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

Learning Objective

Use the exponential decay equation \$x = x_0 e^{-\frac{t}{RC}}\$ to describe the time‑dependence of current \$I\$, charge \$Q\$, and potential difference \$V\$ when a capacitor discharges through a resistor.

Physical Situation

A charged capacitor of capacitance \$C\$ is connected across a resistor of resistance \$R\$. At \$t = 0\$ the switch is closed, allowing the stored charge to flow through the resistor.

Derivation of the Decay Law

1. Apply Kirchhoff’s loop rule: \$VC - VR = 0\$.
2. Express the capacitor voltage in terms of charge: \$V_C = \dfrac{Q}{C}\$.
3. Express the resistor voltage in terms of current: \$V_R = IR\$.
4. Since \$I = -\dfrac{dQ}{dt}\$ (the current is the rate of loss of charge), substitute to obtain

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

5. Rearrange and integrate:

   \$\frac{dQ}{Q} = -\frac{dt}{RC}\;\Longrightarrow\;\ln Q = -\frac{t}{RC} + \ln Q_0.\$

   Exponentiating gives the charge decay law

   \$Q(t) = Q_0 e^{-\frac{t}{RC}}.\$

6. Using \$V = Q/C\$ and \$I = -\dfrac{dQ}{dt}\$, the corresponding expressions for voltage and current are

   \$V(t) = V0 e^{-\frac{t}{RC}}, \qquad I(t) = I0 e^{-\frac{t}{RC}}.\$

Key Parameters

Interpretation of the Time Constant \$\tau = RC\$

• After a time \$t = \tau\$, the quantity \$x\$ (charge, voltage or current) has fallen to \$x = x0 e^{-1} \approx 0.37\,x0\$.
• After \$t = 3\tau\$, \$x \approx 0.05\,x_0\$ (5 % of the initial value).
• Practically, the capacitor is considered fully discharged after \$5\tau\$, when \$x \approx 0.007\,x_0\$ (less than 1 %).

Worked Example

Problem: A \$10\;\mu\text{F}\$ capacitor is charged to \$12\;\text{V}\$ and then discharged through a \$4.7\;\text{k}\Omega\$ resistor. Find the voltage across the capacitor after \$t = 0.02\;\text{s}\$.

Solution:

1. Calculate the time constant:

   \$\tau = RC = (4.7\times10^{3}\,\Omega)(10\times10^{-6}\,\text{F}) = 0.047\;\text{s}.\$

2. Use the voltage decay law:

   \$V(t) = V_0 e^{-t/\tau} = 12\;\text{V}\; e^{-0.02/0.047}.\$

3. Evaluate the exponent: \$-0.02/0.047 \approx -0.426\$.
4. Compute \$e^{-0.426} \approx 0.653\$.
5. Thus \$V(0.02\;\text{s}) \approx 12\;\text{V} \times 0.653 \approx 7.8\;\text{V}.\$

Common Mistakes

• Confusing the sign of the current: during discharge \$I = -\dfrac{dQ}{dt}\$, so the magnitude follows the same exponential form as \$Q\$.
• Using \$RC\$ as a time instead of a time constant; remember it sets the scale for exponential decay.
• Neglecting to convert units (e.g., \$\mu\text{F}\$ to farads, kΩ to ohms) before calculating \$\tau\$.

Summary Checklist

1. Write the loop equation and substitute \$VC = Q/C\$, \$VR = IR\$.
2. Replace \$I\$ with \$-dQ/dt\$ and integrate to obtain \$Q = Q_0 e^{-t/RC}\$.
3. Derive \$V(t)\$ and \$I(t)\$ from \$Q(t)\$.
4. Identify the time constant \$\tau = RC\$ and use it to estimate how quickly the capacitor discharges.
5. Check units and sign conventions in any calculation.