recall and use τ = RC for the time constant for a capacitor discharging through a resistor

Discharging a Capacitor

Learning Objective

Recall and use the time‑constant formula \$\tau = RC\$ for a capacitor discharging through a resistor.

Key Concepts

• A capacitor stores electric charge \$Q\$ and energy \$U = \frac{1}{2}CV^{2}\$.
• When the capacitor is connected across a resistor \$R\$, the charge leaks away – the circuit “discharges”.
• The rate of discharge is exponential and is characterised by the time constant \$\tau\$.

Derivation of the Discharge Equation

Consider a capacitor \$C\$ initially charged to voltage \$V_{0}\$ and then connected to a resistor \$R\$.

1. Apply Kirchhoff’s loop rule: \$V{C} + V{R} = 0\$
2. Express the voltages: \$\frac{Q}{C} + IR = 0\$
3. Since \$I = -\dfrac{dQ}{dt}\$ (current flows opposite to decreasing charge), substitute:

   \$\frac{Q}{C} - R\frac{dQ}{dt}=0\$

4. Rearrange to a first‑order differential equation:

   \$\frac{dQ}{dt} = -\frac{Q}{RC}\$

5. Integrate:

   \$\int{Q{0}}^{Q} \frac{dQ'}{Q'} = -\int_{0}^{t} \frac{dt'}{RC}\$

   \$\ln\!\left(\frac{Q}{Q_{0}}\right) = -\frac{t}{RC}\$

6. Exponentiate to obtain the charge as a function of time:

   \$Q(t) = Q_{0}\,e^{-t/RC}\$

7. Since \$V = Q/C\$, the voltage also decays exponentially:

   \$V(t) = V_{0}\,e^{-t/RC}\$

Time Constant \$\tau\$

The product \$RC\$ is defined as the time constant \$\tau\$:

\$\tau = RC\$

Interpretation:

• At \$t = \tau\$, the voltage (or charge) has fallen to \$e^{-1}\approx 37\%\$ of its initial value.
• After \$5\tau\$, the capacitor is considered effectively discharged (<0.7% of \$V_{0}\$).

Example Calculation

Problem: A \$10\;\mu\text{F}\$ capacitor is discharged through a \$2\;\text{k}\Omega\$ resistor. Find the voltage after \$3\;\text{s}\$ if the initial voltage is \$12\;\text{V}\$.

1. Calculate \$\tau\$:

   \$\tau = RC = (2\times10^{3}\,\Omega)(10\times10^{-6}\,\text{F}) = 0.020\;\text{s}\$

2. Use the discharge equation:

   \$V(t) = V_{0}e^{-t/\tau} = 12\,e^{-3/0.020}\$

3. Evaluate the exponent:

   \$\frac{3}{0.020}=150\$

4. Since \$e^{-150}\$ is extremely small, \$V(3\;\text{s})\approx 0\;\text{V}\$ (practically fully discharged).

Summary Table

Common Mistakes

• Confusing the sign of the exponent – the voltage always decays, so the exponent must be negative.
• Using \$RC\$ without units conversion (e.g., forgetting to convert \$\mu\text{F}\$ to farads).
• Assuming the capacitor is completely discharged after \$1\tau\$; remember it retains \overline{37} % of its initial voltage.

Practice Questions

1. A \$4.7\;\mu\text{F}\$ capacitor discharges through a \$1\;\text{M}\Omega\$ resistor. Calculate \$\tau\$ and the voltage after \$2\tau\$ if \$V_{0}=5\;\text{V}\$.
2. Determine the resistance required to give a time constant of \$0.5\;\text{s}\$ for a \$22\;\mu\text{F}\$ capacitor.
3. Sketch a qualitative graph of \$V(t)\$ on semi‑log paper for a discharge with \$\tau = 0.1\;\text{s}\$.