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

What Happens When a Capacitor Discharges?

Imagine a water tank (the capacitor) filled with water (electric charge). When you open a valve (connect a resistor), the water slowly flows out. The rate at which it empties depends on the size of the valve (resistance) and the amount of water already in the tank (capacitance).

In the water tank analogy, τ (tau) is the time it takes for the water level to fall to about 37% of its initial height. Mathematically:

$$\tau = R \times C$$

Where:

• R = resistance in ohms (Ω)
• C = capacitance in farads (F)

After one τ, the voltage across the capacitor has dropped to e⁻¹ ≈ 0.37 of its starting value.

1. Start with Ohm’s law: V = IR.
2. For a capacitor, current I = C dV/dt.
3. Combine: V = R C dV/dt.
4. Rearrange: dV/V = -dt/(RC).
5. Integrate: ∫dV/V = -∫dt/(RC) → ln(V) = -t/(RC) + ln(V₀).
6. Exponentiate: V(t) = V₀ e^{-t/τ}, where τ = RC.

Suppose a 10 µF capacitor discharges through a 5 kΩ resistor.

Calculate τ:

$$\tau = 5\,000\,\Omega \times 10\,\mu\text{F} = 0.05\,\text{s}$$

After 0.05 s, the voltage is 37% of its initial value. After 5τ (0.25 s), it’s almost zero.

• Always state the formula: τ = RC before plugging in numbers.
• Check units: Ω × F = seconds.
• Remember that after 5τ the capacitor is effectively discharged.
• For multiple resistors in series or parallel, calculate the equivalent R first.
• Use the exponential decay formula if asked for voltage at a specific time.

The time constant τ = RC tells us how fast a capacitor discharges. Think of it as the “leakiness” of the system: larger R or C means a slower discharge.

Keep the formula handy, practice unit conversions, and remember the 5τ rule for quick estimation.