Capacitor Discharging in RC Circuits

When a charged capacitor is connected to a resistor, it does not discharge instantly. Instead, it releases its stored energy gradually, with the voltage (and current) falling in a smooth curve that gets shallower and shallower over time. This pattern is called exponential decay. Think of it like a bath draining: water flows out quickly at first when the pressure is high, then more and more slowly as the level drops. The capacitor behaves identically — high voltage drives a large current early on, but as voltage falls, so does the current, slowing the discharge further. The rate at which this decay happens depends on two circuit components: the resistance R and the capacitance C. Together they form the time constant τ (tau), measured in seconds, defined as τ = RC. After one time constant, the voltage has fallen to about 37% of its initial value. After five time constants, it has effectively reached zero — engineers consider the capacitor fully discharged after 5τ. This relationship lets engineers precisely control timing in circuits, from camera flashes and car indicators to medical defibrillators and computer memory. Understanding RC discharging is essential because it bridges two big ideas: the calculus concept of exponential functions, and the physical behaviour of real circuit components. The equations governing discharge are derived directly from Kirchhoff's voltage law, making this a beautiful example of how fundamental circuit laws predict real, measurable behaviour.

Key takeaways

Setting Up the Circuit
Applying Kirchhoff's Voltage Law
Solving the Differential Equation
The Time Constant τ = RC
Energy Considerations

Setting Up the Circuit

Consider a capacitor with initial charge Q₀ (and initial voltage V₀ = Q₀/C) connected in series with a resistor R. There is no battery — the capacitor itself is the energy source. At t = 0, a switch is closed and charge begins to flow through R.

Applying Kirchhoff's Voltage Law

At any instant, the voltage across the capacitor equals the voltage across the resistor. The capacitor voltage is V = Q/C, and the resistor voltage is V = IR. Since current I = −dQ/dt (charge is leaving the capacitor), we get Q/C = −R(dQ/dt), which rearranges to dQ/dt = −Q/(RC). This is a first-order differential equation whose solution involves an exponential function.

Solving the Differential Equation

Separating variables and integrating gives Q(t) = Q₀ e^(−t/RC). Since V = Q/C, the voltage across the capacitor at time t is V(t) = V₀ e^(−t/RC). The current follows from I = V/R, giving I(t) = I₀ e^(−t/RC), where I₀ = V₀/R is the initial current.

The Time Constant τ = RC

The quantity RC has units of ohms × farads = seconds, and is called the time constant τ. It sets the timescale of the decay. At t = τ, the voltage has dropped to V₀ × e⁻¹ ≈ 0.368 V₀, roughly 37% of its starting value. At t = 2τ it is about 13.5%, at t = 5τ it is less than 1%. A larger R or larger C means a slower discharge — more resistance restricts current flow, and greater capacitance means more charge must move before the voltage drops.

Energy Considerations

The energy stored in a fully charged capacitor is E = ½CV₀². As the capacitor discharges, this energy is entirely converted to heat in the resistor. The rate of energy dissipation decreases exponentially too, following P(t) = V²(t)/R = (V₀²/R) e^(−2t/RC). Doubling R slows the discharge but the total energy dissipated remains the same — it just takes longer to release.

The Key Equation

V(t) = V_0 \, e^{-t/RC}

V(t) — Voltage at time t (Volts (V)): The potential difference across the capacitor at time t after discharge begins.
V_0 — Initial voltage (Volts (V)): The voltage across the capacitor at t = 0, just before discharge starts.
e — Euler's number (Dimensionless): The mathematical constant ≈ 2.718, base of the natural logarithm. It appears naturally in all growth and decay processes.
t — Time (Seconds (s)): The time elapsed since the switch was closed and discharge began.
R — Resistance (Ohms (Ω)): The resistance through which the capacitor discharges. Larger R means slower discharge.
C — Capacitance (Farads (F)): The capacitance of the capacitor. Larger C means more charge stored and therefore a slower discharge.

Worked Example

Problem. A 470 μF capacitor is charged to 9.0 V and then allowed to discharge through a 2200 Ω resistor. (a) Calculate the time constant τ. (b) Find the voltage across the capacitor after 2.0 s. (c) How long does it take for the voltage to fall to 1.0 V?

\tau \approx 1.03\,\text{s};\quad V(2.0\,\text{s}) \approx 1.30\,\text{V};\quad t_{1\text{V}} \approx 2.27\,\text{s}

Answer. τ ≈ 1.03 s; V(2.0 s) ≈ 1.30 V; time to reach 1.0 V ≈ 2.27 s

Figure

Diagram

Origins · 1 minLord Kelvin (William Thomson) and how this was discovered

Lord Kelvin (William Thomson)

1824–1907 · British (Scottish–Irish)

Lord Kelvin made foundational contributions to the understanding of transient electrical behaviour in capacitive and resistive circuits. His analysis of the telegraph cable — essentially a massive RC transmission line — in 1855 required him to solve precisely the equations governing RC time constants. His work explained why signals sent through the Atlantic telegraph cable arrived slowly and distorted, and he devised instruments and signalling methods to compensate. This directly established the practical importance of the RC time constant in electrical engineering. Kelvin was so confident he could retire wealthy from his telegraph work that he invested heavily in the Atlantic telegraph project — and nearly went bankrupt when the first cable failed. He eventually recovered his fortune when a redesigned cable succeeded in 1866.

Capacitor Charging in RC Circuits — The charging process is the direct counterpart to discharging, described by a complementary exponential equation V(t) = V₀(1 − e^(−t/RC)). Understanding both charging and discharging gives a complete picture of RC circuit behaviour.
Kirchhoff's Voltage Law — The discharge equation is derived directly by applying Kirchhoff's Voltage Law around the RC loop. Without KVL, there is no rigorous starting point for deriving V(t) = V₀e^(−t/RC).
Capacitors and Capacitance — A solid understanding of how capacitors store charge (Q = CV) and energy (E = ½CV²) is essential before analysing what happens when they discharge through a resistor.
Resistance and Ohm's Law — The resistor controls the discharge current via V = IR. Without Ohm's Law, the relationship between voltage and current during discharge cannot be established.