Chapter 30 - Induction and Inductance - Problems - Page 899: 50

$T=12.5s$

We know that; $\frac{i}{i_{\circ}}=1-e^{(\frac{-t}{T})}$ We plug in the known values to obtain: $\frac{1}{3}=1-e^{(\frac{-5.00}{T})}$ Next, we rearrange and solve this equation for T: $e^{(\frac{-5.00}{T})}=1-\frac{1}{3}$ $e^{(\frac{-5.00}{T})}=0.66$ Taking log on both sides, we obtain: ${(\frac{-5.00}{T})}=\ln 0.66=-0.40$ $T=\frac{-5.00}{-0.40}=12.5s$