Engineer's Notebook §1.3

RC time constant

A capacitor charging through a resistor approaches the supply voltage exponentially. The time constant sets the pace:

τ = RC
τ
time constant
s
R
resistance
Ω
C
capacitance
F
V
supply voltage
V
vC
capacitor voltage
V
t
time since switch-on
s
i
current
A
L
inductance
H
The charging curve of a capacitor: voltage rises steeply at first and levels off towards the supply voltage. Time is marked in multiples of the time constant, and a point marks the chosen time. TIME tVOLTAGE v, V V = 5 V 63% τ 2τ 3τ 4τ τ = 1 s v = 3.16 V
Fig. 1.3-1Time in multiples of τ. After one τ the capacitor reaches 63% of the supply.
τ = RC1 s
t1 s
i = (V/R)e−t/τ0.184 mA
vC = V(1 − e−t/τ)3.16 V

Worked example

Uses the values in Fig. 1.3-1. Change them and the working updates.

Given
  • Rseries resistance10 kΩ
  • Ccapacitance100 µF
  • Vsupply voltage5 V
  • ttime since switch-on1τ = 1 s
Find

The capacitor voltage vC at time t.

Solution
τ = RC = 10,000 Ω × 1 × 10⁻⁴ F = 1 s vC = V(1 − e−t/τ) = 5 V × (1 − e^(−1)) = 3.16 V
Answer
vC = 3.16 V

After 5τ = 5 s it is above 99% charged.

Related forms

  • Discharge from V0

    vC = V0e−t/τ

  • RL circuit

    τ = LR

  • Time to reach a voltage

    t = −τ ln(1 − vCV)

Assumptions
  • The capacitor starts uncharged, and the supply switches on as an ideal step at t = 0.
  • Components are ideal: no capacitor leakage, and any source resistance is included in R.

Derivation
  1. Round the loop, the supply voltage splits between resistor and capacitor: V = iR + vC.
  2. The capacitor's current is i = C dvC/dt, so RC dvC/dt + vC = V.
  3. With vC(0) = 0, the solution is vC = V(1 − e−t/RC).

At t = τ the capacitor reaches 1 − 1/e ≈ 63.2% of V. After 5τ it is above 99%, usually taken as fully charged.