University Physics II · Kirchhoff's Rules and RC Circuits · 7.7
Energy and Power in RC Transients
Charging a capacitor through a resistor costs twice what the capacitor keeps. Track the three powers instant by instant, integrate them, and the missing half turns up as heat — whatever resistor you use.
Build the model
Connect the measurement to the mechanism.
Kirchhoff's loop rule is a statement about energy per unit charge. Multiply it by the current — the same current flows through every element of a series RC branch — and every volt becomes a watt: εi = i²R + (q/C)i. The three terms are the rate the source does work, the rate the resistor turns energy into heat, and the rate the capacitor stores it, and they balance at every instant, not merely on average.
Integrating each term across a full charging transient gives Cε² supplied, ½Cε² stored, and ½Cε² dissipated. That split is exact and independent of R, C and τ, because a smaller resistor raises every power and shortens the transient by the same factor. Discharging hands the other half to the resistor as well, so a complete charge–discharge cycle converts Cε² of source energy entirely to heat and leaves the capacitor exactly where it started.
- Simple definition
- Energy accounting for an RC transient: multiply each term of the loop equation by the current to get instantaneous powers, then integrate them to compare the source's work, the resistor's heat, and the capacitor's stored energy.
- Example
- Charging a 100 μF capacitor to 12.0 V through any resistor, the battery does Cε² = 14.4 mJ of work, the capacitor keeps ½Cε² = 7.2 mJ, and the resistor dissipates the other 7.2 mJ.
Energy per charge becomes energy per second. The balance holds at every instant, not just overall.
V × A = W · one current through every series element
PR is always a loss; PC is a rate of storage and changes sign on discharge.
UC = q²/(2C) = ½CvC² · P in W, U in J
Uncharged start. The current sets all three powers, so they rise and fade together.
Ω × F = s · 2.0 kΩ × 100 μF = 0.20 s
The last two add to the first for every x — conservation written in closed form.
x = e(−t/τ) · Wbat = εq(t), no integral needed
Exactly half the source work is dissipated, however large or small the resistor is.
R and τ absent · 12.0 V, 100 μF → 14.4, 7.2, 7.2 mJ
All stored energy returns as heat, and energy decays twice as fast as charge.
Energy time constant τ/2 · half the energy gone by 0.35τ
Turn the loop rule into a power balance
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 vC dq/dt = d/dt(q²/2C) = dUC/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.
Watch the two shares trade places
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 vC = 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 vC = 0. As vC rises the current falls, the source power falls with it, and the split shifts. The capacitor's share PC = vC 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 vC = 6.0 V and i = 3.0 mA, so PC = 18 mW = ε²/(4R) — and PR = 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 PR dies as e(−2t/τ) while PC dies only as e(−t/τ): at t = 2τ they are 1.32 mW and 8.43 mW.
Integrate for a running energy audit
Integrate each power from 0 to t. The source term needs no exponential at all: ∫εi dt = ε∫(dq/dt) dt = εq(t), so Wbat = Cε²(1 − x) with x = e(−t/τ). The stored energy is UC = 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 vC² 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%.
Half the source work is lost, whatever R you choose
Let t → ∞ and the totals are clean: Wbat = Cε², UC = ½Cε², WR = 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 PR curve, which is unchanged. Every power doubles, every duration halves. The 50% follows from holding the source at a fixed ε while vC 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.
Discharging returns the stored half as heat
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) and i = (V₀/R)e(−t/τ), the dissipation is PR = (V₀²/R)e(−2t/τ), and ∫₀^∞ PR 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.
Checks that catch a broken audit
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 vC = 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 Wbat = UC + WR 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.
Change one variable at a time
Make the relationship visible.
Push the cursor out to t = 1.20 s and then sweep R: every curve grows taller and narrower, yet the resistor's heat total holds at 7.20 mJ and its share readout settles near 50%, because the battery's full 14.4 mJ always splits in half.
BATTERY WORK εq7.25 mJ
STORED ½CvC²1.82 mJ
RESISTOR HEAT ∫i²R dt5.42 mJ
HEAT'S SHARE SO FAR74.8 %
Live interpretationBATTERY WORK εq: 7.25 mJ. STORED ½CvC²: 1.82 mJ. RESISTOR HEAT ∫i²R dt: 5.42 mJ. HEAT'S SHARE SO FAR: 74.8 %
Catch the common trap
Explain before calculating.
A 100 μF capacitor, initially uncharged, is charged to 12.0 V by an ideal 12.0 V battery through a series resistor R. How much energy does the resistor dissipate over the whole charging process?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn ideal 9.0 V battery charges an initially uncharged 47 μF capacitor through a 3.3 kΩ series resistor until the current stops. Find the work the battery does, the energy the capacitor ends up holding, and the energy the resistor turns into heat.
- Charge delivered: q = Cε = (47 × 10⁻⁶ F)(9.0 V) = 4.23 × 10⁻⁴ C. Every one of those coulombs crosses the battery's full 9.0 V, so Wbat = εq = Cε² = 3.807 × 10⁻³ J = 3.81 mJ.
- Stored at the end: UC = ½Cε² = ½(47 × 10⁻⁶ F)(9.0 V)² = 1.9035 × 10⁻³ J = 1.90 mJ, exactly half of what the battery paid.
- The rest can only be heat: WR = Wbat − UC = 3.807 mJ − 1.904 mJ = 1.90 mJ, since nothing else in the ideal loop can hold or shed energy.
- Notice R is absent from all three answers. It sets τ = RC = (3.3 × 10³ Ω)(47 × 10⁻⁶ F) = 0.155 s, which fixes how long the transfer takes, not how much of it is lost.
AnswerWbat = 3.81 mJ, UC = 1.90 mJ, WR = 1.90 mJ; τ = 0.155 s changes the timing only.
MediumA 24.0 V ideal battery charges an uncharged 20 μF capacitor through a 5.0 kΩ resistor. At t = 0.150 s find the current, the source power, the resistor power and the capacitor power, and show that they balance.
- τ = RC = (5.0 × 10³ Ω)(20 × 10⁻⁶ F) = 0.100 s, so t = 0.150 s is 1.50τ and x = e(−1.50) = 0.22313.
- i = (ε/R)x = (24.0 V ÷ 5.0 × 10³ Ω)(0.22313) = (4.80 mA)(0.22313) = 1.0710 mA.
- vC = ε(1 − x) = (24.0 V)(0.77687) = 18.645 V, leaving vR = ε − vC = 5.355 V across the resistor.
- Pbat = εi = (24.0 V)(1.0710 × 10⁻³ A) = 2.5704 × 10⁻² W = 25.704 mW.
- PR = i²R = (1.0710 × 10⁻³ A)²(5.0 × 10³ Ω) = 5.735 mW, and PC = vC i = (18.645 V)(1.0710 × 10⁻³ A) = 19.969 mW.
- Balance: 5.735 mW + 19.969 mW = 25.704 mW = Pbat. The two shares sit in the ratio vC : vR = 18.645 : 5.355 = 3.48 : 1, because one current runs through both elements.
Answeri = 1.071 mA; Pbat = 25.704 mW splits into PR = 5.735 mW and PC = 19.969 mW, the capacitor taking 3.48 times the resistor's share.
HardTake ε = 12.0 V, C = 100 μF and R = 2.0 kΩ from an uncharged start. How long until the resistor has dissipated 90.0% of all the heat it will ever produce, and how much of its final energy does the capacitor hold at that moment?
- Final heat: WR(∞) = ½Cε² = ½(100 × 10⁻⁶ F)(12.0 V)² = 7.20 mJ, so 90.0% of it is 6.48 mJ.
- The running heat is WR(t) = ½Cε²(1 − x²) with x = e(−t/τ), so 1 − x² = 0.900 gives x² = 0.100 and x = 0.3162.
- τ = RC = (2.0 × 10³ Ω)(100 × 10⁻⁶ F) = 0.200 s, and t = τ ln(1/x) = (0.200 s)(1.1513) = 0.2303 s = 0.230 s.
- Stored by then: UC = ½Cε²(1 − x)² = (7.20 mJ)(0.6838)² = (7.20 mJ)(0.4675) = 3.37 mJ, only 46.8% of its final 7.20 mJ.
- Audit: Wbat = Cε²(1 − x) = (14.4 mJ)(0.6838) = 9.85 mJ, and 6.48 mJ + 3.37 mJ = 9.85 mJ, so the books balance.
- The ordering is the point: the heating is nearly over while the store is not yet half built, because PR decays as e(−2t/τ) but PC only as e(−t/τ).
Answert = 0.230 s, or 1.15τ; the capacitor then holds 3.37 mJ, 46.8% of its final 7.20 mJ, while the battery has done 9.85 mJ of work.