R-L, R-C and R-L-C Transients: what happens in the first few milliseconds
Flip a switch and nothing settles instantly — the time constant tells you how long.
Start here: filling a bucket through a narrow pipe
You are filling a bucket from a tap, through a thin pipe. The bucket does not fill instantly — the pipe only lets water through so fast.
And here is the part everyone has noticed without thinking about it: the filling slows down as it goes. The bucket fills fastest at the start and the last little bit seems to take forever.
A capacitor charging through a resistor does exactly this. The capacitor is the bucket, the resistor is the narrow pipe, and voltage is the water level. It fills quickly at first, then slower and slower as it approaches full.
Why it slows down
Current flows because there is a difference in voltage pushing it. When the capacitor is empty, the full battery voltage is pushing — lots of current, fast charging.
As the capacitor fills, it pushes back. The difference shrinks, so the current shrinks, so the charging slows. Near the end there is almost no difference left, so almost no current flows, and progress crawls.
💡 This is why it never quite reaches full in theory. Each moment it gets closer, and each moment it moves more slowly because it is closer. In practice we call it finished after a while — see below.
The time constant: one number for "how long"
The speed depends on two things, and both are obvious once stated:
⏱️ What makes it slow
A bigger capacitorA bigger bucket takes longer to fill. Bigger C, slower.
A bigger resistorA narrower pipe lets less through. Bigger R, slower.
Multiply them together and you get the time constant, written τ (the Greek letter "tau"):
τ = R × C
Example: 100 kΩ and 10 µF
τ = 100,000 × 0.00001 = 1 second
After 1τ → 63% full
After 2τ → 86% full
After 3τ → 95% full
After 5τ → 99.3% — call it finished
63% is the number to remember. One time constant always gets you 63% of the way there — whatever the circuit, whatever the values. That is what makes τ useful: it is a single number that describes the whole curve.
💡 Engineers use "five time constants" as the practical definition of finished, because 99.3% is close enough for any real purpose. If a question asks "how long until the capacitor is fully charged", the expected answer is 5τ.
An inductor is the same story backwards
A capacitor resists a sudden change in voltage — it takes time to fill up.
An inductor resists a sudden change in current — it takes time to get the current moving, like pushing a heavy trolley. Once it is moving it wants to keep moving, which is why switching an inductor off causes a spark.
Capacitor: τ = R × C
Inductor: τ = L / R ← divided, not multiplied
Example: 2 H with 50 Ω
τ = 2 / 50 = 0.04 s = 40 ms
💡 The formulas look inconsistent until you notice why. For a capacitor, more resistance means slower charging. For an inductor, more resistance means the current settles faster — because the final current is smaller, so there is less to build up to. That is why R is on the bottom.
Connect a capacitor to a battery through a resistor. The voltage doesn't jump to the supply value; it climbs along a curve. Why? Because charging a capacitor means moving charge, and the resistor limits how fast charge can move. The same story with different characters plays out in inductors, and understanding the shape of these curves explains everything from camera flashes to relay snubbers.
The two elements' stubborn rules
v(t) = V(1 − e^(−t/RC)) With t = 10 ms, R = 10 kΩ, C = 1 µF, V = 10 V, tau = 10 ms.
Set the time equal to τ (R × C in ms) and the capacitor reaches 63%, not 100%. It takes about five time constants to be practically full — which is why a circuit that must settle before the next sample is designed around 5τ rather than τ.
⚡ What each element refuses to do
CapacitorVoltage across it cannot change instantly (that would need infinite current). At the instant of switching it behaves like a short circuit if uncharged. After a long time, no current flows — it behaves like an open circuit.
InductorCurrent through it cannot change instantly (that would need infinite voltage). At the instant of switching it behaves like an open circuit. After a long time it behaves like a short circuit (a plain wire).
Remember by"Capacitor holds voltage, inductor holds current." Everything else follows from that one sentence.
Time constants
R-C circuit: τ = RC (seconds)
R-L circuit: τ = L/R (seconds)
Charging (rising toward a final value F):
x(t) = F (1 − e^(−t/τ))
Discharging (decaying from an initial value X₀):
x(t) = X₀ e^(−t/τ)
Percentage complete after n time constants:
1τ → 63.2% 2τ → 86.5% 3τ → 95.0%
4τ → 98.2% 5τ → 99.3% ← treated as "fully settled"
τ is not the time to finish — it's the time to get 63.2% of the way. The practical rule is 5τ ≈ steady state. And note the units work out: for RC, ohms × farads = seconds; for L/R, henries ÷ ohms = seconds. If your τ doesn't come out in seconds, you've used the wrong formula.
Worked numerical 1 — R-C charging
A 100 kΩ resistor charges a 10 µF capacitor from a 12 V supply. Find τ, the voltage after 1.5 s, and the time to reach 9 V.
τ = RC = (100 × 10³)(10 × 10⁻⁶) = 1 s
At t = 1.5 s:
v = 12(1 − e^(−1.5/1)) = 12(1 − 0.2231) = 12 × 0.7769
= 9.32 V
Time to reach 9 V — rearrange:
9 = 12(1 − e^(−t/1))
0.75 = 1 − e^(−t)
e^(−t) = 0.25
−t = ln(0.25) = −1.386
t = 1.386 s
Initial current (t = 0, capacitor acts as a short):
I₀ = V/R = 12/(100 × 10³) = 0.12 mA
Final current (t → ∞, capacitor acts as open): 0 A
Worked numerical 2 — R-L growth
A 2 H inductor with 50 Ω in series is switched onto 100 V. Find τ, the final current, and the current after 0.08 s.
τ = L/R = 2/50 = 0.04 s = 40 ms
Final (steady-state) current — inductor becomes a short:
I_f = V/R = 100/50 = 2 A
At t = 0.08 s (which is exactly 2τ):
i = 2(1 − e^(−0.08/0.04)) = 2(1 − e^(−2))
= 2(1 − 0.1353) = 2 × 0.8647 = 1.729 A
(86.5% of final — matches the 2τ table entry ✔)
Energy stored at steady state:
E = ½LI² = ½ × 2 × 2² = 4 J
R-L-C: three possible behaviours
Add both elements and the circuit can oscillate. Which of three responses you get depends on how the damping compares with the natural frequency.
Series R-L-C:
Damping factor α = R/2L
Natural freq ω₀ = 1/√(LC)
R > 2√(L/C) → OVERDAMPED — slow, no overshoot
R = 2√(L/C) → CRITICALLY DAMPED — fastest with no overshoot
R < 2√(L/C) → UNDERDAMPED — oscillates, rings, overshoots
R_critical = 2√(L/C)
Critical damping is what you want in a measuring instrument — the needle reaches the reading fastest without swinging past it. Underdamping is what you want in an oscillator, because the ringing is the output.
Worked numerical 3 — classifying an R-L-C circuit
A series circuit has R = 100 Ω, L = 0.1 H, C = 10 µF. Which response is it?
💡 Exam angle: the two most reliable marks are the initial and final conditions (capacitor: short then open; inductor: open then short) — state them explicitly even if not asked. For numericals, compute τ first and always show it with units. The 63%/86.5%/95% table lets you sanity-check any answer in seconds.
Syllabus points
Transient response; time constant (numerical)
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