$A$ 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 ...... $s$.

The resistance $R$ is the same as that of the coil that makes $L$. Which of the following statements gives the correct description of the happenings when the switch $S$ is closed?

An inductor $(L = 0.03 \; H)$ and a resistor $(R = 0.15 \; k\Omega)$ are connected in series to a battery of $15 \; V$ $EMF$ in the circuit shown below. The key $K_1$ has been kept closed for a long time. Then at $t = 0$,$K_1$ is opened and key $K_2$ is closed simultaneously. At $t = 1 \; ms$,the current in the circuit will be ........... $mA$. $(e^5 \cong 150)$

An inductor $(L)$ - resistance $(R)$ battery circuit is switched on at $t = 0$. If the $e.m.f.$ of the battery is $E$,what is the charge passing through the battery in time constant $T$?

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

$A$ coil of self-inductance $50\,H$ is connected to the terminals of a battery of $e.m.f.$ $2\,V$ through a resistance of $10\,\Omega$,and a steady current is flowing through the circuit. If the battery is now disconnected,the time in which the current will decay to $1/e$ of its steady value is $...\,s$.