A coil of self inductance 10 mH and resistance 0.1 Ω is connected through a switch to a battery of internal resistance 0.9 Ω. After the switch is closed, what is the time taken for the current to attain 80% of its maximum value?

Correct option is B. 0.016s
i = i₀(1 - e⁻ᵗ⁄τ)
80/100 i₀ = i₀(1 - e⁻ᵗ⁄τ)
0.8 = 1 - e⁻ᵗ⁄τ
e⁻ᵗ⁄τ = 0.2 = 1/5
-t/τ = ln(1/5)
-t/τ = -ln(5)
t = τ ln(5)
τ = L/R_{eq} = 10 × 10⁻³ / (0.1 + 0.9) = 10 × 10⁻³ / 1 = 10 × 10⁻³ s
t = 10 × 10⁻³ × ln(5)
t ≈ 10 × 10⁻³ × 1.6
t = 1.6 × 10⁻² s
t = 0.016 s