🔋 Discharging a Capacitor
What happens? A charged capacitor releases its stored energy through a resistor, just like a water tank slowly draining through a pipe.
Key Relationships
- Voltage across the capacitor: \$V(t) = V_0\,e^{-t/RC}\$
- Charge on the capacitor: \$Q(t) = Q_0\,e^{-t/RC}\$
- Current through the resistor: \$I(t) = I_0\,e^{-t/RC}\$
- Time constant: \$\tau = RC\$ (the time for the voltage/charge/current to fall to 37 % of its initial value)
Graph Features
- Exponential decay – all three curves (V, Q, I) drop smoothly, never crossing zero.
- Same slope at the start – the initial rate of change is proportional to the initial value.
- Half‑life – time for the value to halve is \$t_{1/2} = \tau \ln 2\$.
- Time constant marker – at \$t=\tau\$, each curve is at 37 % of its starting value.
Practical Example
Suppose a capacitor of 10 µF is charged to 12 V and then connected to a 2 kΩ resistor.
| Parameter | Value |
|---|
| Capacitance \$C\$ | 10 µF |
| Resistance \$R\$ | 2 kΩ |
| Time constant \$\tau\$ | \$RC = 20\$ s |
At \$t = 20\$ s, the voltage, charge and current will each be 37 % of their initial values.
Analogy: The Water Tank
Imagine the capacitor as a water tank filled to a height \$h_0\$. When you open a valve (the resistor), water flows out. The height \$h(t)\$ decreases exponentially, just like the voltage:
\$h(t) = h_0\,e^{-t/\tau}\$
Here, the valve’s size is analogous to the resistance: a wider valve (lower \$R\$) lets water out faster, reducing \$\tau\$.
Exam Tips 📚
- Always identify the time constant first: \$\tau = RC\$.
- When asked for the voltage after a time \$t\$, use \$V(t) = V_0\,e^{-t/\tau}\$.
- For half‑life questions, remember \$t_{1/2} = \tau \ln 2 \approx 0.693\,\tau\$.
- Check units: \$R\$ in Ω, \$C\$ in F, so \$\tau\$ comes out in seconds.
- Sketch the graph: exponential decay, never negative, asymptotic to zero.
Quick Quiz 🚀
- What is the value of \$I(t)\$ at \$t = 3\tau\$ if \$I_0 = 6\$ mA?
- How long will it take for the charge to drop to 10 % of its initial value?
- Explain why the current starts at a maximum and then decreases.
Use the equations above to solve – you’ve got this! 💪