Result

Charging

q (t) = C ε (1 - e^{- t / R C})

Statement

The charge climbs towards its final value Cε while the current dies away.

Symbols and units

$q (t)$	charge on the capacitor at time t	C
$ε$	emf of the source	V
$R$	series resistance	Ω
$C$	capacitance	F
$R C$	time constant τ	s

When it applies

Series R and C on a constant emf, switched at t = 0.

Don't use when

Assumes a series R and C on a constant EMF switched on at t = 0 — a time-varying source requires solving the differential equation directly instead.

Derivation

Apply the loop rule at any instant, with the capacitor voltage q/C:

ε - \frac{q}{C} - I R = 0

The current is the rate at which charge arrives, which turns it into a differential equation:

ε - \frac{q}{C} - R \frac{d q}{d t} = 0

Separate the variables:

\frac{d q}{C ε - q} = \frac{d t}{R C}

Integrate from q = 0 at t = 0:

- ln (\frac{C ε - q}{C ε}) = \frac{t}{R C}

Exponentiate and solve for q:

q (t) = C ε (1 - e^{- t / R C})

Differentiate to get the current:

I (t) = \frac{d q}{d t} = \frac{ε}{R} e^{- t / R C}

Worked examples

Timing

R = 10 kΩ with C = 100 µF:

τ = R C = 1.0 s

q (τ) = 0.63 C ε, q (3 τ) = 0.95 C ε, q (5 τ) = 0.993 C ε

Energy balance over a full charge

W_{source} = C ε^{2}, U_{C} = \frac{1}{2} C ε^{2}, W_{R} = \frac{1}{2} C ε^{2}

Exactly half the energy supplied is dissipated in the resistor, whatever its value.

Watch out for

Two limits solve most problems with no algebra: at switch-on an uncharged capacitor acts like a plain wire, and long afterwards like an open circuit.

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