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

Quantity	Symbol	Expression	Typical \cdot alue
Capacitance	$C$	given	$10\;\mu\text{F}$
Resistance	$R$	given	$2\;\text{k}\Omega$
Time constant	$\tau$	$RC$	$0.020\;\text{s}$
Voltage after time $t$	$V(t)$	$V_{0}e^{-t/\tau}$	see example

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 $\tau = 0.1\;\text{s}$.