Energy and Power in RC Transients

Around the charging loop, ε = iR + q/C, where every term is an energy per unit charge. The same current passes through the source, the resistor and the capacitor leads, so multiply through by i: εi = i²R + (q/C)i. Every term is now an energy per unit time in watts, because V × A = J C⁻¹ × C s⁻¹ = J s⁻¹. Read them in turn. εi is the rate at which the source does work driving charge around the loop, positive whenever conventional current leaves its positive terminal. i²R is the rate at which the resistor converts that energy into internal energy, positive for either current direction. And (q/C)i is v_C dq/dt = d/dt(q²/2C) = dU_C/dt, so the third term is not a loss at all — it is the rate at which stored energy climbs. The balance holds at every instant of the transient, which makes it a sharper tool than any start-to-finish energy statement.

Take ε = 12.0 V, R = 2.0 kΩ and C = 100 μF, so τ = RC = 0.200 s. From an uncharged start, i = (ε/R)e^(−t/τ) and v_C = ε(1 − e^(−t/τ)). At t = 0 the capacitor behaves as a short: i = 6.0 mA, the source delivers εi = 72 mW, and all of it lands on the resistor because v_C = 0. As v_C rises the current falls, the source power falls with it, and the split shifts. The capacitor's share P_C = v_C i = (ε²/R)(e^(−t/τ) − e^(−2t/τ)) is zero at both ends and peaks where e^(t/τ) = 2, at t = τ ln 2 = 0.139 s. There v_C = 6.0 V and i = 3.0 mA, so P_C = 18 mW = ε²/(4R) — and P_R = i²R = 18 mW too, so the source's 36 mW splits evenly at that one instant. Before it the resistor takes the larger share, after it the capacitor does. Late in the transient P_R dies as e^(−2t/τ) while P_C dies only as e^(−t/τ): at t = 2τ they are 1.32 mW and 8.43 mW.

Integrate each power from 0 to t. The source term needs no exponential at all: ∫εi dt = ε∫(dq/dt) dt = εq(t), so W_bat = Cε²(1 − x) with x = e^(−t/τ). The stored energy is U_C = q²/(2C) = ½Cε²(1 − x)². The resistor's is ∫i²R dt = (ε²/R)(τ/2)(1 − x²) = ½Cε²(1 − x²), using τ/R = C. The three satisfy ½Cε²[(1 − x)² + (1 − x²)] = Cε²(1 − x) for every x, which is the conservation check in closed form. Numbers make the lag visible. At t = τ, x = 0.368, so the charge stands at 63.2% of its final value, but the stored energy goes as v_C² and is only ½ × 100 μF × (7.59 V)² = 2.88 mJ, or 40.0% of the final 7.2 mJ. The battery has by then done 9.10 mJ of work and the resistor has already absorbed 6.23 mJ of it — 68%.

Let t → ∞ and the totals are clean: W_bat = Cε², U_C = ½Cε², W_R = ½Cε². For the example, 14.4 mJ supplied, 7.2 mJ stored, 7.2 mJ dissipated. Notice what is missing from the resistor's total: R itself. The cancellation is structural. The resistor's power scales as ε²/R, so halving R doubles it, but τ = RC halves at the same time, and the total heat is the area ε²τ/(2R) under the P_R curve, which is unchanged. Every power doubles, every duration halves. The 50% follows from holding the source at a fixed ε while v_C climbs from 0 to ε: averaged over the transferred charge Cε, the drop across the resistor is exactly ε/2. Charge in N equal voltage steps instead, so the resistor never sees more than ε/N, and the loss falls to ½Cε²/N. That is the principle behind stepwise and resonant charging.

Disconnect the source and close the loop through R with the capacitor at V₀. The loop rule is now q/C + iR = 0, and multiplying by i gives (q/C)i = −i²R: stored energy falls at exactly the rate the resistor heats. With q = CV₀e^(−t/τ) the current is i = dq/dt = −(V₀/R)e^(−t/τ), negative because charge is leaving, so the dissipation is P_R = i²R = (V₀²/R)e^(−2t/τ), and ∫₀^∞ P_R dt = (V₀²/R)(τ/2) = ½CV₀². Everything stored comes back out as heat; none of it returns to the battery. The exponent matters when reading data. Charge, current and voltage decay with time constant τ, but energy goes as the square of the voltage and so decays with τ/2. In the example the capacitor is at half its charge at τ ln 2 = 0.139 s but at half its energy already at (τ/2)ln 2 = 0.069 s. A full charge–discharge cycle therefore converts all Cε² of source work into heat and leaves the capacitor where it started.

Four checks are worth running on any RC energy problem. Units first: V × A = W, Ω × F = s, F × V² = J, so ½Cε² with 100 μF and 12.0 V is 7.2 mJ and nothing else. Endpoints next: at t = 0 the capacitor takes no power because v_C = 0, and as t → ∞ every power goes to zero because i does — a result predicting a nonzero steady power has an error in it. Then the instantaneous balance at one chosen time, which is faster than any integral: at t = 0.139 s the example reads 36 mW = 18 mW + 18 mW. Finally the totals, which must give W_bat = U_C + W_R exactly, since nothing else in the ideal model can hold or shed energy. When a measured audit fails, suspect the model before the arithmetic: a real source has internal resistance r, which claims r/(R + r) of the dissipated half and never reaches the external resistor, and a real capacitor has leakage, series resistance, and dielectric absorption of its own.