Universal Formula — applies to ALL first-order RC and RL circuits
x(t) = X_f + (X₀ − X_f) · e^(−t/τ)
x = any voltage or current | X₀ = initial value (at t=0⁺) | X_f = final value (at t→∞) | τ = time constant
Time Constants
RC circuit
τ = R · C [seconds]
R in Ω, C in Farads → τ in seconds
RL circuit
τ = L / R [seconds]
L in Henrys, R in Ω → τ in seconds
💡 How to find R for τ
Kill all independent sources (V sources → short, I sources → open), then find the Thevenin resistance seen by L or C at its terminals.
The 5τ Table
Time
% of final value
% remaining
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%
Charging curve: reaches 63.2% at τ, essentially complete at 5τ. Discharging curve is mirrored (starts at X₀, decays to X_f=0).
DC Steady-State Rules (Critical!)
Inductor at DC (t→∞)
Replace with
SHORT CIRCUIT (wire)
v_L = L·di/dt = 0 when i = constant
📌 Why?
At DC steady state, current is constant → di/dt=0 → v_L=0 → inductor behaves as a wire (short). Current flows freely through it.
Capacitor at DC (t→∞)
Replace with
OPEN CIRCUIT (no connection)
i_C = C·dv/dt = 0 when v = constant
📌 Why?
At DC steady state, voltage is constant → dv/dt=0 → i_C=0 → no current flows → acts like an open gap.
⚠️ Initial vs Final — Key Difference!
t=0⁻ (just before switch): Old circuit, DC steady state → use rules above to find X₀
t=0⁺ (just after switch): i_L(0⁺) = i_L(0⁻) and v_C(0⁺) = v_C(0⁻) — they CANNOT change instantaneously
t→∞ (after switch): New circuit, DC steady state again → use rules above to find X_f
5-Step Solution Procedure
Find X₀ — initial value at t=0⁺ Analyze circuit at t=0⁻ (just before switching) with DC steady state rules (L=short, C=open). The value of i_L or v_C you find carries over: i_L(0⁺)=i_L(0⁻), v_C(0⁺)=v_C(0⁻).
Find X_f — final value at t→∞ Analyze the NEW circuit (after switching) with DC steady state rules again. This is X_f in the formula.
Find τ — time constant Kill all independent sources. Find Thevenin R seen by L or C: τ=RC or τ=L/R.
Write the complete expression: x(t) = X_f + (X₀−X_f)·e^(−t/τ)
Find any other requested quantity If you have v_C(t), get i_C(t) = C·dv_C/dt. If you have i_L(t), get v_L(t) = L·di_L/dt.
RC Circuit — Lecture Examples
RC Charging — Time Constant Calculation
Given: R = 47 kΩ, C = 1000 μF. VS = 5V. Find: τ, voltage at t=0.7τ, time to fully charge.
At t=τ: VC = VS·e^(−1) = 0.368·VS → 36.8% of initial (= 1−63.2%)
Current discharges in the same direction as charging current (capacitor acts as source)
VC(t) = VS · e^(−t/τ) (discharging) At t=τ: VC = 0.368×VS (37% remaining)
RC & RL Waveform Sketches
RC charging/discharging waveforms for v_C(t). At t=τ: charging reaches 63.2%, discharging drops to 36.8%. RL i_L has identical shape with τ=L/R.
RL Circuit — Lecture Examples
RL Series Circuit — L=40mH, R=2Ω, VS=20V DC
Given: L=40mH, R=2Ω, VS=20V DC. Find: (a) final steady state current, (b) time constant τ, (c) transient time.
(a) Final steady state current (t→∞):
At DC steady state: L = short circuit → all voltage across R
I_f = VS/R = 20/2 = 10 A
(b) Time constant:
τ = L/R = 40×10⁻³ / 2 = 20 ms
(c) Transient time (until steady state):
t_transient = 5τ = 5 × 20×10⁻³ = 100 ms
I_f = 10 A | τ = 20 ms | Transient time = 100 ms
RL Discharge — Find time for current to reach given value
Given: RL circuit, switch was closed for a long time (I₀=1.2A). L=10mH, R=10Ω. Find time for I to drop from 1.2A to 0.8A after switch opens.
Set iL = 0.8A and solve for t:
0.8 = 1.2·e^(−t/10⁻³)
0.8/1.2 = e^(−t/10⁻³)
0.667 = e^(−1000t)
ln(0.667) = −1000t
−0.4055 = −1000t
t = 0.4055/1000 = 0.41 ms
Lec 12/13 — Inductor current i=10t·e^(−5t) A, L=100mH
Given: i=0 for t<0, i=10t·e^(−5t) A for t≥0. L=100mH.
(a) Time of maximum current:
di/dt = 10e^(−5t) + 10t·(−5)e^(−5t) = 10e^(−5t)(1 − 5t)
Set di/dt = 0: 1 − 5t = 0 → t = 0.2 s = 1/5 s
(c) Voltage across inductor:
v = L·di/dt = 0.1 × 10e^(−5t)(1−5t)
v = (1 − 5t)e^(−5t) V
(e) Instantaneous voltage change?
YES — at t=0, voltage jumps from 0 to 1V. (Voltage across L CAN change instantaneously!)
(f) Same time for max v and max i?
NO — v is proportional to di/dt, not i. Max v at t=0, max i at t=0.2s.
(g) Voltage changes polarity when:
v=0 → (1−5t)=0 → t = 0.2 s
i_max at t = 0.2s | v(t) = (1−5t)e^(−5t) V | voltage polarity changes at t=0.2s
Lec 12/13 — Power & Energy in inductor (Example 3)
Given: Same i=10t·e^(−5t) A, L=100mH.
v = (1−5t)e^(−5t) V
p = v·i = (1−5t)e^(−5t) · 10t·e^(−5t) = 10t(1−5t)e^(−10t) W
w = ½Li² = ½ × 0.1 × (10t·e^(−5t))² = 5t²e^(−10t) J
(b) Energy stored interval: p>0 → 0 < t < 0.2 s (c) Energy extracted interval: p<0 → 0.2 s < t < ∞ (d) Maximum energy stored:
At t=0.2s: w_max = 5×(0.2)²×e^(−10×0.2) = 5×0.04×e^(−2) = 0.2×0.1353
Energy stored: 0 < t < 0.2s | Energy extracted: t > 0.2s w_max = 27.07 mJ at t = 0.2s
Capacitor Dynamics — Lecture Examples
Lec 12/13 — 2mH Inductor: current pulse → find voltage
Given: L=2mH. iL=0 for t<1μs, iL=10³(t−10⁻⁶) for 1μs≤t≤2μs, iL=1mA for t>2μs.
v = L·di/dt
For t<1μs: di/dt=0 → v=0 V
For 1μs≤t≤2μs: di/dt = 10³ A/s → v = 2×10⁻³ × 10³ = 2 V
For t>2μs: di/dt=0 → v=0 V
Note: V spikes to 2V during the 1μs transition. If current were ~1A, spike would be 2000V — this is how inductors boost voltage in ignition coils!
v = 2V for 1μs ≤ t ≤ 2μs, zero otherwise
Lec 12/13 — Example 4: Capacitor C=0.2μF with triangular current pulse
Given: C=0.2μF, uncharged. i(t): 0 (t≤0), 5000t A (0≤t≤20μs), 0.2−5000t A (20≤t≤40μs), 0 (t≥40μs).
v(t) = (1/C)∫i dt + v(0)
0 ≤ t ≤ 20μs:
v = (1/0.2×10⁻⁶) × ∫₀ᵗ 5000t dt = (5×10⁶) × (5000t²/2) = 12.5×10⁹ t² V
p = v·i = 12.5×10⁹t² × 5000t = 62.5×10¹² t³ W
w = ½Cv² = ½×0.2×10⁻⁶×(12.5×10⁹t²)² = 15.625×10¹² t⁴ J
20 ≤ t ≤ 40μs:
v = 10⁶t − 12.5×10⁹t² − 10 V
t ≥ 40μs:
v = 10V (constant — energy trapped)
p = 0 (no current)
w = ½×0.2×10⁻⁶×10² = 10 μJ
Why does voltage remain after current=0?
Ideal source + ideal capacitor → no dissipation path → energy stays trapped in electric field.
v(0→20μs) = 12.5×10⁹ t² V v(t≥40μs) = 10 V (trapped) Final stored energy = 10 μJ
Lec 12/13 — C=0.5μF with given v(t) — full analysis