$A$ coil of self-inductance $10\,mH$ and resistance $0.1\,\Omega$ is connected through a switch to a battery of internal resistance $0.9\,\Omega$. After the switch is closed,the time taken for the current to attain $80\%$ of its saturation value is: [take $\ln\,5 = 1.6$] (in $,s$)

$A$ 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 $\ln 2 = 0.693$,$e^{-3/2} = 0.25$)

$A$ coil of inductance $40 \, H$ is connected in series with a resistance of $8 \, \Omega$ and the combination is joined to the terminals of a $2 \, V$ battery. The time constant of the circuit is ...... $s$.

Find the current in the sliding rod $AB$ (resistance $= R$) for the arrangement shown in the figure. $\vec{B}$ is constant and is directed out of the paper. The parallel wires have no resistance. $v$ is constant. The switch $S$ is closed at time $t = 0$.

When a $DC$ voltage is applied at the two ends of a circuit kept in a closed box, it is observed that the current gradually increases from zero to a certain value and then remains constant. What do you think the circuit contains?

For the given circuit,the current through the battery of $6\,V$ just after closing the switch $S$ will be $.......A$.