Two components that store energy — and remember
A resistor is forgetful. Push 5 V across a 1 kΩ resistor and exactly 5 mA flows — now, instantly, with no history. The relationship V = IR contains no clock. A capacitor and an inductor are different animals entirely: they store energy in a field, and that stored energy means the circuit's present depends on its past. This single fact — memory — is what turns a static DC puzzle into a circuit that *evolves*.
A capacitor is two conductive plates separated by a thin insulator. Connect it to a source and charge piles up on the plates — positive on one, negative on the other — building an electric field across the gap. The more charge Q you stash, the higher the voltage V across it: Q = CV, where C (capacitance, in farads) measures how much charge the part holds per volt. Differentiate that and you get the law that defines a capacitor's *behaviour in time*: i = C dv/dt. Current flows only when the voltage is *changing*. Hold the voltage steady and the current drops to zero — the capacitor is full and refuses to take more.
An inductor is the mirror image. It's usually just a coil of wire, and it stores energy in a *magnetic* field. By Faraday's law, a coil opposes changes in the current through it by generating a back-voltage. Its defining law swaps the roles of voltage and current: v = L di/dt. An inductor resists sudden current changes the way a capacitor resists sudden voltage changes. Where a capacitor wants its voltage to be continuous, an inductor wants its current to be continuous.
The step response and the time constant τ
Now build the simplest time-varying circuit there is: a resistor R in series with a capacitor C, fed from a battery through a switch. This is the RC circuit, and flipping the switch is called applying a *step input* — the source jumps from 0 to V₀ in an instant. What does the capacitor voltage do? It does *not* jump. It can't: i = C dv/dt says an instantaneous voltage jump would demand infinite current. Instead it climbs smoothly, fast at first while the resistor lets lots of current through, then ever more slowly as the capacitor fills and the driving voltage difference shrinks.
Write KVL around the loop and you get a first-order differential equation, RC·(dv/dt) + v = V₀. Its solution is the famous charging exponential. The whole curve is governed by one number, the [[ee-rc-time-constant|time constant]] τ = R·C, measured in seconds (ohms × farads = seconds, a small miracle worth checking). τ is the natural timescale of the circuit: how long it takes to respond.
Charging: v(t) = V0 · ( 1 − e^(−t/τ) ) τ = R·C
Discharging: v(t) = V0 · e^(−t/τ)
t v(t)/V0 (charging) how settled?
----- ------------------- ------------
1τ 0.632 (63.2 %) rising fast
2τ 0.865 (86.5 %)
3τ 0.950 (95.0 %) "close enough" in many specs
4τ 0.982 (98.2 %)
5τ 0.993 (99.3 %) treated as fully settled
v/V0
1.0 | . . . . . . . . . . . . ← V0
| . '
0.63| .' ← reaches 63.2% at exactly t = 1τ
| .
| .
0 |.________________________________________ t
0 1τ 2τ 3τ 4τ 5τRead that table like an engineer. After one τ the capacitor has covered 63.2 % of the gap to its final value — that's the *definition* of τ, the point where the exponential has decayed by a factor of e. The gorgeous part: the curve is *self-similar*. From wherever you are, you'll always cover another 63 % of the *remaining* gap in the next τ. By 5τ you're within 0.7 % of the target, and in practice everyone calls the transient settled at 5τ.
A worked RC charging example
Numbers make it concrete. Take a 5 V supply, R = 10 kΩ and C = 10 µF, with the capacitor starting fully discharged. Close the switch at t = 0 and ask three things: how long is τ, what is the voltage after 15 ms, and when does it reach 4 V?
- Time constant. τ = R·C = (10 × 10³ Ω)(10 × 10⁻⁶ F) = 0.1 s = 100 ms. So this circuit's whole story plays out over a few hundred milliseconds — slow enough to watch on a cheap multimeter.
- Voltage at t = 15 ms. t/τ = 0.015 / 0.1 = 0.15, so v = 5·(1 − e^(−0.15)) = 5·(1 − 0.861) = 0.70 V. Only 15 % of the way there — early on the curve, still climbing briskly.
- Time to reach 4 V. Set 4 = 5·(1 − e^(−t/τ)) → e^(−t/τ) = 0.2 → t = τ·ln(5) = 0.1 × 1.609 = 161 ms. Reaching 80 % takes about 1.6 time constants — exactly what the curve's shape promised.
* SPICE: watch a 10k / 10uF RC charge to 5V
V1 in 0 PWL(0 0 1u 5) ; 5V step at t=0
R1 in out 10k
C1 out 0 10u IC=0 ; cap starts discharged
.tran 1m 600m UIC
.end
measured v(out):
t = 100ms (1 tau) -> 3.16 V (63 %)
t = 200ms (2 tau) -> 4.32 V (86 %)
t = 500ms (5 tau) -> 4.97 V (99 %, settled)Why a capacitor blocks DC but passes AC
Here's the line you'll hear a thousand times in electronics: *"a capacitor blocks DC and passes AC."* It sounds like a special rule to memorise. It isn't — it falls straight out of i = C dv/dt. Apply a steady DC voltage and after the transient dies away (a few τ), dv/dt = 0, so i = 0. No changing voltage means no current. The capacitor sits there fully charged, an open circuit. DC is blocked.
Now wiggle the voltage as a sine wave. The faster you wiggle — the higher the frequency — the larger dv/dt is at every instant, so the larger the current. A capacitor effectively *conducts better at high frequencies*. Quantify this with capacitive reactance X_C = 1/(2πfC): it's huge at DC (f→0 makes X_C→∞, an open) and tiny at high frequency (a near-short). The capacitor is a frequency-dependent gatekeeper. AC is passed.
Reactance of a 100 nF capacitor vs frequency:
frequency X_C = 1/(2πfC) behaves like
---------- --------------- --------------
DC (0 Hz) infinite open circuit (blocks)
100 Hz ~15,900 Ω big resistor
10 kHz ~159 Ω modest resistor
1 MHz ~1.6 Ω nearly a wire (passes)
AC in o──────||──────o AC out ← high f: cap = short, signal through
C
DC in o──────||─────── (no out) ← DC: cap = open, blockedThis one behaviour underwrites a huge amount of practical design. A coupling capacitor sits between amplifier stages to pass the wiggling signal while blocking each stage's DC bias from disturbing the next. Its opposite-job twin, the [[ee-decoupling-capacitor|decoupling capacitor]], sits right beside every chip's power pins: it's a tiny local energy reservoir that shorts high-frequency noise on the supply straight to ground and instantly delivers charge when the chip's current demand spikes — keeping the voltage rail from sagging faster than the distant power supply can react.
The RL mirror, and where transients bite
Everything you just learned has an inductive twin. Put a resistor in series with an inductor and you get the RL circuit, governed by the [[ee-rl-time-constant|RL time constant]] τ = L/R. Notice the flip: the resistor is now in the *denominator*. A bigger R makes an RL circuit settle *faster*, the opposite of RC, because a large resistor lets the current establish quickly. When you close the switch, the inductor current climbs as i(t) = (V/R)·(1 − e^(−t/τ)) — the same exponential shape, now in current rather than voltage.
Transients are not an academic curiosity — they are where circuits live and die. The clean exponential you sketched is the *settling* of a digital logic gate, and if it hasn't settled within one clock period the chip reads the wrong bit. It's the *inrush* current that trips a breaker when you switch on a big supply. It's the ringing you see on an oscilloscope when a real wire's stray inductance and capacitance form an accidental resonant circuit. Wherever something switches — and modern electronics is switching billions of times a second — transients are the whole game.
There's one more reason this rung matters: it's the doorway to AC. We solved the step response by grinding through a differential equation, which is fine once but tedious forever. In the next rung you'll learn that for *sinusoidal* steady state there's a shortcut so good it feels like cheating — the phasor and complex impedance, which let you treat a capacitor or inductor as just another 'resistor' with an imaginary value and solve the whole thing with Ohm's law again. The reactance X_C you met above is the first glimpse of it.