Switch $S$ is closed at $t = 0$ in the circuit shown. The change in magnetic flux in the inductor $(L = 500 \, mH)$ from $t = 0$ to an instant when it reaches steady state is ...... $Wb$.

$A$ $10 \text{ cm}$ 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]

$A$ series $L-R$ circuit is connected to a battery of emf $V$. If the circuit is switched on at $t = 0$,then the time at which the energy stored in the inductor reaches $\left(\frac{1}{n}\right)$ times of its maximum value,is

The unit of $L/R$ is (where $L$ = inductance and $R$ = resistance).

Consider an electrical circuit containing a two-way switch $S$. Initially,$S$ is open and then $T_{1}$ is connected to $T_{2}$. As the current in $R = 6 \, \Omega$ attains a maximum value of steady-state level,$T_{1}$ is disconnected from $T_{2}$ and immediately connected to $T_{3}$. The potential drop across the $r = 3 \, \Omega$ resistor immediately after $T_{1}$ is connected to $T_{3}$ is $.... \, V$. (Round off to the nearest integer)

An $L-R$ circuit has a cell of $e.m.f.$ $E$,which is switched on at time $t = 0$. The current in the circuit after a long time will be