In the circuit shown,the switch $S_1$ is closed at time $t = 0$ and the switch $S_2$ is kept open. At some later time $t_0$,the switch $S_1$ is opened and $S_2$ is closed. The behavior of the current $I$ as a function of time $t$ is given by:

$A$ 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 $,s$)

Three identical resistors with resistance $R = 12\,\Omega$ and two identical inductors with self-inductance $L = 5\,mH$ are connected to an ideal battery with an emf of $12\,V$ as shown in the figure. The current through the battery long after the switch has been closed will be $......A$.

In the given circuit,the switch is in position $a$. For what value of $R$ will the circuit have a time constant of $10\,\mu s$ in $K\Omega$?

$A$ $10 \text{ cm}$ long perfectly conducting wire $PQ$ is moving with a velocity $1 \text{ cm/s}$ on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor $L = 1 \text{ mH}$ and a resistance $R = 1 \ \Omega$ as shown in the figure. The horizontal rails,$L$,and $R$ lie in the same plane with a uniform magnetic field $B = 1 \text{ T}$ perpendicular to the plane. If the key $S$ is closed at a certain instant,the current in the circuit after $1 \text{ millisecond}$ is $x \times 10^{-3} \text{ A}$,where the value of $x$ is. . . . . . [Assume the velocity of wire $PQ$ remains constant $(1 \text{ cm/s})$ after key $S$ is closed. Given: $e^{-1} = 0.37$,where $e$ is the base of the natural logarithm]

When a battery is connected across a series combination of self inductance $L$ and resistance $R$,the variation in the current $i$ with time $t$ is best represented by