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RC Circuits I: Charging & the Differential Equation

You’ll be able to

The setup: a battery, a resistor, and an empty capacitor

Close a switch connecting a battery ε in series with a resistor R and an uncharged capacitor C. Current flows, charge lands on the plates, and the growing capacitor voltage q/C pushes back against the battery. The current must therefore decay as the capacitor fills. Because the loop rule links q and its own derivative dq/dt, the circuit obeys a first-order differential equation — this is where circuits meet calculus.

From loop rule to ODE

Kirchhoff’s loop rule at any instant: ε − iR − q/C = 0. Since the current delivers charge to the plates, i = dq/dt, giving R·dq/dt = ε − q/C. This says the charging rate is proportional to how far q still is from its final value εC — the signature of exponential approach. The same structure will reappear for the LR circuit in Unit 6, and for every “rate proportional to remaining gap” system you meet.

Worked example

Solve the charging equation R·dq/dt = ε − q/C with q(0) = 0 to derive q(t), then find the current i(t).

  1. 1.Rewrite with a common denominator: dq/dt = (εC − q)/(RC).
  2. 2.Separate variables: dq/(εC − q) = dt/(RC).
  3. 3.Integrate both sides from the start (q = 0 at t = 0) to time t: ∫₀^q dq′/(εC − q′) = ∫₀^t dt′/(RC), giving −ln[(εC − q)/(εC)] = t/(RC).
  4. 4.Exponentiate: (εC − q)/(εC) = e^(−t/RC), so q(t) = εC(1 − e^(−t/RC)) = Q_max(1 − e^(−t/RC)) with Q_max = εC.
  5. 5.Differentiate for the current: i(t) = dq/dt = (εC)(1/RC)e^(−t/RC) = (ε/R)e^(−t/RC).
  6. 6.Check the limits: q(0) = 0 and q → εC as t → ∞ ✓; i(0) = ε/R (the empty capacitor offers no opposition) and i → 0 (full capacitor blocks the current) ✓.
Answer: q(t) = Q_max(1 − e^(−t/RC)) with Q_max = εC; i(t) = (ε/R)·e^(−t/RC)
Charging capacitor
q(t) = Q_max(1 − e^(−t/RC)), i(t) = (ε/R)e^(−t/RC), Q_max = εC
Charge grows toward Q_max while current decays from ε/R — the two curves are mirror images in behavior.
Time constant
τ = RC (1 Ω × 1 F = 1 s)
After one τ the capacitor holds 1 − e⁻¹ ≈ 63% of full charge and the current has fallen to e⁻¹ ≈ 37% of its start. After 5τ the circuit is effectively settled.

Reading the time constant

The time constant τ = RC sets the clock for the whole transient. Check the units: Ω·F = (V/A)(C/V) = C/A = s — genuinely a time. Larger R throttles the current, and larger C means more charge to deliver, so either one slows the charging. Every RC quantity — charge, current, resistor voltage, capacitor voltage — changes by the factor e⁻¹ ≈ 0.368 each time one τ elapses.

Worked example

A 12 V battery charges a 500 μF capacitor through a 2.0 kΩ resistor. Find τ, the final charge, and the charge and current at t = 2.0 s.

  1. 1.Time constant: τ = RC = (2.0 × 10³)(500 × 10⁻⁶) = 1.0 s.
  2. 2.Final charge: Q_max = εC = (12)(500 × 10⁻⁶) = 6.0 × 10⁻³ C = 6.0 mC.
  3. 3.At t = 2.0 s = 2τ: q = 6.0(1 − e⁻²) mC = 6.0(1 − 0.135) = 6.0 × 0.865 ≈ 5.2 mC.
  4. 4.Initial current: i₀ = ε/R = 12 / 2000 = 6.0 mA. At 2τ: i = 6.0·e⁻² ≈ 6.0 × 0.135 = 0.81 mA.
  5. 5.Sanity check: after two time constants the capacitor is 86% full and the current is down to 14% — well on the way to the 5τ “fully charged” rule of thumb.
Answer: τ = 1.0 s, Q_max = 6.0 mC; at t = 2.0 s, q ≈ 5.2 mC and i ≈ 0.81 mA
Watch out

No charge ever crosses the gap between the capacitor plates — current flows in the wiring while equal-and-opposite charge accumulates on the two plates. And keep the two exponentials straight: during charging, q grows as (1 − e^(−t/RC)) while i decays as e^(−t/RC). Writing q(t) = Q_max·e^(−t/RC) for a charging capacitor is the classic error — that is the discharge curve.

Checkpoint

One time constant after charging begins, the charge on the capacitor is closest to what fraction of its final value?

Checkpoint

The resistance in an RC charging circuit is doubled and the capacitance is halved. The time constant:

Answer the 2 checkpoints as you read.

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