← CA-5 · CAPACITORS, INDUCTORS & FIRST-ORDER CIRCUITS

RC step response

YOU WILL BE ABLE TO

Define the time constant τ = RC and interpret it.
Write the charging and discharging responses of an RC circuit.
Use the universal first-order formula with any initial and final values.
Solve for the time to reach a given voltage.

Connect a resistor and capacitor to a source through a switch, and close the switch at t = 0. The capacitor cannot jump to the source voltage (Lesson 1's continuity rule), so it charges gradually: fast at first — a big voltage difference drives a big current through R — then ever slower as the gap closes. The result is the exponential curve that governs every first-order system in engineering, and one number sets its speed.

\tau = RC

the time constant, seconds

The step: at t = 0 the switch closes and C charges toward V_s through R, with time constant τ = RC.

Solving the circuit's differential equation (KVL plus i = C dv/dt — a clean separable ODE from Calculus I) gives the charging response from rest, and its mirror, the discharge of an initially charged capacitor through a resistor:

v_C(t) = V_s\left(1 - e^{-t/\tau}\right)

v_C(t) = V_0\, e^{-t/\tau}

charging from 0 · discharging from V_0

The landmarks to internalise: after one time constant, a charging capacitor has covered 63.2% of its journey (1 − e⁻¹), and a discharging one has fallen to 36.8%. After 5τ the transient is 99.3% done, and engineers round that to "finished" — the five-time-constant rule. τ itself scales exactly as intuition demands: more resistance means less charging current, more capacitance means more charge needed; either way, slower.

v_C(t) = v_C(\infty) + \left[v_C(0) - v_C(\infty)\right]e^{-t/\tau}

the universal first-order response

WORKED EXAMPLE

Charging, start to finish

In the figure: V_s = 12 V, R = 1 kΩ, C = 1 µF, capacitor initially uncharged.
τ = RC = 1000 × 10⁻⁶ = 1 ms.
$v_C(t) = 12(1 - e^{-t/1\,\mathrm{ms}})$ . At t = 1 ms: 12 × 0.632 ≈ 7.59 V. At t = 5 ms: ≈ 11.92 V — done, by the 5τ rule.

τ = 1 ms; v_C(1 ms) ≈ 7.59 V

WORKED EXAMPLE

Solving for time

Same circuit: when does v_C reach 6 V — half the supply?
$6 = 12(1 - e^{-t/\tau})$ → $e^{-t/\tau}$ = ½ → t = τ ln 2 ≈ 0.693 ms.
τ ln 2 is the half-way time of every first-order system — the same mathematics as radioactive half-life, and the constant behind the "half-voltage" bench trick from CA-4's practice.

t = τ ln 2 ≈ 0.693 ms

PRACTICE

An RC circuit has R = 10 kΩ and C = 100 nF. Find τ, and how long until a charging transient is considered complete.

A capacitor charged to 9 V discharges through a resistor with τ = 2 ms. Find its voltage at t = 2 ms.

A first-order RC circuit has v_C(0) = 2 V, v_C(∞) = 10 V, and τ = 1 ms. Write v_C(t) and evaluate it at t = 1 ms.

You need a delay of roughly τ = 0.1 s using a 10 µF capacitor. What resistor do you need? (This is the timing behind button debouncing, coming in ME-6.)