In an RL circuit, the growth of current I(t) to its maximum value I₀ is given by
[tex]I(t) = I_{0}e^{-\lambda t} \\ where \\ \lambda = \frac{R}{L} [/tex]
For the given problem,
R = 0.0100 Ω
L = 5.00 H
Therefore
λ = 0.01/5 = 0.002
When I = 99.999% of I₀, obtain
[tex]0.99999I_{0} = I_{0}e^{-0.002t} \\ ln(0.99999) = -0.002t \\
t = \frac{ln(0.99999)}{-0.002} = 0.005 \, s[/tex]
Answer: 0.005 s
