A resistor-capacitor circuit does not generally jump instantly from one DC state to another. The capacitor voltage changes as charge flows, producing an exponential transient.

Consider a series resistor $R$, capacitor $C$, and ideal source of EMF $\mathcal E$.

Charging from an initially uncharged capacitor

Let $q(t)$ be the charge magnitude on the capacitor and let current $i=dq/dt$ flow during charging. Kirchhoff's loop law gives

$$\mathcal E-iR-\frac{q}{C}=0.$$

Using $i=dq/dt$,

$$R\frac{dq}{dt}+\frac{q}{C}=\mathcal E.$$

Multiplying by $C$,

$$\boxed{RC\frac{dq}{dt}+q=C\mathcal E}.$$

For initial condition

$$q(0)=0,$$

the solution is

$$\boxed{q(t)=C\mathcal E\left(1-e^{-t/(RC)}\right)}.$$

The capacitor voltage is

$$\boxed{v_C(t)=\mathcal E\left(1-e^{-t/(RC)}\right)},$$

while the current is

$$\boxed{i(t)=\frac{\mathcal E}{R}e^{-t/(RC)}}.$$

Initially the uncharged capacitor has zero voltage, so the full source voltage appears across the resistor and

$$i(0)=\frac{\mathcal E}{R}.$$

At long times,

$$v_C\to\mathcal E,\qquad i\to0.$$

The ideal capacitor therefore approaches an open-circuit condition in steady DC after charging.

The RC time constant

Define

$$\boxed{\tau=RC}.$$

After one time constant,

$$v_C(\tau)=\mathcal E(1-e^{-1})\approx0.632\mathcal E,$$

and the charging current has fallen to

$$i(\tau)=i(0)e^{-1}\approx0.368i(0).$$

The time constant sets the characteristic response time, not a sharp completion time. The exponential approaches its final value asymptotically.

Discharging a capacitor through a resistor

Suppose a capacitor initially has voltage $V_0$ and charge

$$Q_0=CV_0,$$

then is disconnected from the source and allowed to discharge through $R$.

Kirchhoff's law gives

$$iR+\frac{q}{C}=0$$

with current sign chosen consistently with decreasing $q$. The charge magnitude decays as

$$\boxed{q(t)=Q_0e^{-t/(RC)}},$$

so

$$\boxed{v_C(t)=V_0e^{-t/(RC)}}.$$

The current magnitude is

$$\boxed{|i(t)|=\frac{V_0}{R}e^{-t/(RC)}}.$$

Capacitor voltage cannot jump in the ideal finite-current model

Because

$$i=C\frac{dv_C}{dt},$$

an instantaneous finite jump in $v_C$ would require an unbounded current impulse. In ordinary ideal-circuit problems with finite currents,

$$\boxed{v_C(0^+)=v_C(0^-)}.$$

This continuity condition is often the fastest way to determine the initial state after a switch changes position.

Worked example

Let

$$R=10,\mathrm{k\Omega},\qquad C=100,\mu\mathrm F,$$

so

$$\tau=RC=(10^4)(100\times10^{-6})=1.0,\mathrm s.$$

With source

$$\mathcal E=12,\mathrm V,$$

after $2.0,\mathrm s=2\tau$ the charging capacitor voltage is

$$v_C=12(1-e^{-2})\approx10.4,\mathrm V.$$

The current is

$$i=\frac{12}{10^4}e^{-2} \approx1.62\times10^{-4},\mathrm A =0.162,\mathrm{mA}.$$

The circuit has not stopped changing completely after two time constants, but most of the transition has already occurred.

RC transients are a physical example of a first-order dynamical system: stored electric-field energy and resistive dissipation determine an exponential approach between circuit states.