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Capacitor discharge, energy one eighth

Short answer

During discharge, stored energy U = U₀ e^(−2t/RC). Setting U = U₀/8 gives e^(−2t/RC) = 1/8 ⇒ t = (ln 8)/2 RC ≈ 1.04 time constants.

A capacitor of capacitance C is discharging through a resistor of resistance R. After how many time constants is the stored energy 1/8 of its initial value?

During discharge the charge falls as q = q₀ e^(−t/RC), so the stored energy U = q²/2C falls as

U = U₀ e^(−2t/RC)

Let t = x RC, where x is the number of time constants. We need U/U₀ = 1/8.

e^(−2x) = 1/8
  2x    = ln 8 = 2.079
   x    = 1.04

Answer: after about 1.04 time constants (t ≈ 1.04 RC).

Note that the energy decays twice as fast as the charge: it falls to 1/e² of its initial value after 1 time constant.

✓ Answer checked by Claude Opus 5.5 (high effort)

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