The time constant of an $LR$ circuit represents the time in which the current in the circuit

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

An inductor $(L = 100 \, mH)$,a resistor $(R = 100 \, \Omega)$ and a battery $(E = 100 \, V)$ are initially connected in series as shown in the figure. After a long time,the battery is disconnected by short-circuiting the points $A$ and $B$. The current in the circuit $1 \, ms$ after the short circuit 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)

The time constant of an $LR$ circuit is $10 \, s$. When a resistance of $10 \, \Omega$ is connected in series with the existing circuit,the time constant becomes $2 \, s$. The self-inductance of the circuit is ..... $H$.