The figure shows three circuits with identical batteries,inductors,and resistors. Rank the circuits according to the currents through the battery just after the switch is closed,greatest first.

The current $(i)$ at time $t = 0$ and $t = \infty$ respectively for the given circuit is

$A$ coil has an inductance of $2.5\,H$ and a resistance of $0.5\,\Omega$. If the coil is suddenly connected across a $6.0\,V$ battery,then the time required for the current to rise to $0.63$ of its final value is.....$s$.

$A$ $20 \, H$ inductor coil is connected to a $10 \, \Omega$ resistance in series as shown in the figure. The time at which the rate of dissipation of energy (Joule's heat) across the resistance is equal to the rate at which magnetic energy is stored in the inductor is:

The figure shows an $L-R$ circuit. When the switch $S$ is closed,the currents through resistors $R_1, R_2$ and $R_3$ are $I_1, I_2$ and $I_3$ respectively. The values of $I_1, I_2$ and $I_3$ at $t=0 \, s$ are:

$A$ coil of inductance $1\,H$ and resistance $100\,\Omega$ is connected to a battery of $6\,V$. Determine approximately:
$(a)$ The time elapsed before the current acquires half of its steady-state value.
$(b)$ The energy stored in the magnetic field associated with the coil at an instant $15\,ms$ after the circuit is switched on. (Given $\ln 2 = 0.693$,$e^{-3/2} = 0.25$)