RC circuit transient response

By the end of this topic you will be able to

Write the voltage and current equations for charging and discharging an RC circuit.
Calculate and interpret the time constant τ = RC.
Estimate how long a capacitor takes to reach a given fraction of its final voltage.

A capacitor stores charge

A capacitor stores energy in an electric field. Its voltage depends on the charge stored, $v_C = q/C$ , so its voltage cannot change instantly: charge must flow in or out through the rest of the circuit. With a resistor in the way, that takes time, and the voltage follows a smooth exponential curve.

Charging

Connect a capacitor, initially empty, through a resistor $R$ to a supply $V$ . At the start the capacitor is like a short circuit and the current is $V/R$ . As charge builds, the capacitor voltage rises and the current falls, until the capacitor voltage equals the supply and the current stops.

v_C(t) = V\left(1 - e^{-t/\tau}\right), \qquad i(t) = \frac{V}{R}e^{-t/\tau}, \qquad \tau = RC

The time constant

The time constant $\tau = RC$ (seconds, with $R$ in ohms and $C$ in farads) measures how quickly the circuit responds.

Time	Charging reaches
$1\tau$	63.2% of V
$2\tau$	86.5%
$3\tau$	95.0%
$5\tau$	99.3% (regarded as fully charged)

Discharging

If a charged capacitor discharges through a resistor, the voltage decays by the same exponential: $v_C(t) = V\,e^{-t/\tau}$ . After one time constant 36.8% of the original voltage remains.

Interactive modelChange R, C and the time. See the 63% mark at one time constant.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

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Worked example

A $100\ \mu\text{F}$ capacitor is charged from a 10 V supply through $10\ \text{k}\Omega$ . Find the time constant, the capacitor voltage after 1 s, and the initial current.

Time constant. $\tau = RC = 10\,000\times100\times10^{-6} = 1.0$ s.

Voltage after 1 s $(t = \tau)$ : $v_C = 10(1 - e^{-1}) = 6.32$ V.

Initial current. $i(0) = V/R = 10/10\,000 = 1.0$ mA, decaying to 0.37 mA after $\tau$ .

Time to reach 9 V (90%): $1 - e^{-t/\tau} = 0.9$ gives $t = \tau\ln10 = 2.30$ s.

Energy at full charge: $\tfrac12\times100\times10^{-6}\times10^2 = 5$ mJ.

Practice problems

Find the time constant for $R = 47\ \text{k}\Omega$ and $C = 22\ \mu\text{F}$ .

Show answer

$\tau = 47\,000\times22\times10^{-6} = 1.03$ s.
A discharging capacitor starts at 12 V with $\tau = 2$ s. What is the voltage after 4 s?

Show answer

$12e^{-2} = 1.62$ V.
How long does a charging RC circuit take to reach 99% of the supply voltage?

Show answer

$e^{-t/\tau} = 0.01$ , so $t = 4.6\tau$ (about 5τ).
In the model, doubling R doubles what?

Show answer

The time constant, so the curve takes twice as long. The final voltage and the stored energy are unchanged.

Key formulae

Time constant: $\tau = RC$
Charging voltage: $v_C(t) = V\left(1 - e^{-t/\tau}\right)$
Discharging voltage: $v_C(t) = V\,e^{-t/\tau}$
Current while charging: $i(t) = \frac{V}{R}\,e^{-t/\tau}$
Energy stored: $E = \tfrac{1}{2} C v_C^2$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (40 min). 5 min: camera flash example; 10 min: derive charging from KVL (or state the result); 10 min: time constant table; 15 min: model, discharging, and worked example.

Activity. Give students a target (voltage at a time) and ask them to choose R and C in the model to achieve it.

Discussion. Where are RC circuits used? (Timers, filters, debouncing switches, camera flash.)

Assessment idea. A two-stage RC with a switch that changes the time constant.