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

$A$ coil of inductance $8.4 \, mH$ and resistance $6 \, \Omega$ is connected to a $12 \, V$ battery. The current in the coil is $1.0 \, A$ at approximately the time

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

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

In the circuits $(a)$ and $(b)$,switches $S_1$ and $S_2$ are closed at $t = 0$ and are kept closed for a long time. The variations of current in the two circuits for $t \ge 0$ are roughly shown by which figure? (Figures are schematic and not drawn to scale.)

Two resistances $20\,\Omega$ and $50\,\Omega$ and a pure inductance of $50\,H$ are connected to a $10\,V$ battery through a key as shown in the figure. The key is closed at $t = 0$. Find the final value of current in the $50\,\Omega$ resistor.