RC circuits: charging a capacitor

Physics · Senior school physics and first-year engineering · about 10 min

Connect a capacitor to a battery through a resistor and it doesn't charge instantly: the voltage rises quickly at first, then more and more slowly. That curve shows up in camera flashes, timer circuits and sensor filters, and it's all controlled by one number, the time constant.

The key idea

For a capacitor $C$ charging through a resistor $R$ from a supply $V_0$ :

V_C(t) = V_0\left(1 - e^{-t/\tau}\right) \qquad \tau = R C

and when discharging, $V_C(t) = V_0\, e^{-t/\tau}$ .

A circuit diagram in thin black lines with standard symbols. A rectangular loop with a battery on the left side, an open switch and a zigzag resistor along the top, and a capacitor on the right side. — A capacitor charging through a resistor.

What the time constant means

After one time constant ( $t = \tau$ ), a charging capacitor reaches $1 - e^{-1} \approx 63\%$ of the supply voltage. After $5\tau$ it's above $99\%$ , which is usually treated as fully charged.

$\tau = RC$ has units of seconds: a $10\,\text{k}\Omega$ resistor with a $100\,\mu\text{F}$ capacitor gives $\tau = 1$ s.

A graph with capacitor voltage on the vertical axis and time on the horizontal axis. A smooth curve rises from the origin, steeply at first, flattening towards a horizontal thin dashed line at the supply voltage. Thin dashed lines mark the point where t = τ: a vertical line up from the time axis and a horizontal line across from the voltage axis, at 63% of the height of the supply line. A second tick on the time axis at five times that distance, where the curve almost reaches the dashed line. — After one time constant the capacitor reaches about 63% of the supply voltage.

Why it slows down

At the start the capacitor is empty, so the full supply voltage drives a big current through $R$ . As it charges, its own voltage pushes back, leaving less voltage across the resistor, so the current, and the rate of charging, keep falling. That's what produces the exponential curve.

Worked example: charging through 10 kΩ

A $100\,\mu\text{F}$ capacitor charges through a $10\,\text{k}\Omega$ resistor from a $5$ V supply. Find the time constant and the capacitor voltage after $2$ s.

Show the solution

Time constant:

\tau = RC = 10\,000 \times 100 \times 10^{-6} = 1 \text{ s}

Voltage after $2$ s:

V_C = 5\left(1 - e^{-2/1}\right) = 5\,(1 - 0.1353) = 4.32 \text{ V}

That's $86\%$ of the supply after two time constants.

Common mistakes

Forgetting the unit prefixes

$\mu$ F is $10^{-6}$ F and k $\Omega$ is $10^3\,\Omega$ . Getting these wrong changes $\tau$ by a factor of a million.

Thinking it's fully charged after one time constant

After $\tau$ it's only at $63\%$ . Use about $5\tau$ for "fully" charged.

Using the charging formula for discharging

Discharging is $V_0 e^{-t/\tau}$ : it falls towards zero, without the $1 - \dots$ .

Check your understanding

1. What is the time constant for $R = 2\,\text{k}\Omega$ and $C = 50\,\mu\text{F}$ ?

2. After one time constant, a charging capacitor has reached about what fraction of the supply voltage?

3. What happens to the charging current as the capacitor fills up?

Still not sure? Talk it through.

Waypoint Mentor's tutor explains this out loud, works through your own problems on a whiteboard, and checks you've got it. Free to start.

Explain this to me