Circuits with resistors and batteries have time-independent solutions: the current doesn't change as time goes by. Adding one or more capacitors changes this. The solution is then time-dependent: the current is a function of time.

Consider a series RC circuit with a battery, resistor, and capacitor in series. The capacitor is initially uncharged, but starts to charge when the switch is closed. Initially the potential difference across the resistor is the battery emf, but that steadily drops (as does the current) as the potential difference across the capacitor increases. If t=0 represents the time when the switch is closed:

The capacitor voltage is given by V_C(t) = e [ 1 - e^-t/t ]

where t = RC is known as the time constant in the circuit.

The resistor voltage is V_R(t) = e e^-t/t

The current in the circuit is given by:

I(t)

=

e R

e-t/t

=

Io e-t/t

where

Io

=

e R

is the maximum current possible in the circuit.

The time constant t = RC determines how quickly the capacitor charges. If RC is small the capacitor charges quickly; if RC is large the capacitor charges more slowly.

What happens if the capacitor has a charge Q_o and is then discharged through the resistor? Now the potential difference across the resistor is the capacitor voltage, but that decreases (as does the current) as time goes by.

The capacitor voltage is given by

VC(t)

=

Qo C

e-t/t

The resistor voltage is

VR(t)

=

-Qo C

e-t/t

The current in the circuit is:

I(t)

=

-Qo RC

e-t/t

=

-Io e-t/t

where

Io

=

Qo RC

Note that, except for the minus sign, this is basically the same expression for current we had when the capacitor was charging. The minus sign simply indicates that the charge flows in the opposite direction.