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Capacitor discharge, energy one fourth

Short answer

Energy U = U_0 e^(−2t/RC). Set U/U_0 = 1/4 gives e^(−2t/RC) = 1/4 → t/RC = (ln4)/2 = ln2 ≈ 0.693. So after 0.693 time constants.

A capacitor of capacitance 'C' is discharged through a resistor of resistance 'R'. After how many time constants is the energy stored becomes one fourth of initial value?

Charge on a discharging capacitor: q = q₀ e^(−t/RC). Stored energy U = q²/2C, so

U = U₀ e^(−2t/RC)

Let t = x RC and U/U₀ = 1/4.

e^(−2x) = 1/4
  2x    = ln 4 = 1.386
   x    = 0.693 = ln 2

Answer: after 0.693 time constants (t = RC ln 2 ≈ 0.69 RC).

This is the same time at which the charge falls to 1/2 of its initial value, because (1/2)² = 1/4.

✓ Answer checked by Claude Opus 5.5 (high effort)

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