Capacitor Discharge through a Resistor

A capacitor charged to $V_0$ discharges through a resistor $R$ in a smooth exponential fall. The voltage $V_C(t) = V_0 e^{-t/\tau}$ drops to 37 percent (that is $1/e$) at the time constant $\tau = RC$, the characteristic time of every timing circuit. The current is largest at the very start and dies away with the same exponential, so the resistor dissipates the most power, $P = I^2 R$, at $t = 0$ and cools as the current fades. Watch the charge circulate through the glowing resistor at the top, the voltage decay against the $\tau$ markers in the middle, and the power curve at the bottom: its height is the heating rate (it falls, like the glow), while the shaded area swept out under it is the energy delivered as heat, growing toward the capacitor's initial store.

Figure 1. Capacitor discharging through a resistor. Method: closed-form $V(t) = V_0 e^{-t/\tau}$, $\tau = RC$. Source: Griffiths, Introduction to Electrodynamics, 5e, Ch. 7.

V_0 (V)10.0

R (kOhm)10.0

C (uF)10.0

WHAT TO TRY

Increase $R$ or $C$ and the decay stretches out: $\tau = RC$ grows in proportion.
Watch the voltage curve cross the $\tau$ line at 37 percent, the same fraction whatever $V_0$ is.
Follow the power curve: it peaks at the start (the resistor is hottest at $t=0$) and falls as the current dies, while the shaded area, the heat already delivered, keeps growing toward the capacitor's full initial energy.