# A coil inductance 8.44 mH and resistance 6 O is connected to a. 12 V battery. The current in the coil is 1.0  A at approximately the time

$\\for current in LR circuit you should remember this formulae \\I=\frac{V}{R}(1-e^{-\frac{Rt}{L}}) \\=>I=\frac{12}{6}(1-e^{-\frac{6*t}{8.44*10^{-3}}}) \\=>I=2(1-e^{-710.9t}) \\substituting I=1 \\=>710.9t=ln2 \\=>t=\frac{ln2}{710.9}$