Charging and discharging: RC circuits

Charging a capacitor through a resistor

When a capacitor is connected through a resistor to a supply of voltage $V_0$:

• At the instant of connection ($t = 0$):
  • the capacitor is uncharged, so $V_C = 0$,
  • the full supply voltage appears across the resistor,
  • the current is therefore at its maximum, $I_0 = \dfrac{V_0}{R}$.
• As charging proceeds:
  • charge accumulates, so $V_C$ rises,
  • less voltage remains across the resistor, so the current falls,
  • the falling current means charge accumulates more slowly — which is why the curves are exponential rather than straight.
• After a long time ($t \to \infty$):
  • $V_C \to V_0$ and the capacitor is fully charged,
  • the current falls to zero, and no more charge flows.

The shapes of the three graphs

Quantity	Charging	Discharging
Charge $Q$	rises from 0, levelling off at $Q_0$	falls from $Q_0$ toward 0
Voltage $V_C$	rises from 0, levelling off at $V_0$	falls from $V_0$ toward 0
Current $I$	falls from $I_0$ toward 0	falls from $I_0$ toward 0, in the opposite direction

• Note that the current decays in both cases — this catches people out. During discharge the current flows the other way, but its magnitude still falls from a maximum toward zero.
• $Q$ and $V_C$ always have the same shape, because $Q = CV$ with $C$ constant.
• The curves approach their final values asymptotically — strictly they never quite arrive.

Watch the three curves develop together below, and change $R$ and $C$ to see the timescale stretch and shrink:

The time constant

$$\tau = RC$$

• $\tau$ — the time constant (seconds)

• $R$ — resistance (Ω), $C$ — capacitance (F)

• Check the units: (Ω)(F) $=$ seconds. If your answer is not in seconds, a prefix has been missed.

• The time constant is the single number that sets the timescale of the whole process. It does not change during charging or discharging.

What $\tau$ means:

• Charging: after one time constant the capacitor has reached $63\%$ of its final charge and voltage.
• Discharging: after one time constant the charge and voltage have fallen to $37\%$ of their initial values.
• The current, which always decays, falls to $37\%$ of its initial value after one $\tau$ in both cases.

Time elapsed	Charging (% of final)	Discharging (% of initial)
$1\tau$	63%	37%
$2\tau$	86%	14%
$3\tau$	95%	5%
$5\tau$	99%	1%

• A capacitor is usually treated as fully charged or discharged after about $5\tau$.

Finding the time constant from a graph

• From a discharge curve: find the time at which the value has fallen to $37\%$ of its initial value. That time is $\tau$.
• From a charging curve: find the time to reach $63\%$ of the final value.
• Using the initial gradient: the tangent to the curve at $t = 0$, extended, meets the final value at $t = \tau$. This is often more accurate than reading a percentage off a flattening curve.

What changes the timescale

• Larger $R$ → larger $\tau$ → slower charging, because the resistor limits the current.
• Larger $C$ → larger $\tau$ → slower charging, because more charge is needed to reach a given voltage.
• The final charge and voltage depend on $C$ and $V_0$ but not on $R$. Changing the resistor changes how fast, never how much.