Two snapshots and one exponential

To change $V_C$ you have to move charge onto the plates, and moving charge takes current, which takes time. A discontinuous jump in $V_C$ would require infinite current. So across the instant a switch flips:

$$V_C(t_{\text{after}}) = V_C(t_{\text{before}}).$$

That freezes one voltage in the circuit, and freezing one voltage makes the rest a plain resistor problem. It gives the two snapshots that carry most of the marks:

1. At $t = 0$ with an UNCHARGED capacitor, $V_C = 0$, so the capacitor behaves as a plain WIRE. Replace it by a short and solve the resulting resistor network for every current.
2. As $t \to \infty$, no more charge is arriving, so the capacitor's branch carries no current and behaves as an OPEN switch. Replace it by a gap, solve again, and the voltage that appears across that gap is the capacitor's final voltage.

Those two replacements are the ones students most often swap, and the swap is not a small error: the two resistor networks are usually genuinely different circuits, so every current and every voltage in both snapshots comes out wrong at once.

If the capacitor starts with charge already on it, the $t = 0$ replacement is a battery of that voltage, not a wire. The wire is the special case $V_C(0) = 0$.

For a source $\varepsilon$ in series with $R$ and $C$, walk the loop with $q$ on the capacitor and $i = dq/dt$ in the resistor:

$$\varepsilon - iR - \frac{q}{C} = 0 \qquad \Longrightarrow \qquad R\frac{dq}{dt} = \varepsilon - \frac{q}{C}.$$

Separating and integrating from $q(0) = 0$ gives the charging solution and its current:

$$q(t) = C\varepsilon\left(1 - e^{-t/RC}\right), \qquad i(t) = \frac{\varepsilon}{R}\,e^{-t/RC}.$$

With the source removed and an initial charge $Q_0$, the same equation without the $\varepsilon$ term gives discharge:

$$q(t) = Q_0 e^{-t/RC}, \qquad i(t) = \frac{Q_0}{RC}\,e^{-t/RC}.$$

Two structural facts are worth reading off these directly. The current in BOTH cases is a pure decaying exponential, and it is largest at $t = 0$, which is another way of saying the capacitor starts out looking like a wire. And $q$ approaches its final value asymptotically: the process never terminates, it only becomes uninteresting.

The exponent's denominator, $\tau = R_{\text{eq}}C$, has units of seconds and sets the only clock in the problem. The $R$ in it is not "the resistor drawn next to the capacitor" and not "the total resistance in the circuit". It is the resistance looking OUT from the capacitor's two terminals, with ideal sources killed:

1. Remove the capacitor, leaving its two terminals exposed.
2. Replace every ideal battery by a wire and every ideal current source by a gap.
3. Compute the equivalent resistance between those two terminals. That is $R_{\text{eq}}$.

Worked example. A battery drives $R_1 = 6\ \Omega$ into node $A$; from $A$, a resistor $R_2 = 3\ \Omega$ runs to the bottom rail and the capacitor runs to the bottom rail alongside it. Shorting the battery puts $R_1$ from $A$ to the bottom rail as well, so the capacitor sees $R_1 \parallel R_2 = 2\ \Omega$, and with $C = 100\ \mu$F, $\tau = 200\ \mu$s. Grabbing $R_2$ alone would predict a transient half again too slow; grabbing $R_1 + R_2$ would predict one four and a half times too slow.

The reason the battery becomes a wire is that an ideal source holds its voltage no matter what, so it presents zero resistance to the small changes the transient consists of. The steady offset it supplies is already accounted for in the final value.

Exponentials close a fixed FRACTION of the remaining gap per time constant, never a fixed amount:

$$1 - e^{-1} = 0.632, \qquad 1 - e^{-2} = 0.865, \qquad 1 - e^{-3} = 0.950, \qquad 1 - e^{-5} = 0.993.$$

1. At $t = \tau$ the capacitor holds $63\%$ of its final charge and the current still runs at $37\%$ of its initial value. It is not finished.
2. $\tau$ is not a half-life. The half point is $0.693\tau$, and treating $\tau$ as a halving time predicts $25\%$ remaining at $2\tau$ when the true figure is $13.5\%$.
3. Extrapolating the initial slope in a straight line predicts completion at exactly $t = \tau$, which is where the mistaken 100-percent claim comes from. The curve bends away from that tangent immediately.
4. $5\tau$ is the working definition of done, at $99.3\%$.

The energy accounting has a result that surprises people. Charging $C$ to $\varepsilon$ through any resistance:

$$W_{\text{battery}} = Q\varepsilon = C\varepsilon^2, \qquad U_C = \tfrac12 C\varepsilon^2, \qquad W_R = \int_0^\infty i^2R\,dt = \tfrac12 C\varepsilon^2.$$

Exactly half the delivered energy is dissipated, and $R$ cancels out of that result completely. A smaller resistance does not save the lost half; it loses the same half faster. The reason is that the battery pushes every charge through the full $\varepsilon$, while the capacitor's own voltage averaged only $\varepsilon/2$ over the filling.