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.

Applying Kirchoff's loop rule:

e - IR - Q/C = 0

As Q increases I decreases, but Q changes because there is a current I. As the current decreases Q changes more slowly.

I = dQ/dt, so the equation can be written:

e - R (dQ/dt) - Q/C = 0

This is a differential equation that can be solved for Q as a function of time. The solution (derived in the text) is:

Q(t) = Q_{o} [ 1 - e^{-t/t} ]

where Q_{o} = C e and the time constant t = RC.

Differentiating this expression to get the current as a function of time gives:

I(t) = (Q_{o}/RC) e^{-t/t} = I_{o} e^{-t/t}

where I_{o} = 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.

time | current |
---|---|

0 | I_{o} |

1*t | I_{o}/e = 0.368 I_{o} |

2*t | I_{o}/e^{2} = 0.135 I_{o} |

3*t | I_{o}/e^{3} = 0.050 I_{o} |

What happens if the capacitor is now fully charged 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.

Applying Kirchoff's loop rule:

-IR - Q/C = 0

I = dQ/dt, so the equation can be written:

R (dQ/dt) = -Q/C

This is a differential equation that can be solved for Q as a function of time. The solution is:

Q(t) = Q_{o} e^{-t/t}

where Q_{o} is the initial charge on the capacitor and the time constant t = RC.

Differentiating this expression to get the current as a function of time gives:

I(t) = -(Q_{o}/RC) e^{-t/t} = -I_{o} e^{-t/t}

where I_{o} = Q_{o}/RC

Note that, except for the minus sign, this is 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.

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