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RC Circuits II: Discharge, Energy & Transients

You’ll be able to

Discharge: the capacitor becomes the battery

Disconnect the battery and let a charged capacitor drive current through a resistor. Now the capacitor’s own voltage q/C is the only push, and it weakens as charge drains away — the rate of loss is proportional to the charge remaining, so the decay is exponential. Every quantity in the circuit (q, i, V_C, V_R) decays with the same time constant τ = RC.

Worked example

A capacitor with initial charge Q₀ discharges through a resistor R. Derive q(t) and i(t).

  1. 1.Loop rule with the capacitor driving the current: q/C − iR = 0. The plate charge is decreasing, so the discharge current is i = −dq/dt.
  2. 2.Substitute: q/C + R·dq/dt = 0, i.e. dq/dt = −q/(RC).
  3. 3.Separate variables: dq/q = −dt/(RC), and integrate from Q₀ at t = 0: ln(q/Q₀) = −t/(RC).
  4. 4.Exponentiate: q(t) = Q₀e^(−t/RC).
  5. 5.Current: i = −dq/dt = (Q₀/RC)e^(−t/RC) = I₀e^(−t/RC), where I₀ = Q₀/(RC) = V₀/R is set by the capacitor’s initial voltage.
  6. 6.Check: q(0) = Q₀ ✓, and both q and i decay smoothly to zero as the stored energy is dissipated in R ✓.
Answer: q(t) = Q₀e^(−t/RC), i(t) = (V₀/R)e^(−t/RC) — pure exponential decay with τ = RC
Discharging capacitor
q(t) = Q₀e^(−t/RC), i(t) = I₀e^(−t/RC), I₀ = V₀/R
Half of the charge remains after t = RC·ln 2 ≈ 0.69 RC — the same half-life logic as radioactive decay.

The two limiting behaviors of a capacitor

Transient problems in larger circuits reduce to two snapshots. At t = 0, an uncharged capacitor has q = 0 so V_C = q/C = 0: it behaves like a bare wire (a short circuit), and current flows as if it were not there. As t → ∞, the capacitor is full, no more charge flows onto it, and its branch carries zero current: it behaves like an open circuit (a break in the wire). Analyze the circuit twice — once with each replacement — and you have the start and end states without solving any ODE.

Where the energy goes

A charged capacitor stores U = q²/(2C) = ½CV². During discharge, all of it becomes heat in the resistor — integrate P = i²R over the decay and you recover exactly Q₀²/(2C). Charging holds a famous surprise: the battery supplies Q_max·ε = Cε² of energy, but the capacitor stores only ½Cε². Exactly half the battery’s energy is dissipated in the resistor during charging — independent of the value of R.

Worked example

A 12 V battery connects through R₁ = 4.0 Ω to a parallel combination of R₂ = 8.0 Ω and an uncharged 500 μF capacitor. Find the battery current just after the switch closes, the current a long time later, and the final charge on the capacitor.

  1. 1.At t = 0 the uncharged capacitor acts like a wire, short-circuiting R₂ (all current takes the zero-resistance path). The circuit is just ε and R₁: I₀ = 12 / 4.0 = 3.0 A.
  2. 2.As t → ∞ the full capacitor acts like an open circuit, so current flows through R₁ and R₂ in series: I_∞ = 12 / (4.0 + 8.0) = 1.0 A.
  3. 3.The capacitor’s final voltage equals the voltage across R₂ (they are in parallel): V_C = I_∞·R₂ = (1.0)(8.0) = 8.0 V.
  4. 4.Final charge: Q = CV_C = (500 × 10⁻⁶)(8.0) = 4.0 × 10⁻³ C = 4.0 mC.
Answer: I₀ = 3.0 A, I_∞ = 1.0 A, Q_final = 4.0 mC
On the exam

Transient shortcut worth memorizing for the exam: t = 0 → replace each uncharged capacitor with a plain wire; t → ∞ → replace each capacitor with an open break, then find V_C from the resistors around it. Most multiple-choice RC questions are exactly these two substitutions, no exponentials required.

Tip

To find a discharge time from a target fraction, invert the exponential with a logarithm: q/Q₀ = e^(−t/RC) gives t = RC·ln(Q₀/q). Keep the argument of the log greater than 1 and the time comes out positive automatically.

Checkpoint

Two time constants after a capacitor begins discharging, the fraction of the initial charge remaining is closest to:

Checkpoint

How long does it take a discharging capacitor to lose half of its charge?

Answer the 2 checkpoints as you read.

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