The recurring confusion goes like this: you read "time constant," reason that τ = RC is the time the capacitor takes to charge, and conclude that at t = τ the capacitor sits at the full supply voltage. So for a 5 V supply you expect 5 V at one time constant. Then you plug into the formula, get 3.16 V, and assume you made an arithmetic slip. You didn't — 63.2% is exactly the right answer. The name is what's misleading, not your math.
Short answer: no, it never finishes — not in the mathematical sense. A capacitor charging through a resistor follows V(t) = Vsource·(1 − e^(−t/τ)), and that curve approaches the supply voltage but never equals it. At one time constant (t = τ) you are at 63.2% of the supply, not 100%. The famous "5 time constants = fully charged" rule isn't the moment it hits the top. It's a convention: at 5τ you're at 99.3%, close enough for any real circuit. Both textbook claims you've seen — "it never fully charges" and "it's done in 5τ" — are true at the same time. One is the math; the other is engineering rounding.
Why one time constant feels like it should be the finish line
The trap is right there in the name. It's called the time constant, which reads like "the time it takes." So when you first meet τ = RC, the natural assumption is: that's how long charging lasts, and at t = τ the capacitor sits at the supply voltage.
Then you plug in and get 63%, and it looks like you botched the arithmetic. You didn't. 63.2% is exactly what's supposed to happen at one time constant.
The second half of the confusion comes from the shape of the curve. A beginner expects charging to be a straight ramp — voltage climbing at a steady rate until it hits the ceiling and stops. If it were linear, "finish time" would be a real, well-defined moment. But charging isn't linear. It's exponential, governed by 1 − e^(−t/τ), and that function rises fast at first and then bends over, creeping toward the top without ever touching it.
So you end up holding two statements that look like a contradiction: "it never fully charges" and "5 time constants and it's done." They only clash if you think both are claims about the same kind of fact. They aren't. The first is a mathematical limit. The second is a practical cutoff.
What τ actually is
τ is the e-folding time, not the finish time. After one τ, the gap between where you are and the supply voltage shrinks by a factor of e (about 2.718). After another τ, the remaining gap shrinks by e again. And again. The gap keeps getting cut down, but cutting something by a factor of e over and over never drives it to exactly zero.
That's the whole reason charging is asymptotic. Each time constant closes a fixed fraction of whatever distance is left — not a fixed amount. A fixed fraction of a shrinking gap is always a smaller step, so you approach the target forever without arriving.
A worked example you can check
Take a concrete circuit: R = 10 kΩ, C = 100 µF, charged from Vsource = 5 V.
First the time constant:
τ = RC = 10,000 Ω × 100×10⁻⁶ F = 1.0 s
So one time constant here is exactly one second, which makes the numbers easy to read. The voltage across the capacitor is:
V(t) = 5·(1 − e^(−t/1))
Now walk it forward:
- t = 1 s (1τ): V = 5·(1 − e⁻¹) = 5·0.632 = 3.16 V — that's 63.2%, not 5 V.
- t = 2 s (2τ): V = 5·0.865 = 4.32 V (86.5%).
- t = 3 s (3τ): V = 5·0.950 = 4.75 V (95.0%).
- t = 5 s (5τ): V = 5·0.993 = 4.97 V (99.3%) — this is where "fully charged" comes from.
- t = 10 s (10τ): V = 4.99977 V (99.995%) — still not exactly 5 V.
Notice what happens after 5τ. You're at 99.3%, and doubling the wait to 10τ only buys you another 0.7%. Keep going and you'll claw out fractions of fractions of a volt. The capacitor is always almost charged and never exactly charged. That's the asymptote, made of real numbers.
Why the curve slows down
The reason the rise stalls is the charging current. The current into the capacitor is:
I(t) = (Vsource/R)·e^(−t/τ)
At t = 0 the resistor sees the full 5 V across it (the empty capacitor offers no opposing voltage yet), so the current starts at:
I(0) = 5 V / 10,000 Ω = 0.5 mA
As the capacitor voltage climbs, less voltage is left across the resistor, so the current falls — decaying with the same τ. Less current means charge arrives more slowly, which means the voltage rises more slowly, which means even less current. The capacitor effectively chokes off its own supply as it fills. The current heads toward zero but, like the voltage, never quite gets there. That dwindling current is the mechanism behind the asymptote. (If you're hazy on how the charge actually piles onto the plates in the first place, the companion piece on which way charge moves when a battery is connected untangles that.)
So which statement is right?
Both. Strictly, the capacitor never reaches 100% — that's the limit of an exponential, the same way 0.999… approaches 1 without you ever writing a "last 9." Practically, at 5τ you're within 0.7% of the supply, which is below the noise and tolerance of any real component, so engineers call it done. "5τ = fully charged" is a rounding convention dressed up as a deadline.
The fix for the original misread is just this: τ is not the finish time. At t = τ you're at 63.2%. The finish line, in the textbook sense, doesn't exist — only an agreed-upon "close enough." The same exponential-approach logic governs the dual problem in an RL circuit, where the question is whether an inductor blocks current because it opposes change — same τ-driven asymptote, current instead of voltage.
Check yourself
A capacitor charges through a resistor with time constant τ = 2 s from a 12 V supply. How long until it reaches exactly 12 V, and what's the voltage at t = 2 s?
A) 2 s to reach 12 V; voltage at t = 2 s is 12 V. B) Never reaches exactly 12 V; voltage at t = 2 s is about 7.6 V. C) 10 s to reach exactly 12 V; voltage at t = 2 s is 6 V. D) Never reaches exactly 12 V; voltage at t = 2 s is 12 V.
Correct answer: B.
At t = 2 s you're at one time constant, so V = 12·(1 − e⁻¹) = 12·0.632 = 7.58 V — about 63%, not the full 12 V. And it never reaches exactly 12 V, because the curve is asymptotic. A reads τ as the finish line and assumes a linear hit. C treats 5τ as an exact arrival (it's 99.3%, and 6 V is just half of 12, which the curve never sits at as a milestone here). D gets the asymptote right but forgets that t = τ is only 63%.
Close the gap
The two beliefs here — "τ is the finish time" and "5τ is exactly 100%" — come apart the moment you see one is a limit and the other is a practical cutoff. The hard part isn't the algebra; it's noticing which of your two true-sounding rules is the math and which is the rounding. That's the kind of split a tutor can catch in the moment, by asking you to compute 63.2% yourself before naming it. That adaptive correction is what Gradual Learning is built to deliver.