In the circuit shown,the key $(K)$ is closed at $t = 0$. Find the current through the key at the instant $t = 10^{-3} \ln 2 \, s$.

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

In the figure below,the switches $S_1$ and $S_2$ are closed simultaneously at $t=0$ and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) $V$ and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current $I$ in the middle wire reaches its maximum magnitude $I_{\max}$ at time $t=T$. Which of the following statements is (are) true?
$(A)$ $I_{\max}=\frac{V}{2R}$
$(B)$ $I_{\max}=\frac{V}{4R}$
$(C)$ $T=\frac{L}{R} \ln 2$
$(D)$ $T=\frac{2L}{R} \ln 2$

In an $LR$ circuit,the time constant is the time in which the current grows from zero to the value (where ${I_0}$ is the steady-state current):

In the given $L-R$ circuit,which of the following statements is correct?

An emf of $20\; V$ is applied at time $t=0$ to a circuit containing in series $10\; mH$ inductor and $5\; \Omega$ resistor. The ratio of the currents at time $t=\infty$ and at $t=40\; ms$ is close to: (Take $e^{2}=7.389$)