use equations of the form x = x0 e–(t / RC) where x could represent current, charge or potential difference for a capacitor discharging through a resistor

🔋 Discharging a Capacitor

When a charged capacitor is connected across a resistor, it loses its stored energy. The charge, voltage, and current all decrease exponentially with time. This behaviour is described by the same mathematical form:

📉 The exponential decay equation

For any quantity x that decays during discharge we can write:

\$x = x_0 \, e^{-\frac{t}{RC}}\$

  • \$x_0\$ – initial value at t = 0 (charge, voltage or current)

  • \$R\$ – resistance in ohms (Ω)

  • \$C\$ – capacitance in farads (F)

  • \$RC\$ – time constant, often denoted τ (tau)

  • \$t\$ – time elapsed since the discharge began (seconds)

After one time constant (t = RC) the quantity has fallen to about 37 % of its initial value (e⁻¹ ≈ 0.37). After five time constants it is less than 1 % of the start value – effectively fully discharged.

🔢 Applying the formula to charge, voltage and current

  1. Charge (\$Q\$): \$Q = Q_0 e^{-t/(RC)}\$

  2. Voltage across the capacitor (\$V\$): Since \$V = Q/C\$, we get \$V = V0 e^{-t/(RC)}\$ where \$V0 = Q_0/C\$

  3. Current through the resistor (\$I\$): From Ohm’s law \$I = V/R\$, thus \$I = I0 e^{-t/(RC)}\$ with \$I0 = V_0/R\$

All three quantities share the same exponential decay curve; they only differ in their initial values.

🧮 Example calculation

A capacitor of C = 100 µF is charged to V₀ = 12 V and then discharged through a resistor of R = 2 kΩ.

  • Time constant: \$\tau = RC = (2000\,\Omega)(100\times10^{-6}\,F) = 0.2\,s\$

  • Initial charge: \$Q0 = CV0 = (100\times10^{-6})(12) = 1.2\times10^{-3}\,C\$

  • Initial current: \$I0 = V0/R = 12/2000 = 0.006\,A = 6\,mA\$

Find the voltage after t = 0.5 s:

\$V = V_0 e^{-t/\tau} = 12\,e^{-0.5/0.2} = 12\,e^{-2.5} \approx 12 \times 0.082 = 0.98\,V\$

So after half a second the voltage has dropped to roughly 1 V.

📊 Values at multiples of the time constant

Time tFraction remaining \$e^{-t/\tau}\$Example voltage (if \$V_0=12\,V\$)
01.00012.0 V
\$\tau\$0.3684.4 V
2\$\tau\$0.1351.6 V
3\$\tau\$0.0500.6 V
5\$\tau\$0.0070.08 V

💡 The table shows how quickly the voltage falls – after just two time constants it is already below 2 V.

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Summary

During discharge a capacitor’s charge, voltage and current all follow the same exponential law:

\$x = x_0 e^{-t/(RC)}\$

Knowing the resistance and capacitance lets you predict how fast the capacitor will lose its energy – a key skill for circuits involving timers, filters and pulse generators.

📚 Cambridge A‑Level Physics 9702 – Topic: Discharging a capacitor