Time Constant τ — The Golden Table
Capacitor Charging — Voltage vs Time (V_f = V_S)
| Multiples of τ | % of Final Value (Charging) | % Remaining (Discharging) |
|---|---|---|
| 1τ | 63.2% | 36.8% |
| 2τ | 86.5% | 13.5% |
| 3τ | 95.0% | 5.0% |
| 4τ | 98.2% | 1.8% |
| 5τ ✅ | 99.3% ≈ "Fully charged" | 0.7% ≈ "Fully discharged" |
RC Circuit Analysis
RC Charging Circuit
RC Time Constant
τ = R·C [seconds]
R in ohms, C in farads → τ in seconds. Represents the "speed" of the transient.
General RC Formula
Vc(t) = Vf + (V₀ − Vf) · e^(−t/τ)
V₀ = initial voltage (just before switch). Vf = final voltage (at t→∞, steady state).
RC Charging
Vc(t) = VS · (1 − e^(−t/τ))
i(t) = (VS/R) · e^(−t/τ)
i(t) = (VS/R) · e^(−t/τ)
At t=0: Vc=0, i=V_S/R (max). At t=∞: Vc=V_S, i=0.
RC Discharging
Vc(t) = VS · e^(−t/τ)
i(t) = −(VS/R) · e^(−t/τ)
i(t) = −(VS/R) · e^(−t/τ)
At t=0: Vc=V_S (full), i=−V_S/R. At t=∞: both → 0.
🔧 How to Solve Any RC/RL Problem
1
Find
V₀ (or I₀): The initial value just before the switch. If initially open/uncharged → V₀=0.2
Find
Vf (or If): Replace C with open circuit (or L with short circuit) and solve the DC steady-state circuit.3
Find
τ: Kill all sources, find R seen by C (or L). Then τ=RC or τ=L/R.4
Apply formula:
x(t) = Xf + (X₀ − Xf) · e^(−t/τ)5
Find current/power/energy from the result using v=L·di/dt or i=C·dv/dt.
📝 Solved RC Example
Switch closes at t=0. VS=12V, R=1kΩ, C=5μF. Find Vc at t=10ms.
V₀ = 0 V (initially uncharged)
Vf = 12 V (DC: C open, all voltage across C)
τ = RC = 1×10³ × 5×10⁻⁶ = 5ms
Vc(10ms) = 12 + (0−12)e^(−10/5) = 12 − 12e^(−2) = 12 − 1.62 = 10.38 V
Vf = 12 V (DC: C open, all voltage across C)
τ = RC = 1×10³ × 5×10⁻⁶ = 5ms
Vc(10ms) = 12 + (0−12)e^(−10/5) = 12 − 12e^(−2) = 12 − 1.62 = 10.38 V
RL Circuit Analysis
RL Time Constant
τ = L/R [seconds]
L in henrys, R in ohms → τ in seconds. Larger L or smaller R = slower response.
General RL Formula
iL(t) = If + (I₀ − If) · e^(−t/τ)
I₀ = initial current. If = final current (DC: L = short circuit, find I from circuit).
RL Charging (Energizing)
iL(t) = (VS/R) · (1 − e^(−t/τ))
vL(t) = VS · e^(−t/τ)
vL(t) = VS · e^(−t/τ)
At t=0: iL=0, vL=VS. At t=∞: iL=VS/R, vL=0.
💡 RL is "opposite" to RC: in RL, current builds up slowly; in RC, voltage builds up slowly.
RL Discharging
iL(t) = IS · e^(−t/τ)
vL(t) = −IS·R · e^(−t/τ)
vL(t) = −IS·R · e^(−t/τ)
At t=0: iL=IS (max), vL=−IS·R. At t=∞: both → 0.
📝 Solved RL Example
RL discharge: L=10mH, R=10Ω. At t=0, current=1.2A. Find t when i drops to 0.8A.
I₀ = 1.2A, If = 0A (discharging)
τ = L/R = 10×10⁻³ / 10 = 1ms
iL(t) = 1.2 · e^(−t/τ) = 0.8
e^(−t/0.001) = 0.8/1.2 = 0.667
−t/0.001 = ln(0.667) = −0.4055
t = 0.4055 × 0.001 = 0.41 ms
τ = L/R = 10×10⁻³ / 10 = 1ms
iL(t) = 1.2 · e^(−t/τ) = 0.8
e^(−t/0.001) = 0.8/1.2 = 0.667
−t/0.001 = ln(0.667) = −0.4055
t = 0.4055 × 0.001 = 0.41 ms
⚡ Key Rules Summary
RC Circuit
- Max current at t=0 (C acts like wire)
- Current → 0 at t=∞ (C acts like open)
- Voltage cannot jump instantly
- Fully charged at 5τ
RL Circuit
- Max current at t=∞ (L acts like wire)
- Current = 0 at t=0 (L opposes change)
- Current cannot jump instantly
- Fully energized at 5τ