AI Tutor
When you turn off a desktop computer or unplug a camera flash, the status LED often fades slowly over several seconds rather than snapping off instantly.
That lingering glow comes from a capacitor emptying its stored electrical charge through a resistor β€” a process called an exponential discharge.

The Discharging Formula

The voltage VV across a discharging capacitor at any time tt depends on its starting voltage V0V_0, its capacitance CC, and the discharge resistance RR.
V=V0eβˆ’tRCV = V_0 e^{-\frac{t}{RC}}
The term time constant (symbolized by Ο„\tau, the Greek letter tau) equals RΓ—CR \times C in seconds, defining how fast the circuit discharges.
Why is the decay curved rather than a straight line? As charge leaves the plates, the capacitor's voltage drops, which reduces the current (I=VRI = \frac{V}{R}), causing the rate of charge loss to slow down continuously.
After exactly one time constant (t=RCt = RC), the voltage always falls to eβˆ’1β‰ˆ0.368e^{-1} \approx 0.368, or 36.8% of its initial starting value.
How do we predict the exact voltage remaining at any random second during discharge?

Calculating Instantaneous Voltage

Let's find the voltage left in a 470Β ΞΌF470\text{ }\mu\text{F} capacitor charged to 12.0Β V12.0\text{ V} after discharging through a 10.0Β kΞ©10.0\text{ k}\Omega resistor for 3.0Β s3.0\text{ s}.
First, calculate the circuit's time constant Ο„=RC=(10.0Γ—103Β Ξ©)(470Γ—10βˆ’6Β F)=4.70Β s\tau = RC = (10.0 \times 10^3\text{ }\Omega)(470 \times 10^{-6}\text{ F}) = 4.70\text{ s}.