Consider the circuit below of a cell, switch, resistor and capacitor:

Let's assume that the circuit initially starts with the switch open, and that at a certain time t_{0} the switch is abruptly closed.

What potential difference will initially be present across the capacitor terminals? What will the potential difference be across the capacitor terminals after a long time has passed?

At time t_{0} when the switch is first closed we can assume that the capacitor has no charge, and so the potential difference across the capacitor terminals will be 0 Volts. The switch is assumed to be "perfect", in the sense that it can instantly close. The initial current will be (E/R) Amps at t_{0} but will fall as time progresses. In other words, the current is a maximum at t_{0}, and falls from that point on. For the very briefest of moments (infinitessimally small), the current is (E/R).

MathML formula | Formula Image | Straight text formula |
---|---|---|

$$v\left(t\right)=E\left(1-{e}^{-\left(\frac{1}{RC}\right)t}\right)$$ | v(t) = E(1-e^(-(1/(RC))t)) |

Where

v(t) = potential difference across capacitor at time t seconds after switch has been closed, Volts

E = Cell Voltage = Voltage that would be present across the capacitor terminals if the resistor was not present and the switch was closed, Volts

C = capacitance, Farads

R = resistance, Ohms

# "Real World" RC Circuits

Consider a "real world" example. The cell, interconnecting wires, switch and capacitor plates will all have some resistance. So long as these resistances are small in comparison to that of the resistor, however, then the formulas already given will give results that are accurate enough for all practical circuits.