 12 Ω and it dissipates 48 W. Now hold the voltage at 12 V instead: the 6.0 Ω takes 24 W and the 12 Ω takes only 12 W. Same two resistors, opposite ranking, because a series chain shares current while a parallel bank shares voltage. So the first question in any heating problem is not which formula looks tidier, it is which quantity the connection forces the components to share.

Watts alone say nothing about how much energy moved; you need the interval too. In general W = ∫₀ᵗ I V dt, which collapses to W = P t only while P is steady. Take a 20 Ω resistor carrying a ramped current I(t) = 0.50 t amperes for 6.0 s. Then P(t) = I² R = 20 × 0.25 t² = 5.0 t² watts, rising from zero to 180 W at the end. The energy is W = ∫₀⁶ 5.0 t² dt = 5.0 × 6.0³/3 = 360 J. Grabbing the final power and multiplying by 6.0 s would have claimed 1080 J, three times too much; using the initial power would have claimed nothing at all. Dividing back, the mean power is 360 J ÷ 6.0 s = 60 W, exactly one third of the peak, which is what averaging a quadratic should give. Utility bills use the same integral in different clothes: 1 kW h is 1000 W × 3600 s = 3.6 × 10⁶ J.

A loop with a source of emf ε and internal resistance r driving an external R obeys ε = I (R + r). Multiply every term by the current and each one becomes a power: ε I = I² R + I² r. With ε = 12.0 V, r = 0.50 Ω and R = 5.5 Ω the current is 2.00 A, so the chemistry converts 24.0 W in total, of which 2.0 W stays inside heating the cell and 22.0 W reaches the load. The efficiency of the delivery is P_load/Pₜₒₜₐₗ = R/(R + r) = 5.5/6.0 = 91.7%. There is a cross-check worth turning into a habit: the terminal voltage is V = ε − I r = 12.0 − 1.0 = 11.0 V, and applying P = I V at the terminals gives 2.00 A × 11.0 V = 22.0 W, the same load power by a different route. When a power audit refuses to balance, one of the currents or drops in the network is wrong.

Write the load power as a function of the load alone: P_load = ε² R/(R + r)². Differentiate with the quotient rule; the numerator gives ε²[(R + r)² − 2R(R + r)] = ε²(R + r)(r − R), so dP/dR = ε²(r − R)/(R + r)³, which vanishes at R = r and changes there from positive to negative. With ε = 12.0 V and r = 0.50 Ω, matching at R = 0.50 Ω drives I = 12.0 A and delivers 72 W to the load — more than three times the 22.0 W above. But the cell now dissipates 72 W internally as well, so the efficiency has collapsed to 50% and the cell is cooking itself. Go the other way, making R large, and the efficiency approaches 100% while the delivered power approaches zero. A transmitter matched to its antenna accepts that 50% because the scarce thing there is signal power, not energy; power distribution does the opposite, raising the voltage so that the current, and with it the I²R loss in the lines, falls.

The number stamped on a resistor is the power it can shed before its temperature runs away, and it translates into both a current ceiling and a voltage ceiling. For a 0.25 W, 470 Ω part, P = I²R gives Iₘₐₓ = √(0.25/470) = 0.023 A, and P = V²/R gives Vₘₐₓ = √(0.25 × 470) = 10.8 V. Those are not two independent ceilings: Ohm's law ties them together, since Vₘₐₓ = Iₘₐₓ R = 0.023 A × 470 Ω = 10.8 V, so the part meets both at the same instant and you simply use whichever quantity the circuit hands you. Designers then derate, often running at half the rating, because the quoted figure assumes a lone part in still air at room temperature, and a part buried in a dense board has neither. The same reasoning sizes a cable, since 4.0 W lost in a supply lead is 4.0 W heating its insulation. To turn a wattage into a temperature use P t = m c ΔT: a 2.2 kW kettle raising 0.50 kg of water by 80 K needs 0.50 × 4180 × 80 ÷ 2200 = 76 s if nothing escapes. And because resistivity rises with temperature, a heavily loaded part heats, its R shifts, and the heating shifts with it.