Q 12-06-023NEETJEE MainMedium
An inductor of $2$ H and a $4\ \Omega$ resistor are connected in series to a $12$ V battery at $t = 0$. The time constant and the final current are
Answer: (A) $0.5$ s and $3$ A
$\tau = \dfrac{L}{R} = 0.5$ s. Final current $= \dfrac{12}{4} = 3$ A, reached as $I = 3(1 - e^{-t/0.5})$.
Solution by Sreeraj P, M.Sc Physics