Q 12-06-106JEE MainJEE Main 2021 (25 Feb, Shift 1)Medium
The current $(i)$ at time $t = 0$ and $t = \infty$ respectively for the given circuit is:
Answer: (B) $\frac{5E}{18},\ \frac{10E}{33}$
At $t = 0$ the inductor carries no current (open). The current from E splits into two paths: $5 + 1 = 6\ \Omega$ (left) and $5 + 4 = 9\ \Omega$ (right), in parallel:
$$R_0 = \frac{6\times9}{15} = \frac{18}{5}\ \Omega,\quad i_0 = \frac{5E}{18}$$
At $t = \infty$ the inductor is a plain wire joining the left and right ends. Then the two $5\ \Omega$ are in parallel ($2.5\ \Omega$), in series with $1\ \Omega \parallel 4\ \Omega = 0.8\ \Omega$:
$$R_\infty = 3.3\ \Omega,\quad i_\infty = \frac{E}{3.3} = \frac{10E}{33}$$
Solution by Sreeraj P, M.Sc Physics