Charging, discharging and the time constant

Key ideas

A capacitor charging or discharging through a resistor changes exponentially — fast at first, then ever slower, because the current depends on how much voltage is still to change.
The pace is set by the time constant:

$\tau = RC$

$\tau$ — the time constant (s)
$R$ — the resistance in the charging/discharging path (Ω)
$C$ — the capacitance (F)
In one time constant:
- a charging capacitor's voltage (and charge) rises to 63% of its final value
- a discharging capacitor's voltage, charge and current all fall to 37% of where they started
After about 5τ the process is essentially complete.
The graphs (all exponential):
- charging: $V$ and $Q$ rise (63% at $\tau$ ); the current starts maximum and decays — a full capacitor takes no current
- discharging: $V$ , $Q$ and $I$ all decay together (37% at $\tau$ )

Two graphs for a capacitor charging through a resistor: the voltage rises along an exponential curve reaching 63 percent of the supply at one time constant, while the current starts at its maximum and decays exponentially toward zero.

Why it's exponential

Early on, the capacitor is empty, the full supply voltage sits across $R$ , and the current is large — charge arrives quickly.
As the capacitor fills, its voltage opposes the supply, less voltage remains across $R$ , and the current shrinks.
Less current → slower charging → the curve flattens. The rate is always proportional to what's left to do.

Worked ExampleA timing circuit

A $470$ µF capacitor charges through a $10$ kΩ resistor from a $9.0$ V supply. Find the time constant, and the voltage across the capacitor one time constant after switch-on.

Step 1 — Time constant

$\tau = RC = (10\times10^{3})(470\times10^{-6}) = 4.7 \text{ s}$

Step 2 — Voltage after one time constant (charging → 63%)

$V = 0.63 \times 9.0 = 5.7 \text{ V}$

Step 3 — Interpret

After $\tau = 4.7$ s the capacitor is at $5.7$ V; after about $5\tau \approx 24$ s it is effectively full at $9.0$ V.

Tips

Ask "is this quantity rising or decaying?" before reaching for 63% or 37%. Charging voltage rises (63%); charging current decays (37%); discharging everything decays (37%).
A bigger $R$ or bigger $C$ both slow the process — useful for camera flashes (charge slowly, dump quickly through a tiny resistance).

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A $100$ µF capacitor discharges through a $22$ kΩ resistor.

Show that the time constant is $2.2$ s.

Merit

A capacitor charged to $12$ V discharges through a resistor with time constant $3.0$ s.

Calculate the voltage after $3.0$ s and after $6.0$ s.

Excellence

Explain why the current in a charging RC circuit is largest at the instant of switch-on and then decreases, and why the capacitor's voltage–time graph has its steepest slope at the start.

Charging, discharging and the time constant

Covers AS91526

Key ideas

A capacitor charging or discharging through a resistor changes exponentially — fast at first, then ever slower, because the current depends on how much voltage is still to change.
The pace is set by the time constant:

$\tau = RC$

$\tau$ — the time constant (s)
$R$ — the resistance in the charging/discharging path (Ω)
$C$ — the capacitance (F)
In one time constant:
- a charging capacitor's voltage (and charge) rises to 63% of its final value
- a discharging capacitor's voltage, charge and current all fall to 37% of where they started
After about 5τ the process is essentially complete.
The graphs (all exponential):
- charging: $V$ and $Q$ rise (63% at $\tau$ ); the current starts maximum and decays — a full capacitor takes no current
- discharging: $V$ , $Q$ and $I$ all decay together (37% at $\tau$ )

Why it's exponential

Early on, the capacitor is empty, the full supply voltage sits across $R$ , and the current is large — charge arrives quickly.
As the capacitor fills, its voltage opposes the supply, less voltage remains across $R$ , and the current shrinks.
Less current → slower charging → the curve flattens. The rate is always proportional to what's left to do.

Worked ExampleA timing circuit

A $470$ µF capacitor charges through a $10$ kΩ resistor from a $9.0$ V supply. Find the time constant, and the voltage across the capacitor one time constant after switch-on.

Step 1 — Time constant

$\tau = RC = (10\times10^{3})(470\times10^{-6}) = 4.7 \text{ s}$

Step 2 — Voltage after one time constant (charging → 63%)

$V = 0.63 \times 9.0 = 5.7 \text{ V}$

Step 3 — Interpret

After $\tau = 4.7$ s the capacitor is at $5.7$ V; after about $5\tau \approx 24$ s it is effectively full at $9.0$ V.

Tips

Ask "is this quantity rising or decaying?" before reaching for 63% or 37%. Charging voltage rises (63%); charging current decays (37%); discharging everything decays (37%).
A bigger $R$ or bigger $C$ both slow the process — useful for camera flashes (charge slowly, dump quickly through a tiny resistance).

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A $100$ µF capacitor discharges through a $22$ kΩ resistor.

Show that the time constant is $2.2$ s.

Merit

A capacitor charged to $12$ V discharges through a resistor with time constant $3.0$ s.

Calculate the voltage after $3.0$ s and after $6.0$ s.

Excellence

Explain why the current in a charging RC circuit is largest at the instant of switch-on and then decreases, and why the capacitor's voltage–time graph has its steepest slope at the start.