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Time for current to reach 80 percent

Short answer

Time constant τ = L/R = 5.25 ns. Time to reach 80% of final current is t = τ ln(1/0.2) ≈ 8.45 ns.

A solenoid having an inductance of 6.3 μH is connected in series with a 1.2 kΩ resistance. If a 14 V battery is connected across the pair, how long will it take for the current through the resistor to reach 80% of its final value?

For an LR circuit connected to a battery, the current grows as I = I_max(1 − e^(−t/τ)), where τ = L/R and I_max = ε/R.

Given: L = 6.3 μH, R = 1.2 kΩ, ε = 14 V.

τ     = L/R = 6.3×10⁻⁶ / 1.2×10³ = 5.25×10⁻⁹ s
I_max = ε/R = 14 / 1200 = 1.167×10⁻² A = 11.67 mA

I = 0.80 I_max  →  1 − e^(−t/τ) = 0.80
                   e^(−t/τ)     = 0.20
                   t/τ = ln 5   = 1.609
t = 1.609 × 5.25×10⁻⁹ = 8.45×10⁻⁹ s

At that time the current is 0.8 × 11.67 mA = 9.33 mA.

Answer: t ≈ 8.45 ns (about 1.61 time constants)

✓ Answer checked by Claude Opus 5.5 (high effort)

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