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Q.
A coil of inductance $8.4\, mH$ and resistance $6\, \Omega$ is connected to a $12\, V$ battery. The current in the coil is $1.0\, A$ at approximately the time
Electromagnetic Induction
Solution:
Peak current in the circuits, $i_{0}=\frac{12}{6}=2\, A$
Current decreases from $2\, A$ to $1\, A$
i.e., becomes half in time $t=0.693 \frac{L}{R}$
$=0.693 \times \frac{8.4 \times 10^{-3}}{6}=1\, ms$