Learn › Topic 6

Capacitors, Inductors and RC and RL Circuits: The Time Constant τ

8 min read · bitGateLab Team

Resistors respond instantly: you change the voltage and the current changes at that same moment. Capacitors and inductors do not; they store energy, so they take time to react. That delay is the basis of timers, filters, switching power supplies and button debouncing. Here you'll learn how each one behaves, what the time constant τ is, and how to calculate voltages and currents at any instant during a charge or discharge.

The capacitor

A capacitor consists of two conducting plates separated by an insulator. When you apply a voltage to it, positive charge builds up on one plate and negative charge on the other. Its capacitance C (in farads, F) tells you how much charge it stores per volt: Q = C · V. In practice, µF, nF and pF are used.

The current in a capacitor depends on how fast its voltage is changing:

i = C · dv/dt        E = ½ · C · V²

Two important consequences follow from this:

The voltage across a capacitor cannot change abruptly; that would take an infinite current.
At DC, once it is charged, dv/dt = 0 and no current flows: the capacitor behaves like an open circuit.

Capacitors in parallel add (C_eq = C₁ + C₂); in series they combine like resistors in parallel. A 100 µF capacitor charged to 5 V stores ½ × 100×10⁻⁶ × 25 = 1.25 mJ.

The inductor

An inductor is a coil of wire. The current flowing through it creates a magnetic field that stores energy. Its inductance L (in henries, H) relates the voltage to the rate of change of the current:

v = L · di/dt        E = ½ · L · I²

Its behavior is the dual of the capacitor's:

The current in an inductor cannot change abruptly. If you try to cut it off suddenly, the inductor generates a very high voltage spike; that is why relay coils have a diode connected in antiparallel across them.
At DC, in steady state, di/dt = 0 and there is no voltage: the inductor behaves like a wire.

Inductors combine like resistors: in series they add, and in parallel it is "product over sum".

Charging and discharging an RC circuit

Connect a source V_s, a resistor R and an uncharged capacitor in series, and close the switch at t = 0. At first the capacitor is at 0 V, so the entire source voltage drops across R and the current is at its maximum, V_s/R. As the capacitor charges, the voltage across R shrinks, and so does the current. The result is an exponential curve:

Charging:    v_C(t) = V_s · (1 − e^−t/τ)        i(t) = (V_s/R) · e^−t/τ
Discharging: v_C(t) = V₀ · e^−t/τ
             τ = R · C

The time constant τ (tau) is measured in seconds (ohms times farads gives seconds). It is the time the capacitor takes to cover 63.2 % of the way to its final value.

Charging (solid line) and discharging (dashed line) of a capacitor. At t = τ the charging curve reaches 63.2 % and the discharging curve has fallen to 36.8 %.

These percentages come up in every RC or RL circuit, so it is worth keeping them handy:

Time	Charging (% of final)	Discharging (% of initial)
1τ	63.2 %	36.8 %
2τ	86.5 %	13.5 %
3τ	95.0 %	5.0 %
4τ	98.2 %	1.8 %
5τ	99.3 %	0.7 %

By convention, the circuit is said to have reached steady state after 5τ.

Example 1. A 100 µF capacitor charges through 10 kΩ from a 5 V source (a step from 0 to 5 V at t = 0).

τ = R · C = 10 000 Ω × 100×10⁻⁶ F = 1 s.
Initial current: 5 V / 10 kΩ = 0.5 mA.
v_C(1 s) = 5 × 0.632 = 3.16 V; v_C(2 s) = 5 × 0.865 = 4.32 V; v_C(3 s) = 5 × 0.950 = 4.75 V.
Time to reach the halfway point (2.5 V): 1 − e^−t/τ = 0.5, so t = τ · ln 2 = 0.693 s.
Practical steady state: 5τ = 5 s, with v_C ≈ 4.97 V and practically zero current.

Example 2. A 47 µF capacitor charged to 12 V discharges through 22 kΩ. How long does it take to fall to 1 V?

τ = 22 000 × 47×10⁻⁶ = 1.034 s.
From 1 = 12 · e^−t/τ: t = τ · ln(12/1) = 1.034 × 2.485 = 2.57 s.

The RL circuit

With an inductor in series with a resistor, something very similar happens, but now the variable that cannot jump is the current. When a voltage step V_s is applied, the current starts at zero and rises toward V_s/R, while the voltage across the inductor starts at V_s and falls to zero:

i(t) = (V_s/R) · (1 − e^−t/τ)        v_L(t) = V_s · e^−t/τ        τ = L / R

Watch out with the formula: in the RL circuit, a larger resistance makes the response faster, the opposite of the RC circuit.

Example 3. A 100 mH inductor in series with 50 Ω is connected to 10 V.

τ = L / R = 0.1 H / 50 Ω = 2 ms.
Final current: 10 V / 50 Ω = 0.2 A = 200 mA.
At t = 1 ms (half of τ): i = 0.2 × (1 − e^−0.5) = 0.2 × 0.3935 = 78.7 mA.
At t = 2 ms: i = 0.2 × 0.632 = 126.4 mA. By 10 ms (5τ) it is practically at 200 mA.
Energy stored at the end: ½ × 0.1 × 0.2² = 2 mJ.

Key point: to solve any first-order circuit with a single equivalent resistance, you only need three things: the initial value, the final value and τ. The response is always x(t) = x_final + (x_initial − x_final) · e^−t/τ. If there are more resistors, use the Thévenin resistance seen from the capacitor or the inductor.

Common mistakes

Forgetting to convert units: 100 µF is 100×10⁻⁶ F and 100 mH is 0.1 H.
Using τ = R/L or τ = L·R in the RL circuit; it is L/R.
Assuming the capacitor is "full" after 1τ; at that point it is only at 63.2 %.
Thinking that the current in a capacitor follows its voltage; in fact it is greatest when the voltage is changing fastest, at the start of the charge.
Opening a circuit that has an energized inductor without leaving a path for its current; the voltage spike can damage a transistor or a switch.

Try it in the simulator

With the virtual oscilloscope you can see the exponential curve and measure τ directly on it. Use a square wave as the source to watch the charge and discharge over and over.

Place a square-wave source from 0 to 5 V with a period of at least 10τ (for example, 10 s for Example 1, or use smaller values, such as 10 kΩ and 100 nF, with τ = 1 ms and a period of 10 ms).
Connect the resistor and the capacitor in series.
Connect one oscilloscope channel to the source and another to the capacitor.
Look at the charging curve and measure the time it takes to reach 3.16 V; that is τ.
Double the resistance and notice that the curve takes twice as long.
Repeat with an inductor and a resistor, watching the current instead of the voltage.

Exercises

Calculate τ and the time to reach steady state (5τ) for R = 1 kΩ and C = 470 µF.

Show answer

τ = 1 000 × 470×10⁻⁶ = 0.47 s. 5τ = 2.35 s.
An RC circuit with τ = 2 s charges from 0 V with a 9 V source. What is the capacitor voltage at t = 3 s?

Show answer

v_C = 9 × (1 − e^−1.5) = 9 × (1 − 0.2231) = 9 × 0.7769 ≈ 6.99 V.
A 20 mH inductor in series with 100 Ω is connected to 12 V. Find τ, the final current and the current at t = τ.

Show answer

τ = 0.02 / 100 = 0.2 ms. Final current: 12 / 100 = 120 mA. At t = τ: 0.632 × 120 ≈ 75.9 mA.
How much energy does a 1 000 µF capacitor charged to 25 V store?

Show answer

E = ½ × 1×10⁻³ × 25² = ½ × 10⁻³ × 625 = 0.3125 J.

To calculate the equivalent resistance seen from the capacitor, review Thévenin, Norton and superposition.