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Energy stored in coil, time constants

Short answer

Maximum current I_max = 12 V / 0.35 Ω = 34.3 A, energy U_max = 0.5 × 0.053 × (34.3)^2 ≈ 31.2 J. Half energy when I = I_max/√2 → 1−e^(−t/τ) = 1/√2 → t/τ ≈ 1.23.

A coil has an inductance of 53 mH and a resistance of 0.35Ω. If a 12V emf is applied across the coil, how much energy is stored in the magnetic field after the current has built up to its equilibrium value? After how many time constants will half this equilibrium energy be stored?

Given: L = 53 mH, R = 0.35 Ω, ε = 12 V.

Energy at equilibrium

At equilibrium the current is steady and limited only by R: I_max = ε/R.

I_max = 12 / 0.35 = 34.29 A
U_max = ½ L I_max² = ½ × 0.053 × (34.29)²
      = ½ × 0.053 × 1175.5
      = 31.2 J

Time for half this energy

The current grows as I = I_max(1 − e^(−t/τ)), with τ = L/R = 0.053/0.35 = 0.1514 s. Energy U = ½LI² is half of U_max when I² = I_max²/2, that is, I = I_max/√2.

1 − e^(−t/τ) = 1/√2 = 0.7071
e^(−t/τ)     = 0.2929
t/τ          = −ln 0.2929 = 1.228
t = 1.228 × 0.1514 = 0.186 s

Answer: U_max ≈ 31.2 J; half of it is stored after about 1.23 time constants (t ≈ 0.19 s).

✓ Answer checked by Claude Opus 5.5 (high effort)

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